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b0505938806fc67ffa6f318f02a9b0a7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$ | First, we set a bound by writing $1000$ in base- $16$ $1000_{10}=3E8_{16}$ . Therefore, we are considering numbers with a maximum of $3$ digits, and a maximum of $3$ in the $256$ ths-place (the first place in a $3$ -digit number).
Case $1$ $1$ -digit numbers:
There are evidently $9$ numbers that fit this category.
Case $2$ $2$ -digit numbers:
There are $9\cdot10=90$ numbers that fit this category.
Case $3$ $3$ -digit numbers:
There are $3\cdot10\cdot10=300$ numbers that fit this category
Adding these up, we get $9+90+300=399$ numbers. $3 + 9 + 9 = \boxed{21}$ ~sosiaops | E | 21 |
b0505938806fc67ffa6f318f02a9b0a7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$ | We can quickly see that $400$ in hexadecimal = $0+0+256*4$ = 1024. If we go down to 399 in hexadecimal, we have $9+9*16+3*256$ which is $921$ , which is obviously less than 1000. Therefore, the answer is $3+9+9$ $\boxed{21}$ | E | 21 |
12d9a94c537f9b47b7de8b5f6ddf8ead | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_20 | A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$
$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$ | Let the rectangle's length be $a$ and its width be $b$ . Its area is $ab$ and the perimeter is $2a+2b$
Then $A + P = ab + 2a + 2b$ . Factoring, we have $(a + 2)(b + 2) - 4$
The only one of the answer choices that cannot be expressed in this form is $102$ , as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$ , keeping $a$ and $b$ as positive integers.
Our answer is then $\boxed{102}$ | B | 102 |
2b7381cb1665b8dbf5046c54367a4e6d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.
Because the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields \[(a - 4)^2 = k^2 + 16.\] Therefore $(a-4)^2 - k^2 = 16$ and \[((a-4) - k)((a-4) + k) = 16.\] Let $(a-4) - k = u$ and $(a-4) + k = v$ ; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$ . Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$ ), $(2, 8), (4, 4), (-2, -8), (-4, -4)$ , yields $a = 9, 8, -1, 0$ . These $a$ sum to $16$ , so our answer is $\boxed{16}$ | C | 16 |
2b7381cb1665b8dbf5046c54367a4e6d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$ , the quadratic can be written as \[(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2\]
By comparing this with $x^2 - ax + 2a$ \[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]
Plugging the first equation in the second, \[r_1r_2 = 2 (r_1 + r_2).\] Rearranging gives \[r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.\] These factors can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$
We want the number of distinct $a = r_1 + r_2$ , and these factors gives $a = -1, 0, 8, 9$ . So the answer is $-1 + 0 + 8 + 9 = \boxed{16}$ | C | 16 |
b4ff2516045e46b534dfd03aa8d6ef39 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{31}$ | B | 31 |
b4ff2516045e46b534dfd03aa8d6ef39 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this down and testing, it appears that this holds for all $x$ . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers.
In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{31}$ | B | 31 |
2dd7298d0c1630d5895f6d28ac47eab2 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_25 | Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$
$\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$ . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least $0.5$ apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be up to $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is \[\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.\]
(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$ , i.e. $(x, y)$ is outside the unit circle with radius $0.5.$
Thus, averaging the probabilities gives \[P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).\]
Thus our answer is $\boxed{59}$ | A | 59 |
2dd7298d0c1630d5895f6d28ac47eab2 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_25 | Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$
$\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases:
Case 1: Point B is on the same side as Point A: This setup occurs with probability $\dfrac{1}{4}.$ This is the standard geometric probability problem. Since Point A and Point B can be anywhere on the side, we can't really "count" all of the possibilities. Hence, we translate the problem to a problem involving areas, which can be a finite value while still "containing" infinitely many points. Let side of a square be a number line from $0$ to $1,$ and $a$ and $b$ be the values representing the positions of Point A and Point B respectively. Our problem now asks for the probability that $|a-b| \geq 1/2.$ Graphing the inequality on a coordinate plane with $a$ and $b$ as the $x$ and $y$ -axes gives us a "good area" of $1/4$ out of a "total area" of $1.$ Hence, the probability the inequality is satisfied is $1/4 \div 1 = 1/4.$
Case 2: Point B is on a side adjacent to the side with Point A: This setup occurs with probability $\dfrac{1}{2}.$ This is a slight deviation of the same geometric probability principle. This time, let the common vertex of the two sides be $0,$ and the sides have side length $1.$ Again, let $a$ and $b$ be the values representing the positions of Point A and Point B respectively, so the Pythagorean Theorem yields $\sqrt{a^2+b^2}\geq 1/2,$ graphing into a quarter circle of radius $1/2,$ and a total area that is $1.$ Hence, our probability is $1-\frac{\pi(1/2)^2}{4} = \frac{16-\pi}{16}.$
Case 3: Point B is on a side opposite to the side with Point A: This setup occurs with probability $\dfrac{1}{4}.$ Clearly, the distance between Point A and Point B are at least 1, so it must be at least $1/2.$ The probability in this case is $1.$
Now taking the probabilities of the setups into account, our final probability is \[\frac{1}{4}\cdot \left( \frac{1}{4} \right)+ \frac{1}{2}\cdot \left(\frac{16-\pi}{16}\right) + \frac{1}{4}\cdot \left(1\right)=\frac{26-\pi}{32}.\] Thus $(a,b,c)=(26,1,32),$ so $a+b+c= \boxed{59}$ | A | 59 |
b3d28b09d1386a7f5d6fffb404b81270 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_3 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is $100$ , and one of the numbers is $28.$ What is the other number?
