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76c87cee4a37736b964dae4280b47222 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_21 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$ . The other two sides are of lengths $10$ and $14$ . The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$ | [asy] size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",E,N); [/asy]
The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$ . This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$ . One may now proceed as in Solution $1$ to obtain an answer of $\boxed{25}$ | B | 25 |
77e0fabb4f8f99b4274e980292ddff5d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_24 | problem_id
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$ . By choosing the full circle, we can obtain $15$ . By choosing everything except for $1, 2, 3, 4,$ and $5$ , we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$
This means that we now only need to check for $6, 7, 8,$ and $9$ . However, once we have found a set summing to $6$ , we can choose all remaining numbers and obtain a set summing to $15-6=9$ , and similarly for $7$ and $8$ . Thus, we only need to check each case for whether or not we can obtain $6$ or $7$
We can make $6$ by having $4, 2$ , or $3, 2, 1$ , or $5, 1$ . We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$x$", A, N); label("$y$", C, SW); label("$z$", D, SE); [/asy]
Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$ , because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$2$", C, SW); label("$1$", D, SE); [/asy]
We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$ , so we must place $4$ next to the $1$ and $5$ next to the $2$
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$5$", B, NW); label("$2$", C, SW); label("$1$", D, SE); label("$4$", E, NE); [/asy]
This is the only solution to make $6$ "bad."
Next we move on to $7$ , which can be made by $3, 4$ , or $5, 2$ , or $4, 2, 1$ . We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$1$", A, N); label("$3$", B, NW); label("$2$", C, SW); label("$4$", D, SE); label("$5$", E, NE); [/asy]
This is the only solution to make $7$ "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{2}$ | B | 2 |
79c841fbea7b7f24b3adf6a47eb279d6 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_1 | A taxi ride costs $$1.50$ plus $$0.25$ per mile traveled. How much does a $5$ -mile taxi ride cost?
$\textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75$ | There are five miles which need to be traveled. The cost of these five miles is $(0.25\cdot5) = 1.25$ . Adding this to $1.50$ , we get $\boxed{2.75}$ | C | 2.75 |
ceabc70afc9c3fd06e4cf8b0c1a48694 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_2 | Alice is making a batch of cookies and needs $2\frac{1}{2}$ cups of sugar. Unfortunately, her measuring cup holds only $\frac{1}{4}$ cup of sugar. How many times must she fill that cup to get the correct amount of sugar?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20$ | To get how many cups we need, we realize that we simply need to divide the number of cups needed by the number of cups collected in her measuring cup each time. Thus, we need to evaluate the fraction $\frac{2\frac{1}{2}}{\frac{1}{4}}$ . Simplifying, this is equal to $\frac{5}{2}(4) = \boxed{10}$ | B | 10 |
d2b8ce0ce3386c2781870d9e05dbadab | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_4 | A softball team played ten games, scoring $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?
$\textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55$ | We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$ . Thus, from this, we know that the five games where they lost by one run were when they scored $1$ $3$ $5$ $7$ , and $9$ runs, and the others are where they scored twice as many runs. We can make the following chart:
$\begin{tabular}{|l|l|} \hline Them & Opponent \\ \hline 1 & 2 \\ 2 & 1 \\ 3 & 4 \\ 4 & 2 \\ 5 & 6 \\ 6 & 3 \\ 7 & 8 \\ 8 & 4 \\ 9 & 10 \\ 10 & 5 \\ \hline \end{tabular}$
The sum of their opponent's scores is $2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{45}$ | C | 45 |
3cc2a3bada1a38fd1489ffd8a4d08574 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | The total amount paid is $105 + 125 + 175 = 405$ . To get how much each should have paid, we do $405/3 = 135$
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that $t-d = 30 - 10 = \boxed{20}$ | B | 20 |
3cc2a3bada1a38fd1489ffd8a4d08574 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | The difference in the money that Tom paid and Dorothy paid is $20$ . In order for them both to have paid the same amount, Tom must pay $20$ more than Dorothy. The answer is $\boxed{20}$ | B | 20 |
3cc2a3bada1a38fd1489ffd8a4d08574 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had $200$ dollars to spend. This means that after the vacation we want the amount of money, they currently have. After the trip, Tom would've $95$ dollars, Dorothy would've $75$ dollars, and Sammy had $25$ dollars. This gives us a total of $95+75+25=195$ dollars.
We want to equally split this money, as that is what happens after splitting the cost equally. This means that we want Dorothy, Tom, and Sammy to each have $65$ dollars. We know that Tom gave Sammy $t$ dollars meaning that we want to split this money first. As Tom gives money to no one else, we want him to reach $65$ dollars in this trade, meaning that as Tom has $95$ dollars and Sammy has $25$ dollars, we can do a trade of $30$ so $t=30$ . After this trade, we get that Tom has $65$ dollars, Sammy has $55$ dollars, and Dorothy has $75$ dollars.
Next trade is where Dorothy gives $d$ dollars to Sammy. Dorothy has $75$ dollars and Sammy has $55$ dollars. As both of these don't have $65$ dollars and this is the last trade, we need to make sure both have $65$ dollars at the end. This is possible if $d=10$
We want to find $t-d=30-10=\boxed{20}$ | B | 20 |
c7e1478ebfc2e0b849d76c2d76f2ad31 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_6 | Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$ | Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ . Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ . Thus, the $11$ -year-old must have stayed home, so Joey is $\boxed{11}$ | D | 11 |
c7e1478ebfc2e0b849d76c2d76f2ad31 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_6 | Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$ | There are only $4$ kids who are under $10$ but since the $5$ -year old stayed home, the only possible ages who went to play baseball are the brothers who are $3,7,9$ , either $13+3$ or $7+9$ is $16$ but since we need $2$ kids to go to baseball who are under $10$ $13,3$ must have been the pair to go to the movies and $9,7$ must have went to baseball, so only the $11$ -year old is left, which is answer choice $\boxed{11}$ | D | 11 |
1f19aef6f29cf6b2eb98d8196508b4c5 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$ | Let us split this up into two cases.
Case $1$ : The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$ : The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by $2$ , and then add to the previous.