$\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18$ | Let the first number be $x$ and the second be $y$ . We have $2x+3y=100$ . We are given one of the numbers is $28$ . If $x$ were to be $28$ $y$ would not be an integer, thus $y=28$ $2x+3(28)=100$ , which gives $x=\boxed{8}$ | A | 8 |
732c24174db66df61056c30fcecd4e92 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_10 | What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$ $\textbf{(A) }$ It is a negative number ending with a 1. $\textbf{(B) }$ It is a positive number ending with a 1. $\textbf{(C) }$ It is a negative number ending with a 5. $\textbf{(D) }$ It is a positive number ending with a 5. $\textbf{(E) }$ It is a negative number ending with a 0. | Since $-5>-2015$ , the product must end with a $5$
The multiplicands are the odd negative integers from $-1$ to $-2013$ . There are $\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$ , the product is negative.
Therefore, the answer must be $\boxed{5.}$ | C | 5. |
df38cc8ba35e2c999de23fc0a06911b7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_14 | Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$ | Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formula the sum of the roots is $\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let $b=9$ $a=8$ , and $c=7$ . Then the answer is $\dfrac{8+18+7}{2}=\boxed{16.5}$ | D | 16.5 |
df38cc8ba35e2c999de23fc0a06911b7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_14 | Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$ | Factoring out $(x-b)$ from the equation yields $(x-b)(2x-(a+c))=0 \Rightarrow (x-b)\left(x-\frac{a+c}{2}\right)=0$ . Therefore the roots are $b$ and $\frac{a+c}{2}$ . Because $b$ must be the larger root to maximize the sum of the roots, letting $a,b,$ and $c$ be $8,9,$ and $7$ respectively yields the sum $9+\frac{8+7}{2} = 9+7.5 = \boxed{16.5}$ | D | 16.5 |
f396e4ddf3defcae0fcdde5838dd6643 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$ | Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ .
We know \[3h=p\] \[4c=s\] \[3p=d\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\times3h)$ . This is equivalent to $13h+5c$ . Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$ . So the answer is $\boxed{47}$ | B | 47 |
f396e4ddf3defcae0fcdde5838dd6643 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$ | As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$ \[13\cdot 5-13-5=47,\] so our answer is $\boxed{47}$ | B | 47 |
e76bf24160ca8531bac88fa17937d190 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]
$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$ | The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\frac{3}{2}$ , so the volume of one pyramid is $\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\cdot5=\boxed{10}$ | B | 10 |
e76bf24160ca8531bac88fa17937d190 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]
$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$ | The "base" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{10}$ | null | 10 |
93ca36180b7e118a06ff0b88e277295d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$ | We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),white); filldraw(circle((-2,0),0.35),white); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
Then, after the second (new heads in blue);
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
And after the third (new head in green);
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),green); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
So in total, $7$ of the $8$ coins resulted in heads. Now we have the ratio of $\frac{7}{8}$ of the total coins will end up heads. Therefore, we have $\frac{7}{8}\cdot64=\boxed{56}$ | D | 56 |
93ca36180b7e118a06ff0b88e277295d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$ | Every time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is $32$ , on the second flip is $16$ , and on the third flip, it is $8$ . Adding these gives $\boxed{56}$ | D | 56 |
93ca36180b7e118a06ff0b88e277295d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$ | Every time the coins are flipped, each of them has a $1/2$ probability of being tails. Doing this $3$ times, $1/8$ of them will be tails. $64-64*1/8=$ $\boxed{56}$ | D | 56 |
93ca36180b7e118a06ff0b88e277295d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$ | (Similar to solution 2)
Notice how:
$E(\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\text{Heads on 1st flip}) + E(\text{Heads on 2nd flip}) + E(\text{Heads on 3rd flip})$
The expected number of heads for the first flip is simply $64 \cdot \frac{1}{2}$ , since each coin has a 1 in 2 chance of being heads. Then, we are left with $64 - 32 = 32$ coins. Then, half of these coins will be heads again, which leaves us with $32 - 16 = 16$ coins. Then, half of these coins will be heads again, which leaves us with $16 - 8 = 8$ coins.
Hence, the expected number of heads is simply:
$32 + 16 + 8 = \boxed{56} \implies D$ | null | 56 |
18bf638a0cb797d222d008b988c16b6d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | [asy]import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); label("2",(0,0,0),S); label("A",(1,0,0),W); label("B",(0,0,1),N); label("1",(1,0,1),NW); label("3",(1,1,0),S); label("C",(0,1,0),E); label("D",(1,1,1),SE); label("4",(0,1,1),NE); [/asy]
We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$
If we define a "move" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from $A$ (since Erin cannot crawl back to $A$ ), this vertex must be $4$
Therefore, we now just need to count the number of paths from $A$ to $4$ . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$ , and 2 choices for which vertex Erin was at 2 moves before $4$ . All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \cdot 2 = \boxed{6}$ | A | 6 |
18bf638a0cb797d222d008b988c16b6d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | Lets say that this cube is an unit cube and the given corner is $(0,0,0)$ . Because Erin cannot return back to her starting point, she cannot be on $(0,0,1)$ $(0,1,0)$ , or $(1,0,0)$ . She cannot be on $(1,1,0)$ $(1,0,1)$ , or $(0,1,1)$ , because after $7$ moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at $(1,1,1)$ . Now, we draw a net and see that there are $3$ choices for the first move, $2$ for the second, and the rest are forced. Thus the answer is $3\cdot2 = \boxed{6}$ | A | 6 |
18bf638a0cb797d222d008b988c16b6d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | Let's suppose the given corner on the cube is $(0,0,0)$ . Erin has 3 identical ways to proceed. Suppose she goes to $(0,1,0)$ . She now has two more identical ways to go. Let's say she goes to $(0,1,1)$ . She has to go to $(0,0,1)$ , otherwise, she will end up on that point after 7 moves. This is because once if she chooses the other path to $(1,1,1)$ , the endpoint is certain to be $(0,0,1)$ , which is directly connected to $(0,0,0)$ . After she goes to $(0,0,1)$ , the only option she has is to go to $(1,0,1)$ , then $(1,0,0)$ , after that $(1,1,0)$ , and finally $(1,1,1)$ . She is forced to go this way because she cannot end up on $(1,0,0)$ . At the start, she had 3 ways to choose, and after that 2 ways to choose, so $3\cdot2=\boxed{6}$ | A | 6 |
efb7a968e717346fe35f1abc02524379 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left). Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$ | Let $n$ be the number of steps. We have
\[\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case $1$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}$
Our equation becomes $\frac{n}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=1$ and $j=4$ yield $n=64$ and $n=66$ , respectively.
Case $2$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}+\frac{1}{2}$
Our equation becomes $\frac{n}{2} +\frac{1}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=2$ yields $n=63$ . Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$ | D | 13 |
efb7a968e717346fe35f1abc02524379 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left). Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$ | We know from the problem that Dash goes $3$ steps further than Cozy per jump (assuming they aren't within $4$ steps from the top). That means that if Dash takes $19$ fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least $3 \cdot 19=57$ steps high. We then start using guess-and-check:
$57$ steps: $\left \lceil {57/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {57/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$58$ steps: $\left \lceil {58/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {58/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$59$ steps: $\left \lceil {59/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {59/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$60$ steps: $\left \lceil {60/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {60/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$\vdots$
By the time we test $61$ steps, we notice that when the number of steps exceeds a multiple of $2$ , the difference in jumps increases. So, we have to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$ $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward.