Assume the mathematics course is algebra. This means that we can choose $2$ of History, Art, and Latin, which is simply $\dbinom{3}{2} = 3$ . If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.
Thus, overall, we can choose a program in $6 + 3 = \boxed{9}$ ways | C | 9 |
1f19aef6f29cf6b2eb98d8196508b4c5 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$ | We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving $3$ remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out of the $5$ is $\binom{5}3$ . The total number of ways of choosing only non-mathematical classes is $\binom{3}3$ . Therefore the amount of ways you can pick classes with at least one math class is $\binom{5}3-\binom{3}3=10-1=\boxed{9}$ ways. | C | 9 |
1f19aef6f29cf6b2eb98d8196508b4c5 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$ | Similar to Solution 1, note that for Case 1 of solution the answer is simply $\binom{4}2$ and for the second case it is $\binom{3}2$ hence $\boxed{9}$ ways | C | 9 |
bd15bde569f68a0308ed613eaef20021 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_9 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$ | Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$ . We know that $x+y=30$ , and we need to evaluate $3(0.2x) + 2(0.3y)$ , as we know that the three-point shots are worth $3$ points and that she made $20$ % of them and that the two-point shots are worth $2$ and that she made $30$ % of them.
Simplifying, we see that this is equal to $0.6x + 0.6y = 0.6(x+y)$ . Plugging in $x+y=30$ , we get $0.6(30) = \boxed{18}$ | B | 18 |
bd15bde569f68a0308ed613eaef20021 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_9 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$ | The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{18}$ points. If we assume Shenille only attempts two-pointers, then she makes $0.3 \cdot 30 = 9$ shots, which are worth $9 \cdot 2 = 18$ points. | B | 18 |
c8597a79380bd9e25152947f7d4956d1 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_10 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70$ | Let the total amount of flowers be $x$ . Thus, the number of pink flowers is $0.6x$ , and the number of red flowers is $0.4x$ . The number of pink carnations is $\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\frac{3}{4}(0.4x) = 0.3x$ . Summing these, the total number of carnations is $0.4x+0.3x=0.7x$ . Dividing, we see that $\frac{0.7x}{x} = 0.7 = \boxed{70}$ | E | 70 |
c8597a79380bd9e25152947f7d4956d1 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_10 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70$ | We have $\dfrac15$ for pink roses, $1-\dfrac6{10}=\dfrac4{10}=\dfrac25$ red flowers, $\dfrac6{10}-\dfrac15=\dfrac35-\dfrac15=\dfrac25$ pink carnations, $\dfrac25\cdot \dfrac34=\dfrac6{20}=\dfrac3{10}$ red carnations we add them up to get $\dfrac25+\dfrac3{10}=\dfrac7{10}=70\%$ so our final answer is 70% or $\boxed{70}$ | E | 70 |
1f87d6f2cfccd8c5e785465b64fa3be9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_11 | A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$ | Let the number of students on the council be $x$ . To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person committee. However, this accounts for the order of committees. To understand this, suppose that Alice and Bob are two students in the council. If we choose Alice and then Bob, that is the same as choosing Bob and then Alice and so latter and former arrangements would be considered the same. Therefore, we have to divide by $2$ to account for overcounting. Thus, there are $\dfrac{x(x-1)} 2=10$ ways to choose the two-person committee. Solving this equation, we find that $5$ and $-4$ are integer solutions. $-4$ is a ridiculous situation, so there are $5$ people on the student council. The solution is $\dbinom 5 3=10\implies \boxed{10}$ | A | 10 |
1f87d6f2cfccd8c5e785465b64fa3be9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_11 | A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$ | To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\binom{n}{2}=10$ which we can easily see $n=5$ , so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$ , so there are $\binom{5}{3}=\boxed{10}$ ways. | A | 10 |
d52b3de505066266c2037437f527ad1d | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_13 | How many three-digit numbers are not divisible by $5$ , have digits that sum to less than $20$ , and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$ .( $\overline{abc}$ denotes the number $100a+10b+c$ ). We see that $x\neq 0$ and $x\neq 5$ , as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that the sum $2x+y<20$ . When $x$ is $1$ $2$ $3$ , or $4$ $y$ can be any digit from $0$ to $9$ , as $2x<10$ . This yields $10(4) = 40$ numbers.
When $x=6$ , we see that $12+y<20$ so $y<8$ . This yields $8$ more numbers.
When $x=7$ $14+y<20$ so $y<6$ . This yields $6$ more numbers.
When $x=8$ $16+y<20$ so $y<4$ . This yields $4$ more numbers.
When $x=9$ $18+y<20$ so $y<2$ . This yields $2$ more numbers.
Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{60}$ | B | 60 |
fcebd3e842982025490fb8ea94148509 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_14 | A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ . How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$ | We can use Euler's polyhedron formula that says that $F+V=E+2$ . We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$ . In addition, each cube creates $7$ new vertices while taking away the original $8$ , yielding $8(7) = 56$ vertices. Thus $E+2=56+30$ , so $E=\boxed{84}$ | D | 84 |
fcebd3e842982025490fb8ea94148509 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_14 | A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ . How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$ | The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \times 8 = \boxed{84}$ | D | 84 |
c52e2ccb47097be9ab51ce922f48d8a7 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\boxed{12}$ | D | 12 |
c52e2ccb47097be9ab51ce922f48d8a7 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | Let the height to the side of length $15$ be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$
Because the area of a triangle is $\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$ . From the problem, we know that $2h_{3} = h_{1} + h_{2}$ . Substituting, we get that \[h_{3} = 1.25h_{1}.\] Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{12}$ | D | 12 |
c52e2ccb47097be9ab51ce922f48d8a7 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$ . This means the height for when the triangles has the $3$ rd side length, the height would be $\dfrac{(x+y)}{2}$ giving us the following equation:
$\dfrac{10x}{2}=\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ For $n$ , we can plug in the answer choices and check when the ratio of $\dfrac{x}{y}$ for $\dfrac{10x}{2}=\dfrac{n(x+y)}{2}$ and $\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that $\boxed{12}$ meaning that $12$ is the $3$ rd base.