Therefore, the possible numbers of steps in the staircase are $63$ $64$ , and $66$ , giving a sum of $193$ . The sum of those digits is $13$ , so the answer is $\boxed{13}$ | D | 13 |
efb7a968e717346fe35f1abc02524379 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left). Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$ | We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$
Let's call $x = 10a + b$ . We now have $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$ , or
$19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$
Obviously, since $b \le 10$ , this will not work for any value of $a$ under $6$ . In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$ , this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$
This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $\left \lceil{\frac{b}{2}}\right \rceil = 2$ , or when $\left \lceil{\frac{b}{5}}\right \rceil = 2$ and $\left \lceil{\frac{b}{2}}\right \rceil = 3$
Case One:
We have $b \le 5$ and $3 \le b \le 4$ , so $b = 3, 4$
Case Two:
We have $6 \le b \le 9$ and $5 \le b \le 6$ , so $b = 6$
We then have $63 + 64 + 66 = 193$ , which has a digit sum of $\boxed{13}$ | null | 13 |
efb7a968e717346fe35f1abc02524379 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left). Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$ | Translate the problem into following equation:
$n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}$
Since $C = D + 19$ , we have
$5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}$
i.e.,
$3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}$
We then have $D = 13$ when $\{1\} - \{0\}$ or $\{2\} - \{1\}$ (the dog's last jump has $2$ steps and the cat's last jump has $1$ step), which yields $n = 64$ and $n = 63$ respectively.
Another solution is $D = 14$ when $\{4\} - \{0\}$ , which yields $n = 66$
Therefore, with $63 + 64 + 66 = 193$ , the digit sum is $\boxed{13}$ | null | 13 |
5cdbf4afa2c7782be8b1f57d384b329e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\]
We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$ $(2n)!$ has only $2$ zeros. But for $n=8,9$ $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work.
Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$ $(2n)!$ has only $4$ zeros. But for $n=13,14$ $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$ , which sum to $44$ . The sum of the digits of $44$ is $\boxed{8}$ | B | 8 |
5cdbf4afa2c7782be8b1f57d384b329e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | By Legendre's Formula and the information given, we have that $3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor$
We have $n<100$ as there is no way that if $n>100$ $(2n)!$ would have $3$ times as many zeroes as $n!$
First, let's plug in the number $5$ .
We get that $3(1)=1$ , which is obviously not true. Hence, $n>5$
After several attempts, we realize that the RHS needs $1$ to $2$ more "extra" zeroes than the LHS. Hence, $n$ is greater than a multiple of $5$
We find that the least four possible $n$ are $8,9,13,14$
$8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{8}$ | B | 8 |
5cdbf4afa2c7782be8b1f57d384b329e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | Let $n=5m+k$ for some natural numbers $m$ $k$ such that $k\in\{0,1,2,3,4\}$ . Notice that $n<5^3=125$ . Thus \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor\] For smaller $n$ , we temporarily let $\left\lfloor\frac{2n}{125}\right\rfloor=0$ \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{2(5m+k)}{5}\right\rfloor+\left\lfloor\frac{2(5m+k)}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{10m+2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[3m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=2m+\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] To minimize $n$ , we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor\] Since $k<5$ $m>0$ , the only integral value of $m$ is $1$ , from which we have $k=3,4\Longrightarrow n=8,9$
Now we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=0$ and $\left\lfloor\frac{10m+2k}{25}\right\rfloor=1$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] Since $k<5$ $10m>15\Longrightarrow m\ge2$
If $m>2$ , then \[m>\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] which is a contradiction.
Thus $m=2\Longrightarrow\left\lfloor\frac{2k}{5}\right\rfloor=1\Longrightarrow n=13,14$
Finally, the sum of the four smallest possible $n=8+9+13+14=44$ and $4+4=8$ $\boxed{8}$ | B | 8 |
5cdbf4afa2c7782be8b1f57d384b329e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | We first note that the number of 0's in $n!$ is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
$\textbf{CASE ONE: } 5\leq 2n < 25.$
The only way we can fulfill the requirements is if $\lfloor{\dfrac{n}{5}}\rfloor = 1$ and $\lfloor{\dfrac{2n}{5}}\rfloor=3$ which means that $5\leq n <10$ and $15\leq 2n < 20$ . The only way this works is if $n = 8 \text{ or } 9.$
$\textbf{CASE TWO: } 25 \leq 2n$
Since we want the smallest values of $n$ , we first try it when $2n<30.$ Thus $(2n)!$ has 6 zeros, which implies that $n!$ must have 2. The only way to do this while maintaining our restrictions for $2n$ is if $n = 13 \text{ or } 14.$
So the sum of the four values is $8+9+13+14=44$ so the digit is sum is $\boxed{8}.$ | B | 8 |
5cdbf4afa2c7782be8b1f57d384b329e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | We will use trial and error to determine the answer to this problem. If $n = 5$ , then $n!$ has $1$ zero, and $2n!$ will have $2$ zeros. But $3 \cdot 1 \neq 2$ so $n = 5$ does not work. Similarly $n = 6, 7$ do not work either. But $n = 8$ works because $8!$ has $1$ zero and $16!$ has $3$ zeros. Note that $n = 9$ also works because $9!$ has $1$ zero and $18!$ has $3$ zeros. After performing trial and error several times, we find that $n = 10, 11, 12$ do not work but $n = 13, 14$ do work. Therefore, the four smallest values of $n$ are $8, 9, 13, 14$ . Therefore adding them together gives $44$ and our answer is $\boxed{8}.$ | B | 8 |
9fe6f95b36e062ef405fe7b61081a89c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_25 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$ | We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions. | B | 10 |
9fe6f95b36e062ef405fe7b61081a89c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_25 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$ | The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions. | B | 10 |
d14a37122148263020e684db8e2fc723 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_3 | Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $\textdollar 0.75$ for her to make. In dollars, what is her profit for the day?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$ | She first sells one-half of her $48$ loaves, or $\frac{48}{2}=24$ loaves. Each loaf sells for $\textdollar 2.50$ , so her total earnings in the morning is equal to \[24\cdot \textdollar 2.50 = \textdollar 60\]
This leaves 24 loaves left, and Bridget will sell $\dfrac{2}{3}\times 24=16$ of them for a price of $\textdollar\frac{2.50}{2}=\textdollar 1.25$ . Thus, her total earnings for the afternoon is \[16\cdot \textdollar 1.25 = \textdollar 20\]
Finally, Bridget will sell the remaining $24-16=8$ loaves for a dollar each. This is a total of $\textdollar 1\cdot 8 = \textdollar 8$
The total amount of money she makes is equal to $60+20+8=\textdollar 88$
However, since Bridget spends $\textdollar 0.75$ making each loaf of bread, the total cost to make the bread is equal to $\textdollar 0.75\cdot48=\textdollar 36$
Her total profit is the amount of money she spent subtracted from the amount of money she made, which is \[88-36=52\implies\boxed{52}\] | E | 52 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\text{B}-\text{O}-\text{R}-\text{Y}$
$\text{O}-\text{B}-\text{R}-\text{Y}$
The answer is $1+2=\boxed{3}$ | B | 3 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\cdot\frac{1}{2}\cdot\frac{1}{2}= \boxed{3}$ | B | 3 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
Therefore, our answer is $\boxed{3}$ | B | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$
Thus, the solution is \[90-87=3\implies\boxed{3}\] | C | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $70$ or $80$ ).The mean is $10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87$ .
So, our solution is $90-87=3\Rightarrow\boxed{3}$ ~sosiaops | C | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | The $\le 80$ -point scores make up $10\%+35\% = 45\% < 50\%$ of the scores, but the $\le 90$ -point scores make up $45\%+30\% = 75\% > 50\%$ of the scores, so the median is $90$
$10\%$ of scores were $70-90 = -20$ more than the median, $35\%$ were $-10$ more, $100\%-75\% = 25\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\left|-3\right| = \boxed{3}$ .
~ emerald_block | C | 3 |
e5fce70b1196e25cb297b450431159e7 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_7 | Nonzero real numbers $x$ $y$ $a$ , and $b$ satisfy $x < a$ and $y < b$ . How many of the following inequalities must be true?
$\textbf{(I)}\ x+y < a+b\qquad$
$\textbf{(II)}\ x-y < a-b\qquad$
$\textbf{(III)}\ xy < ab\qquad$
$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$ . We can write that $x + y < x + y + k + l \implies x + y < a + b$
It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$
$\textbf{(II)}$ states that $x-y<a-b \implies -3 - (-4) < 1 - 4 \implies 1 < -3$ .Therefore, $\textbf{(II)}$ is false.
$\textbf{(III)}$ states that $xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4$ . Therefore, $\textbf{(III)}$ is false.
$\textbf{(IV)}$ states that $\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-4} < \frac{1}{4} \implies 0.75 < 0.25$ . Therefore, $\textbf{(IV)}$ is false.
One of our four inequalities is true, hence, our answer is $\boxed{1}$ | B | 1 |
f262e99b30ccaf7979e0affb9f4a4a02 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.