~ Batmanstark | null | 12 |
12263710cf2174231fb29e0f1909e3f8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_17 | Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$ | The $365$ -day time period can be split up into $6$ $60$ -day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$ $4$ , and $5$ ).
You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.
Remember to subtract $1$ , because you do not wish to count the $60$ th day, when all three visit.
A and B visit $\frac{60}{3 \cdot 4}-1 = 4$ times.
A and C visit $\frac{60}{3 \cdot 5}-1 = 3$ times.
B and C visit $\frac{60}{4 \cdot 5}-1 = 2$ times.
This is a total of $9$ visits per $60$ day period.
Therefore, the total number of $2$ -person visits is $9 \cdot 6 = \boxed{54}$ | B | 54 |
12263710cf2174231fb29e0f1909e3f8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_17 | Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$ | From the information above, we get that $A=3x$ $B=4x$ $C=5x$
Now, we want the days in which exactly two of these people meet up
The three pairs are $(A,B)$ $(B,C)$ $(A,C)$
Notice that we are trying to find the LCM of each pair.
Hence, $LCM(A,B)=12x$ $LCM(B,C)=20x$ $LCM(A,C)=15x$
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
Hence, $LCM(A,B,C)=60x$
Now, we add all of the days up(including overcount).
We get $30+18+24=72$ .
Now, because $60(6)=360$ , we have to subtract $6$ days from every pair.
Hence, our answer is $72-18=\boxed{54}$ | B | 54 |
7477d0bb5051224d806a8ed76362d836 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_18 | Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms. What is $p+q+r+s$
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$ | First, we shall find the area of quadrilateral $ABCD$ . This can be done in any of three ways:
Pick's Theorem $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$
Shoelace Theorem : The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$ , or $\dfrac{15}{2}$
$[ABCD] = \frac{15}{2}$ . Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$
Call the point where the line through $A$ intersects $\overline{CD}$ $E$ . We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$ . Furthermore, we know that $b = 4$ , as $AD = 4$ . Thus, solving for $h$ , we find that $2h = \frac{15}{4}$ , so $h = \frac{15}{8}$ . This gives that the y coordinate of E is $\frac{15}{8}$
Line CD can be expressed as $y = -3x+12$ , so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$ . Solving for $x$ , we find that $x = \frac{27}{8}$
From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$ $27 + 15 + 8 + 8 = \boxed{58}$ | B | 58 |
7477d0bb5051224d806a8ed76362d836 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_18 | Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms. What is $p+q+r+s$
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$ | Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$ .
Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$ , and from there split the base into two parts, $x$ , and $4-x$ , such that $x$ is the segment from the origin to the point $F$ , and $4-x$ is the segment from point $F$ to point $D$ . Then, by the Pythagorean Theorem $x=\frac{27}{8}$ , and the answer is $\boxed{58}$ | B | 58 |
c64e4537e4b25efdf63f66ec0c53aa3e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_19 | In base $10$ , the number $2013$ ends in the digit $3$ . In base $9$ , on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$ . For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$ | We want the integers $b$ such that $2013\equiv 3\pmod{b} \Rightarrow b$ is a factor of $2010$ . Since $2010=2 \cdot 3 \cdot 5 \cdot 67$ , it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$ , as these cannot have the digit $3$ in their base representations, our answer is $16-3=\boxed{13}$ | C | 13 |
09c47b96e58ba8159dbd4799e6f2e766 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$ | Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that
$x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} \cdot \frac{8}{12} \cdot \frac{7}{12} \cdot \frac{6}{12} \cdot \frac{5}{12} \cdot \frac{4}{12} \cdot \frac{3}{12} \cdot \frac{2}{12} \cdot \frac{1}{12}$ must be an integer. Simplifying, we get
$x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2} \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}$ . Now, the minimal $x$ is the denominator of this fraction multiplied out, obviously. We mentioned before that this product must be an integer. Specifically, it is an integer and it is the amount that the $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12} = 1 =$ all of what is remaining.
Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{1925}$ | D | 1925 |
09c47b96e58ba8159dbd4799e6f2e766 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$ | Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$ . Note that this is equivalent to $\frac{x \cdot 11!}{12^{11}}$
We can compute the amount of factors of $2$ $3$ $5$ , etc. but this is not necessary. To minimize this expression, we must take out factors of $2$ and $3$ , since $12^{11}=2^{22} \cdot 3^{11}$ $11!$ has neither $22$ factors of $2$ , nor $11$ factors of $3$ . This means that if $11!$ contains $a$ factors of $2$ , then $x$ will contain $22-a$ factors of $2$ . This also holds for factors of $3$
Thus, once simplified, the expression will have no factors of $2$ . It will also have no factors of $3$
Looking at the answer choices, there is only one answer which is not even, which is $\boxed{1925}$ | D | 1925 |
09c47b96e58ba8159dbd4799e6f2e766 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$ | We know that the 11th pirate takes $\frac{11}{12}$ of what is left from the 10th pirate, so we have the 12th pirate taking \[\frac{12}{12}\cdot(1-\frac{11}{12})=1\cdot\frac{1}{12}\] of what is left from the 10th pirate. Similarly, since the 10th pirate takes $\frac{10}{12}$ of what is left from the 9th pirate, we have the 11th pirate taking \[\frac{11}{12}\cdot(1-\frac{10}{12})=\frac{11}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Thus, the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Repeating the method, we can find that the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot\frac{11}{12}\] of what is left from the 1st pirate, or \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot\frac{11}{12}\cdot\frac{12}{12}\] of the total amount of coins.
Now canceling out the denominator with the numerator as possible, we are left with $\frac{1\cdot5\cdot7\cdot5\cdot11}{...}=\frac{1925}{...}$ with some factors of 12 in the denominator. For this fraction to be an integer, the smallest possible number of coins is the same as the denominator, so the numerator is the number of coins taken by the 12th pirate, or $1925 \,\boxed{1925}$ | D | 1925 |
87b8e05decd899f8fc6b5880a5350d4a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | Let $BX = q$ $CX = p$ , and $AC$ meets the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , so $p$ is either $3$ $11$ , or $33$ . We also know that $p>11$ by the triangle inequality on $\triangle ACX$ . Thus, $p$ is $33$ so we get that $BC = p+q = \boxed{61}$ | D | 61 |
87b8e05decd899f8fc6b5880a5350d4a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 y + xy^2 + 86^2 y = 97^2 y$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$ | D | 61 |
87b8e05decd899f8fc6b5880a5350d4a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | Let $CX=x, BX=y$ . Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$ .