Let $h$ be the third height of the triangle. We have $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{3}$ | C | 3 |
f262e99b30ccaf7979e0affb9f4a4a02 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$ . Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is $\boxed{3}$ (We can also check from the other side). | C | 3 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$
Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{4}$ | B | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]
We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]
Thus, the answer is $\boxed{4}$ | B | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We know from experience that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\boxed{4}$ | null | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{4}$ | B | 4 |
bdf1157d287c23856113e8dc3c9bdb32 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_11 | problem_id
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
Name: Text, dtype: object | Let the listed price be $x$ . Since all the answer choices are above $\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:
Coupon 1: $x\times10\%=.1x$
Coupon 2: $20$
Coupon 3: $18\%\times(x-100)=.18x-18$
For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$
The only choice that satisfies such conditions is $\boxed{219.95}$ | C | 219.95 |
bdf1157d287c23856113e8dc3c9bdb32 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_11 | problem_id
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
Name: Text, dtype: object | For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\boxed{219.95}$ is the smallest number over 200. | C | 219.95 |
be7d45583e889c7e334f9d19370030bd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | [asy]//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1) draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1) draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); label("$a$", (0,5),W); label("$a$",(0,-5),W); label("$a$",(3,4),NW); label("$P$",(0,10),SW); label("$Q$",(0,-10),NW); // wanted to import graph and use xaxis/yaxis but w/e label("$x$",(9,0),E); label("$y$",(0,13),N); [/asy] Note that if the $y$ -intercepts have a sum of $0$ , the distance from the origin to each of the intercepts must be the same. Call this distance $a$ . Since the $\angle PAQ = 90^\circ$ , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\sqrt{6^2+8^2} = 10$ , this means $a=10$ , and the length of the hypotenuse is $2a = 20$ . Since the $x$ -coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{60}$ | D | 60 |
be7d45583e889c7e334f9d19370030bd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$ , and the sum of the y-intercepts is equal to 0, hence the $b, -b$
Since both lines contain the point $(6,8)$ , we can plug this into the two equations to obtain \[8=6m+b\] and \[8=-6\frac{1}{m}-b\]
Adding the two equations gives \[16=6m+\frac{-6}{m}\] Multiplying by $m$ gives \[16m=6m^2-6\] \[\implies 6m^2-16m-6=0\] \[\implies 3m^2-8m-3=0\] Factoring gives \[(3m+1)(m-3)=0\]
Plugging $m=3$ into one of our original equations, we obtain \[8=6(3)+b\] \[\implies b=8-6(3)=-10\]
Since $\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$ , or 6, the area of $\bigtriangleup APQ$ is equal to \[\frac{20\cdot6}{2}=\boxed{60}\] | D | 60 |
be7d45583e889c7e334f9d19370030bd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | Since their sum is $0$ , let the y intercepts be P $(0,a)$ and Q $(0,-a)$ . The slope of $AP$ is $\frac{8-a}{6}$ . The slope of AQ is $\frac{8+a}{6}$ . Since multiplying the slopes of perpendicular lines yields a product of $-1$ , we have $\frac{64-a^2}{36}=-1$ , which results in $a^2=100$ . We can use either the positive or negative solution because if we choose $10$ , then the other y-intercept is $-10$ ; but if we choose $-10$ , then the other y-intercept is $10$ . For simplicity, we choose that $a=10$ in this solution.
Now we have a triangle APQ with points A $(6,8)$ , P $(0,10)$ , and Q $(0,-10)$ . By the Pythagorean theorem, we have that $AP=\sqrt{6^2+2^2}=2\sqrt{10}$ , and that $AQ=\sqrt{6^2+18^2}=6\sqrt{10}$ $PQ$ is obviously $10-(-10)=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{60}$ | D | 60 |
be7d45583e889c7e334f9d19370030bd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | Using point-slope form, the first line has the equation \[y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8\] The second line has the equation \[y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8\] At the y-intercept, the value of the x-coordinate is $0$ , hence: the first equation is $y=-6m+8$ and the second is $y=\frac{6}{m}+8$ . Since the y-intercepts sum to $0$ , they are opposites, so: \[-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8\] \[6m-\frac{6}{m}=16\] Multiply both sides by m: $6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0$ . The solution to this quadratic, using the quadratic formula, is: $\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}$ This yields $m=-\frac{1}{3}$ and $m=3$ . Plugging $m=3$ into the second equation, we get $y=\frac{6}{3}+8=10$ . Plugging $m=-\frac{1}{3}$ into the first equation, we get $y=-10$ So the base is $20$ and the height is $6$ , the area is $60 \Longrightarrow \boxed{60}$ | D | 60 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.
Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.
Simplifying gives $7d+525=10d$ , or $d=175$
Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{210}$ miles. | C | 210 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.
Quickly checking, we know that neither choice $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$ , David has $175$ miles left. This then takes him $3.5$ hours at $50 \text{ mph}$ . But $210/35 = 6 \text{ hours}$ . Since $1+3.5 = 4.5 \text{ hours}$ is $1.5$ hours less than $6$ , our answer is $\boxed{210}$ | C | 210 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\cdot 6=210 \Longrightarrow \boxed{210}$ | C | 210 |
f82074e1fe6caad968075124ec6f106c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_16 | In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$ | Denote $D=(0,0)$ . Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$ . Let the intersection of $AF$ and $DH$ be $X$ , and the intersection of $BF$ and $CH$ be $Y$ . Then we want to find the coordinates of $X$ so we can find $XY$ . From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4$ , and its $y$ -intercept is just $2$ . Thus the equation for $AF$ is $y = -4x + 2$ . We can also quickly find that the equation of $DH$ is $y = 2x$ . Setting the equations equal, we have $2x = -4x +2 \implies x = \frac13$ . Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac13$ , so $XY = 1 - 2 \cdot \frac13 = \frac13$ . Now the area of the kite is simply the product of the two diagonals over $2$ . Since the length $HF = 1$ , our answer is $\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{16}$ | E | 16 |
f82074e1fe6caad968075124ec6f106c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_16 | In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$ | From the diagram in Solution 1, let $e$ be the height of $XHY$ and $f$ be the height of $XFY$ . It is clear that their sum is $1$ as they are parallel to $GD$ . Let $k$ be the ratio of the sides of the similar triangles $XFY$ and $AFB$ , which are similar because $XY$ is parallel to $AB$ and the triangles share angle $F$ . Then $k = f/2$ , as 2 is the height of $AFB$ . Since $XHY$ and $DHC$ are similar for the same reasons as $XFY$ and $AFB$ , the height of $XHY$ will be equal to the base, like in $DHC$ , making $XY = e$ . However, $XY$ is also the base of $XFY$ , so $k = e / AB$ where $AB = 1$ so $k = e$ . Subbing into $k = f/2$ gives a system of linear equations, $e + f = 1$ and $e = f/2$ . Solving yields $e = XY = 1/3$ and $f = \frac{2}{3}$ , and since the area of the kite is simply the product of the two diagonals over $2$ and $HF = 1$ , our answer is $\frac{\frac{1}{3} \cdot 1}{2} = \boxed{16}$ | E | 16 |
5ca605c93d26933e8fd8436bfbc5840e | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | Let the points be $A=(x_1,0)$ $B=(x_2,1)$ $C=(x_3,5)$ , and $D=(x_4,4)$
Note that the difference in $y$ value of $B$ and $C$ is $4$ . By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$ . Note that the difference in $y$ value of $A$ and $B$ is $1$ . We now know that $AB$ , the side length of the square, is equal to $\sqrt{1^2+4^2}=\sqrt{17}$ , so the area is $\boxed{17}$ | B | 17 |
5ca605c93d26933e8fd8436bfbc5840e | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | By translation, we can move the square with point $A$ at the origin. Then, $A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)$ . We will use the relationship among the 4 sides of being perpendicular and equal.
The slope of $AB$ is $\frac{1-0}{x_1-0}=\frac{1}{x_1}$
Because $BC$ is perpendicular to $AB$ , the slope of $BC=-x_1$ .
From the information above we could have the equation:
Because $CD$ is perpendicular to $BC$ , the slope of $CD=\frac{1}{x_1}$ .