Use power of a point on point C to the circle centered at A.
So $CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61$
Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$ .
By the Triangle Inequality, only $x+y=61$ yields a possible length of $BX+CX=BC$
Therefore, the answer is $\boxed{61}$ | D | 61 |
87b8e05decd899f8fc6b5880a5350d4a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | [asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label("A",A,S); label("B",B,NE); label("C",C,S); label("D",D,NE); label("E",E,NE); label("86",(A+B)/2,NW); label("86",(A+D)/2,SE); label("97",(A+C)/2,S); label("h",(A+E)/2,N); label("k",(E+D)/2,NE); label("k",(B+E)/2,NE); label("m",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label("$90^\circ$",anglemark(A,E,D),3*S); [/asy]
We first draw the height of isosceles triangle $ABD$ and get two equations by the Pythagorean Theorem .
First, $h^2 + k^2 = 86^2$ . Second, $h^2 + (k + m)^2 = 97^2$ .
Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$ .
We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 2013 + k^2$ .
We then complete the square to get $(k + m)^2 = 2013 + k^2$ . Because $k$ and $m$ are both integers, we get that $2013 + k^2$ is a square number. Simple guess and check reveals that $k = 14$ .
Because $k$ equals $14$ , therefore $m = 33$ . We want $\overline{BC} = 2k + m$ , so we get that $\overline{BC} = \boxed{61}$ | D | 61 |
87b8e05decd899f8fc6b5880a5350d4a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | Let $E$ be the foot of the altitude from $A$ to $BX.$ Since $\triangle ABX$ is isosceles $AX=AB=86,EB=EX,$ and the answer is $EC+EB=EC+EX.$ $(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013$ by the Pythagorean Theorem. Only $EC+EX=\boxed{61}.$ | D | 61 |
15bb05f8ac65dffe35f4d729436f12a9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24 | Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$ | Let us label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$
$\textbf{1}$ . One way of scheduling all six distinct rounds could be:
Round 1: $AX$ $BY$ $CZ$
Round 2: $AX$ $BZ$ $CY$
Round 3: $AY$ $BX$ $CZ$
Round 4: $AY$ $BZ$ $CX$
Round 5: $AZ$ $BX$ $CY$
Round 6: $AZ$ $BY$ $CX$
The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permutating those $6$ rounds and that can be done in $6!=720$ ways.
$\textbf{2}$ . One can also make the schedule in such a way that two rounds are repeated.
(a)
Round 1: $AX$ $BZ$ $CY$
Round 2: $AX$ $BZ$ $CY$
Round 3: $AY$ $BX$ $CZ$
Round 4: $AY$ $BX$ $CZ$
Round 5: $AZ$ $BY$ $CX$
Round 6: $AZ$ $BY$ $CX$
(b)
Round 1: $AX$ $BY$ $CZ$
Round 2: $AX$ $BY$ $CZ$
Round 3: $AY$ $BZ$ $CX$
Round 4: $AY$ $BZ$ $CX$
Round 5: $AZ$ $BX$ $CY$
Round 6: $AZ$ $BX$ $CY$
As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are $\frac{6!}{2!2!2!}$ $90$
So the total number of schedules is $720+90+90$ $\boxed{900}$ | E | 900 |
15bb05f8ac65dffe35f4d729436f12a9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24 | Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$ | Label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$ . We can start by assigning an opponent to person $A$ for all $6$ games. Since $A$ has to play each of $X$ $Y$ , and $Z$ twice, there are $\frac{6!}{2!2!2!} = 90$ ways to do this. We can assume that the opponents for $A$ in the $6$ rounds are $X$ $X$ $Y$ $Y$ $Z$ $Z$ and multiply by $90$ afterwards.
Notice that for every valid assignment of the opponents of $A$ and $B$ , there is only $1$ valid assignment of opponents for $C$ . More specifically, the opponents for $C$ are the leftover opponents after the opponents for $A$ and $B$ are chosen in each round. Therefore, all we have to do is assign the opponents for $B$ . This is the same as finding the number of permutations of $X$ $Y$ , and $Z$ that do not have a $X$ in the first two spots, an $Y$ in the next two spots, and a $Z$ in the final two spots.
We can use casework to find this by using the fact that after we put down the $X$ 's and $Y$ 's first there is $1$ way to put down the $Z$ 's (the two remaining spots).
If $X$ 's are put in the middle $2$ spots, then there is $1$ way to assign spots to $Y$ , namely the last $2$ spots. (If one of the last two spots are left empty, there will have to be a $Z$ there, which which not valid).
If $X$ 's are put in the last $2$ spots, then there is $1$ way to assign spots to $Y$
Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\cdot2$ ways to assign spots to $X$ and $2\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$ ).
There are $1+1+2\cdot2\cdot2\cdot1 = 10$ ways to assign opponents to $B$ and $90\cdot10 = 900$ ways to order the games. $\boxed{900}$ | E | 900 |
2ed73db49c428cdf4b2236315acb8ad8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{49}$ points. | A | 49 |
2ed73db49c428cdf4b2236315acb8ad8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | Let the number of intersections be $x$ . We know that $x\le \dbinom{8}{4} = 70$ , as every $4$ vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract $\dbinom{4}{2} -1 = 5$ from this count, $70-5 = 65$ . Note that diagonals like $\overline{AD}$ $\overline{CG}$ , and $\overline{BE}$ all intersect at the same point. There are $8$ of this type with three diagonals intersecting at the same point, so we need to subtract $2$ of the $\dbinom{3}{2}$ (one is kept as the actual intersection). In the end, we obtain $65 - 16 = \boxed{49}$ | A | 49 |
2ed73db49c428cdf4b2236315acb8ad8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | We know that the amount of intersection points is at most $\dbinom{8}{4} = 70$ , as in solution $2$ . There's probably going to be more than $5$ intersections counted multiple times (to get $\textbf{(B) }65$ ), leading us to the only reasonable answer, $\boxed{49}$ .
-Lcz | A | 49 |
2ed73db49c428cdf4b2236315acb8ad8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region share one side in common and that side contains $2$ intersection points, so in actuality there are $8 - 2 = 6$ points per region. We multiply this by $8$ to get $6\cdot 8 = 48$ and add the one center point to get $48 + 1 = \boxed{49}$ | A | 49 |
2ed73db49c428cdf4b2236315acb8ad8 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | [asy] size(8cm); pathpen = black; // draw the circle pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR((0,0), 1)); // draw the octagon and diagonals // choose pen colors pen[] colors; colors[1] = yellow; colors[2] = purple; colors[3] = green; colors[4] = orange; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } // draw the 3 or more line intersections pointpen = red + 5; D(IP(A[0]--A[4], A[2]--A[6])); // center of the circle for (int x1=0; x1<8; ++x1) { for (int x2=0; x2<8; ++x2) { int y1 = (x1 + 4)%8; int y2 = (x2 + 3)%8; if (x1 != x2 && y1 != y2 && x1 != y1 && x1 != y2 && x2 != y1 && x2 != y2) D(IP(A[x1]--A[y1], A[x2]--A[y2])); } } // draw the 2 line intersections pointpen = blue + 4; for (int x1 = 0; x1 < 8; ++x1) { int x2 = (x1 + 1)%8; D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+2)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+3)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+4)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+5)%8])); D(IP(A[x1]--A[(x1+3)%8], A[x2]--A[(x2+5)%8])); } [/asy]
This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.
Case 2: Blue Dots
The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are $2 \cdot 5+2 \cdot 6+5=27$ dots that come out of 1 vertex, which includes 7 red dots already counted. So there are $27-7 = 20$ blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to $2 \cdot 2 = 4$ vertices. So there are $\frac{20 \cdot 8}{4} = 40$ blue dots.
The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is $40 + 9 = \boxed{49}$ | A | 49 |
5de9f781dbf1b703b75c0dcf1dc70ed2 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_2 | Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green's steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
$\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400$ | Since each step is $2$ feet, his garden is $30$ by $40$ feet. Thus, the area of $30(40) = 1200$ square feet. Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoes expected is $1200 \times \frac{1}{2} = \boxed{600}$ | A | 600 |
8cea74e11223df3528b1bc394b61bb59 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_3 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11$ | Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$ | C | 5 |
83d5a9e2b25f6debf860fa9597f7746a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_4 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{149}$ | D | 149 |
2b93789b9fdfce4366d1e51bef73eb79 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_5 | Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$
$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$ | Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a$ and $b$ are $5$ , so the answer is $5(2 - 5) = \boxed{15}$ | B | 15 |
9990dd026fd7771c2cbe8708f67c1f5b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_6 | The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$ | The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{24.75}$ | C | 24.75 |
4fc50c472c74d7db59d5096c95358f14 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$ | B | 16 |
4fc50c472c74d7db59d5096c95358f14 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\] | B | 16 |
4fc50c472c74d7db59d5096c95358f14 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac{m}{40} + \frac{m}{10} = \frac{m}{8}$ . The total distance that both drive is equal to $2m$ so the total miles per gallon can be expressed as $\frac{2m}{\frac{m}{8}} = \boxed{16}$ | B | 16 |
50419ae77b3bf7029977c07c9ef3978f | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_9 | Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$ | The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{160}$ | D | 160 |
d1de256c5a728b93ff5b123eb70e3ddb | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_10 | A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30$ | Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more two point shots then threes, $x=1.5y$ . Substituting $1.5y$ for $x$ in the first equation gives $1.5y+1.2y=54$ , which equals $2.7y=54$ so $y=$ $\boxed{20}$ | C | 20 |
9498b283f21e5d85ec8d47f3a469edec | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_11 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{2}$ . ~ asdf334 | B | 2 |
9498b283f21e5d85ec8d47f3a469edec | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_11 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ . Equality holds when the value inside the parhentheses is equal to $0$ . We find that \[(x,y) = (5,-3)\] and the sum we are looking for is \[5+(-3)=2 \implies \boxed{2}.\] - Honestly | B | 2 |
0d6c356e7b41cf412677189c8648965f | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_12 | Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length?
$\textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5$ | The problem is simply asking how many ways are there to choose two sides or two diagonals. Hence, the probability is \[\dfrac{\binom{5}{2} + \binom{5}{2}}{\binom{10}{2}} = \dfrac{10+10}{45} = \dfrac{20}{45}=\boxed{49}\] | B | 49 |
5c066e1d280bb7e7a8767fd2823d408e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$ . Since we are starting from 1 every turn, the 53rd number said will be $\boxed{8}$ | E | 8 |
5c066e1d280bb7e7a8767fd2823d408e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$ , and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{8}$ | E | 8 |
5c066e1d280bb7e7a8767fd2823d408e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$ | E | 8 |
af59524f56972a418df44f9bff92f7ae | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]
$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$ | Let us use mass points:
Assign $B$ mass $1$ . Thus, because $E$ is the midpoint of $AB$ $A$ also has a mass of $1$ . Similarly, $C$ has a mass of $1$ $D$ and $E$ each have a mass of $2$ because they are between $B$ and $C$ and $A$ and $B$ respectively. Note that the mass of $D$ is twice the mass of $A$ , so $AP$ must be twice as long as $PD$ . PD has length $2$ , so $AP$ has length $4$ and $AD$ has length $6$ . Similarly, $CP$ is twice $PE$ and $PE=1.5$ , so $CP=3$ and $CE=4.5$ . Now note that triangle $PED$ is a $3-4-5$ right triangle with the right angle $DPE$ . Since the diagonals of quadrilaterals $AEDC$ $AD$ and $CE$ , are perpendicular, the area of $AEDC$ is $\frac{6 \times 4.5}{2}=\boxed{13.5}$ | B | 13.5 |
af59524f56972a418df44f9bff92f7ae | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]
$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$ | Note that triangle $DPE$ is a right triangle, and that the four angles (angles $APC, CPD, DPE,$ and $EPA$ ) that have point $P$ are all right angles. Using the fact that the centroid ( $P$ ) divides each median in a $2:1$ ratio, $AP=4$ and $CP=3$ . Quadrilateral $AEDC$ is now just four right triangles. The area is $\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{13.5}$ | B | 13.5 |
af59524f56972a418df44f9bff92f7ae | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]
$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$ | From the solution above, we can find that the lengths of the diagonals are $6$ and $4.5$ . Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is $\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{13.5}$ | B | 13.5 |
af59524f56972a418df44f9bff92f7ae | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]
$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$ | We know that $[AEDC]=\frac{3}{4}[ABC]$ , and $[ABC]=3[PAC]$ using median properties. So Now we try to find $[PAC]$ . Since $\triangle PAC\sim \triangle PDE$ , then the side lengths of $\triangle PAC$ are twice as long as $\triangle PDE$ since $D$ and $E$ are midpoints. Therefore, $\frac{[PAC]}{[PDE]}=2^2=4$ . It suffices to compute $[PDE]$ . Notice that $(1.5, 2, 2.5)$ is a Pythagorean Triple, so $[PDE]=\frac{1.5\times 2}{2}=1.5$ . This implies $[PAC]=1.5\cdot 4=6$ , and then $[ABC]=3\cdot 6=18$ . Finally, $[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}$ | null | 13.5 |
af59524f56972a418df44f9bff92f7ae | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]
$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$ | As from Solution 4, we find the area of $\triangle DPE$ to be $\frac{3}{2}$ . Because $DE = \frac{5}{2}$ , the altitude perpendicular to $DE = \frac{6}{5}$ . Also, because $DE || AC$ $\triangle ABC$ is similar to $\triangle{DBE}$ with side length ratio $2:1$ , so $AC=5$ and the altitude perpendicular to $AC = \frac{12}{5}$ . The altitude of trapezoid $ACDE$ is then $\frac{18}{5}$ and the bases are $\frac{5}{2}$ and $5$ . So, we use the formula for the area of a trapezoid to find the area of $ACDE = \boxed{13.5}$ | null | 13.5 |
e79c032724cb4dbbfead166514bd47ee | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.
In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.
Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\boxed{103}$ silver tokens, at which point it is clear he cannot use the booths anymore. | E | 103 |
e79c032724cb4dbbfead166514bd47ee | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end up with: $1 \text{R}$ $2 \text{B}$ and $103 \text{S}$ . So, the answer is $\boxed{103}$ | E | 103 |
e79c032724cb4dbbfead166514bd47ee | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has a red coin and $2$ blue coins. We can then create a system of equations: \[1=-2x+y+75\] \[2=x-3y+75\] Solving yields $x=59$ and $y=44$ . Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\boxed{103}$ silver coins in total. | E | 103 |
2db0ab0eff69a07ec1fce896266bc4fb | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_18 | The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$ | We take cases on the thousands digit, which must be either $1$ or $2$ :
If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \dots + 9 = 45$ possible pairs $(b, c)$ , and each pair uniquely determines the value of $d$ , so we get $45$ numbers with the given property.
If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property.
Therefore, the number of numbers with the property is $45 + 1 = \boxed{46}$ | D | 46 |
2db0ab0eff69a07ec1fce896266bc4fb | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_18 | The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$ | Let's start with the case that starts with $200$ . We have only one number, which is $2002$ . If we look at the $1900s$ , we have no solutions because $1+9 = 10$ , and because we can only use digits from $1$ through $9$ , it is impossible. If we looks at the $1800s$ , we do have one solution, which is $1809$ . If we look a the $1700s$ , we have $2$ solutions, namely, $1708$ and $1719$
We can see a pattern here. The pattern is every hundred you go down, you have $1$ more solution. Therefore, we have $1+0+1+2+3+4+5+6+7+8+9$ which is = $\boxed{46}$ | D | 46 |
2ee242ba369a1d8d1bd942a67d943d40 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_20 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{2}.\] | B | 2 |
75263391ee811eadf92d4646fca2ec7a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_21 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$ | Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one of them be $0$ WLOG , assume that $x_{1} = 0$ . Thus, the smallest possible value of $y_{1}$ is $8$ ; and since the sequences are non-decreasing we get $y_{2} \ge 8$ . To minimize, let $y_{2} = 8$ . Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{104}$ | C | 104 |
75263391ee811eadf92d4646fca2ec7a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_21 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$ | WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \[N=5a_1+8a_2=5b_1+8b_2\] or \[5a_1+8a_2=5(a_1+c)+8(a_2-d)\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \[5a_1+8a_2=5(a_1+8)+8(a_2-5).\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)$ . Using the same method, since $b_2\ge b_1$ , we have $b_2\ge8$
Thus the minimum $N=5b_1+8b_2=104\boxed{104}$ | C | 104 |
f6a30c788e9857725a7cf9976d83509a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22 | The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ $BJF$ $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$
[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J",J,SE); [/asy] | First of all, note that $J$ must be $1$ $5$ , or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J $(1, 5, 9)$",J,SE); [/asy]
We also notice that $A+E = B+F = C+G = D+H$
WLOG, assume that $J = 1$ . Thus the pairs of vertices must be $9$ and $2$ $8$ and $3$ $7$ and $4$ , and $6$ and $5$ . There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$ ).
Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{1152}$ | C | 1152 |
f6a30c788e9857725a7cf9976d83509a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22 | The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ $BJF$ $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$
[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J",J,SE); [/asy] | As in solution 1, $J$ must be $1$ $5$ , or $9$ giving us 3 choices. Additionally $A+E = B+F = C+G = D+H$ . This means once we choose $J$ there are $8$ remaining choices. Going clockwise from $A$ we count, $8$ possibilities for $A$ . Choosing $A$ also determines $E$ which leaves $6$ choices for $B$ , once $B$ is chosen it also determines $F$ leaving $4$ choices for $C$ . Once $C$ is chosen it determines $G$ leaving $2$ choices for $D$ . Choosing $D$ determines $H$ , exhausting the numbers. Additionally, there are three possible values for $J$ . To get the answer we multiply $2\cdot4\cdot6\cdot8\cdot3=\boxed{1152}$ | C | 1152 |
328ae7ab8a0b8397dd61219a55966a3b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\tfrac{12}{15} = \tfrac{4}{5}$ , and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$ . It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$
Let $x=DF$ . By Ptolemy's Theorem , we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{21}$ | B | 21 |
328ae7ab8a0b8397dd61219a55966a3b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = 180 - \angle{DFA} = \angle{EFA}$ , and triangles $\triangle AEF \sim \triangle ADB$ . Solving the resulting proportion gives $EF = 4$ . Therefore, $DF = ED - EF = \frac{16}{5}$ $m + n = 16 + 5 = 21$ and our answer is $\boxed{21}$ | B | 21 |
328ae7ab8a0b8397dd61219a55966a3b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Using the Pythagorean theorem, we now get \[BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}\] and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through base times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$ , similar triangles gives $AE = \tfrac{48}5$ ), and that $EF$ is just $\tfrac{36}5 - x$ . From there, \[AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.\] Now, $BF^2 + AF^2 = 169$ , and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$ , so $x = \tfrac{16}5$ . Thus, the answer is $\boxed{21}$ | B | 21 |
328ae7ab8a0b8397dd61219a55966a3b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | First, we find $BD = 5$ $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$
Points $A$ $B$ $D$ , and $F$ all lie on a circle whose diameter is $AB$ . Let the point where the circle intersects $AC$ be $G$ . Using power of a point, we can write the following equation to solve for $AG$ \[DC\cdot BC = CG\cdot AC\] \[9\cdot 14 = CG\cdot 15\] \[CG = 126/15\] Using that, we can find $AG = \frac{99}{15}$ , and using $AG$ , we can find that $GE = 3$
We can use power of a point again to solve for $DF$ \[FE\cdot DE = GE\cdot AE\] \[(\frac{36}{5} - DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}\] \[\frac{36}{5} - DF = 4\] \[DF = \frac{16}{5} = \frac{m}{n}\] Thus, $m+n = 16+5 = 21$ $\boxed{21}$ | B | 21 |
637b27de99d142c8ab26b609ecca244e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_24 | A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$ ) such that the sum of the four divisors is equal to $n$ . How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$ | A positive integer with only four positive divisors has its prime factorization in the form of $a \cdot b$ , where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can easily deduce that none of the numbers are even near a cube so the second case is not possible. We now look at the case of $a \cdot b$ . The four factors of this number would be $1$ $a$ $b$ , and $ab$ . The sum of these would be $ab+a+b+1$ , which can be factored into the form $(a+1)(b+1)$ . Easily we can see that now we can take cases again.
Case 1: Either $a$ or $b$ is 2.
If this is true then we have to have that one of $(a+1)$ or $(b+1)$ is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work.
Case 2: Both $a$ and $b$ are odd primes.
This implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$ . This leaves only $2012$ and $2016$ $2012={4}\cdot{503}$ so either $(a+1)$ or $(b+1)$ both have a factor of $2$ or one has a factor of $4$ . If it was the first case, then $a$ or $b$ will equal $1$ . That means that either $(a+1)$ or $(b+1)$ has a factor of $4$ . That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 \cdot 504$ so we have $(503 + 1)(3 + 1)$ . 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\boxed{1}$ | A | 1 |
637b27de99d142c8ab26b609ecca244e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_24 | A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$ ) such that the sum of the four divisors is equal to $n$ . How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$ | After deducing that $2012$ and case $1$ is impossible, and since there is no option for $0$ $2016$ is obviously a solution and the answer is $\boxed{1}$ | A | 1 |
d9f2c466598859fbc7310acbe82dae60 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that $N \equiv a \pmod{6}$
also that $N \equiv b \pmod{5}$
Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ $N \equiv a \pmod{6}$ $N \equiv a \pmod{5}$ $\implies N=a \pmod{30}$ $0 \le a \le 4$
Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$
Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ;
Also, we have already found which digits of $y$ will add up into the units digits of $2N$
Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work)
Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\]
$33 \ge x \ge 4$
Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\]
Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x)
Then
\[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\]
and this simplifies to
\[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\]
From careful inspection, this is true when
$n=0, m=6$
$n=1, m=6$
$n=2, m=2$
$n=3, m=2$
$n=4, m=4$
This gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \boxed{25}$ | E | 25 |
d9f2c466598859fbc7310acbe82dae60 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$
We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }5,6=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.
\[\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}\]
\[\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}\]
\[\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}\]
As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{25}$ | E | 25 |
d9f2c466598859fbc7310acbe82dae60 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.
Then let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \[0\le a_i<5,0\le b_i<6\] Thus we have \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Now we are given \[2N \equiv S \equiv N_5+N_6\pmod{100}\] \[2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}\] \[1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}\] \[50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}\] Canceling out $a_5$ left with \[50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}\]
Since $a_5$ $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$ . Thus,
\[50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}\] canceling out $10a_4$ , we have \[50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}\] \[5a_1+5a_2+5a_3 \equiv b_3\pmod{10}\] \[5(a_1+a_2+a_3) \equiv b_3\pmod{10}\]
Thus $b_3$ is a multiple of $5$
Now going back to our original equation \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Since $a_5=b_4$ \[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\] \[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\] \[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\]
Since the left side is a multiple of $5$ , then so does the right side. Thus $5\mid6(6b_1+b_2)+b_3$
Since we already know that $5\mid b_3$ , then $5\mid6(6b_1+b_2)$ , from where we also know that $5\mid6b_1+b_2$
For $b_1,b_2<6$ , there is a total of 7 ordered pairs that satisfy the condition. Namely,
\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\]
Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$ $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.
Since we know that $a_i<5$ , for each of the ordered pairs $(b_1,b_2)$ , there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation
\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\]
Thus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$ , we have a total of $5\cdot5=25$ values for $N$ $\boxed{25}$ | E | 25 |
d9f2c466598859fbc7310acbe82dae60 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ .
Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$
If $a < \frac{25}{2}$ $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ $2N \equiv 2a - 25 \pmod {25}$
Using the same logic for $b$ , if $b < 18$ $2N \equiv 2b \pmod {36}$ , and in the other case $2N \equiv 2b - 36 \pmod {36}$
We can do four cases:
Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$
For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$
Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$
Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold.
We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.
Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$
Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold.
We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.
Case 4: $a + b = 2a + 2b$
Trivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.
By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \boxed{25}$ | E | 25 |
a27c5d953d7c62fca94c9c11773be63e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\boxed{25}$ cupcakes. | D | 25 |
a27c5d953d7c62fca94c9c11773be63e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | In $300$ seconds ( $5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{25}$ cupcakes. | D | 25 |
a27c5d953d7c62fca94c9c11773be63e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{25}$ | D | 25 |
45ba8e3ff5e8aef19422fd5534c264a3 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_3 | A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | [asy] draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); [/asy]
Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{15}$ | E | 15 |
5c5f7c6aa72006e98c9070c7d5cff817 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_4 | Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$ . What is the smallest possible degree measure for $\angle CBD$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12$ | $\angle ABD$ and $\angle ABC$ share ray $AB$ . In order to minimize the value of $\angle CBD$ $D$ should be located between $A$ and $C$
$\angle ABC = \angle ABD + \angle CBD$ , so $\angle CBD = 4$ . The answer is $\boxed{4}$ | C | 4 |
89edc5a40d941d87b85597cfac848878 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_5 | Last year 100 adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was 4. What was the total number of cats and kittens received by the shelter last year?
$\textbf{(A)}\ 150\qquad\textbf{(B)}\ 200\qquad\textbf{(C)}\ 250\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 400$ | Half of the 100 adult cats are female, so there are $\frac{100}{2}$ $50$ female cats. Half of those female adult cats have a litter of kittens, so there would be $\frac{50}{2}$ $25$ litters. Since the average number of kittens per litter is 4, this implies that there are $25 \times 4$ $100$ kittens. So the total number of cats and kittens would be $100 + 100$ $\boxed{200}$ | B | 200 |
58fc53a65f572daf3ca59f415dd4639a | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:
$a+b=12$
$b+c=17$
$a+c=19$
Adding these equations together, we get that
$2(a+b+c)=48$ and
$a+b+c=24$
Substituting the original equations into this one, we find
$c+12=24$
$a+17=24$
$b+19=24$
Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{7}$ | D | 7 |
58fc53a65f572daf3ca59f415dd4639a | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:
$a+b=12$
$a+c=17$
$b+c=19$
To isolate $b$ , We add the first and last equations and then subtract the second one.
$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$
Because $b$ is the middle number, the middle number is $\boxed{7}$ | D | 7 |
6be7b77ada8d863b4740e0ad5bae9abb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_9 | A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$ | The total number of combinations when rolling two dice is $6*6 = 36$
There are three ways that a sum of 7 can be rolled. $2+5$ $4+3$ , and $6+1$ . There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the other, so there are a total of 4 ways to roll the combination of 4 and 3. There are two 6's on one die and two 1's on the other, so there are a total of 4 ways to roll the combination of 6 and 1. Add $4 + 4 + 4 = 12$
Thus, our probability is $\frac{12}{36} = \frac{1}{3}$ . The answer is $\boxed{13}$ | D | 13 |
6be7b77ada8d863b4740e0ad5bae9abb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_9 | A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$ | Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is $\boxed{13}$ | D | 13 |
b23344ad1c9cfcd499feadd67794391a | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$ , which leads us to $a_1 + a_{12} = 60$ $a_{12}$ , the largest term of the progression, can also be expressed as $a_1+11d$ , where $d$ is the common difference. Since each angle measure must be an integer, $d$ must also be an integer. We can isolate $d$ by subtracting $a_1$ from $a_{12}$ like so: $a_{12}-a_1=a_1+11d-a_1=11d$ . Since $d$ is an integer, the difference between the first and last terms, $11d$ , must be divisible by $11.$ Since the total difference must be less than $60$ , we can start checking multiples of $11$ less than $60$ for the total difference between $a_1$ and $a_{12}$ . We start with the largest multiple, because the maximum difference will result in the minimum value of the first term. If the difference is $55$ $a_1=\frac{60-55}{2}=2.5$ , which is not an integer, nor is it one of the five options given. If the difference is $44$ $a_1=\frac{60-44}{2}$ , or $\boxed{8}$ | C | 8 |
b23344ad1c9cfcd499feadd67794391a | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | If we let $a$ be the smallest sector angle and $r$ be the difference between consecutive sector angles, then we have the angles $a, a+r, a+2r, \cdots. a+11r$ . Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.
\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 360\\ 2a+11r &= 60\\ a &= \frac{60-11r}{2} \end{align*}
All sector angles are integers so $r$ must be a multiple of 2. Plug in even integers for $r$ starting from 2 to minimize $a.$ We find this value to be 4 and the minimum value of $a$ to be $\frac{60-11(4)}{2} = \boxed{8}$ | C | 8 |
b23344ad1c9cfcd499feadd67794391a | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Starting with the smallest term, $a - 5x \cdots a, a + x \cdots a + 6x$ where $a$ is the sixth term and $x$ is the difference. The sum becomes $12a + 6x = 360$ since there are $360$ degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, $a - 5x > 0$ \[2a + x = 60\] \[x = 60 - 2a\] \[a - 5(60 - 2a) > 0\] \[11a > 300\] Since $a$ is an integer, it must be $28$ , and therefore, $x$ is $4$ $a - 5x$ is $\boxed{8}$ | C | 8 |
664247a80ab2fdbe940ed79a90a7e790 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_11 | Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$ , respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$ . What is $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ $AB=8$ . Also, let $BC=x$ . As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$ $\triangle ADC \sim \triangle BEC$ . From this we can get a proportion.
$\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{12}$ | D | 12 |
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