From the information above we could have the equation:
Because $AD=AB,$
Note that the square with $x_1=-4$ is just the reflection of square with $x_1=4$ over the origin. I will use $x_1=4$ $B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{17}$ | B | 17 |
5ca605c93d26933e8fd8436bfbc5840e | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
Using the fact that the diagonals bisect each other, we get the equation:
Now we use the fact that the diagonals are perpendicular to each other:
Using the fact that the diagonals are equal in length, we get the equation:
Now we have 3 equations with 3 variables:
$\begin{cases} x_2=x_1+x_3 \\ x_2(x_1-x_3)=15 \\ (x_3-x_1)^2-x_2^2=16 \end{cases}$
We substitute $x_2$ into the 2 other equations:
Now we have 2 equations of $x_1$ and $x_3$
$\begin{cases} x_1^2-x_3^2=15 \\ x_1x_3=-4 \end{cases}$
This is the same equation as solution $2$ . So $x_1= \pm 4, AB=\sqrt{17}, [ABCD]=\boxed{17}$ | B | 17 |
5ca605c93d26933e8fd8436bfbc5840e | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | Draw it out, by inspection the coordinates are $(-1, 4)$ $(0, 0)$ $(4, 1)$ , and $(3, 5)$ . The side length is $\sqrt{17}, [ABCD]=\boxed{17}$ ~JH. L | B | 17 |
6a2cc5541e1749d3d0bb739aacabadf4 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_20 | problem_id
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
Name: Text, dtype: object | We can list the first few numbers in the form $8 \cdot (8....8)$
(Hard problem to do without the multiplication, but you can see the pattern early on)
$8 \cdot 8 = 64$
$8 \cdot 88 = 704$
$8 \cdot 888 = 7104$
$8 \cdot 8888 = 71104$
$8 \cdot 88888 = 711104$
By now it's clear that the numbers will be in the form $7$ $k-2$ $1$ 's, and $04$ . We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$ . Solving, we get $k = 991$ , meaning the answer is $\fbox{(D)}$
Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\boxed{991}$ | D | 991 |
d500296532c80f4d469509ca67eb0c54 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_21 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$ | Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that \[0 = ax + 5 \implies x = -\dfrac{5}{a}\] \[0 = 3x+b \implies x= -\dfrac{b}{3}\] Which means that \[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\] The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5), (5,3), (15, 1)$ . These pairs give respective $x$ -values of $-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}$ which have a sum of $\boxed{8}$ | E | 8 |
d500296532c80f4d469509ca67eb0c54 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_21 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$ | Going off of Solution 1, for the first equation, notice that the value of $x$ cannot be less than $-5$ . We also know for the first equation that the values of $x$ have to be $5$ divided by something. Also, for the second equation, the values of $x$ can only be $-\frac13,-\frac23,-\frac33, \dots$ . Therefore, we see that, the only values common between the two sequences are $-1, -5, -\frac13,-\frac53$ , and adding them up, we get for our answer, $\boxed{8}$ | E | 8 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Note that $\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$ .) Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{20}$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$ . By the angle bisector theorem, $\frac{BC}{BF} = \frac{CE}{EF}$ . Since $\triangle BFC$ is a $30-60-90$ right triangle, $CF = \frac{10\sqrt{3}}{3}$ and $BF = \frac{20\sqrt{3}}{3}$ . Additionally, \[CE + EF = CF = \frac{10\sqrt{3}}{3}\] Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{CE}{EF} \Rightarrow \frac{2\sqrt{3}}{3}CE = EF$ and $CE + EF = \frac{10\sqrt{3}}{3}$ . Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}CE + CE = \frac{10\sqrt{3}}{3} \Rightarrow CE = 20 - 10\sqrt{3}$ , so $DE = 10\sqrt{3}$ . Because $\triangle ADE$ is a $30-60-90$ triangle, $AE = \boxed{20}$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Reflect $\triangle{ECB}$ over line segment $\overline{CD}$ . Let the point $F$ be the point where the right angle is of our newly reflected triangle. By subtracting $90 - (15+15) = 60$ to find $\angle ABF$ , we see that $\triangle{ABC}$ is a $30-60-90$ right triangle. By using complementary angles once more, we can see that $\angle{EAD}$ is a $60^\circ$ angle, and we've found that $\triangle{EAD}$ is a $30-60-90$ right triangle. From here, we can use the $1-2-\sqrt{3}$ properties of a $30-60-90$ right triangle to see that $\overline{AE}=\boxed{20}.$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Square.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Reflect rectangle $ABCD$ along line $CD$ . Let the square be $ABFG$ as shown. Construct equilateral triangle $\triangle EFH$
Because $HF=EF$ $GF=BF=20$ , and $\angle GFH=\angle BFE=15^{\circ}$ $\triangle GFH\cong \triangle BFE$ by $SAS$
So, $GH=BE$ $GH=HE=HF$
Because $GH=HE=HF$ $\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}$ $\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$ $\angle GHE=\angle GHF$
$\triangle GHE \cong \triangle GHF$ by $SAS$
So, $GF=GE$ . By the reflection, $AE=GE=GF=AB$ $AE=AB=\boxed{20}$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Rectangle.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Construct equilateral triangle $\triangle BEF$ , and let $GF$ be the height of $\triangle ABF$
$\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}$ $\angle GBF=\angle CBE$ $\angle BGF=\angle BCE=90^{\circ}$ $BF=BE$
$\triangle BGF \cong \triangle BCE$ by $AAS$
$BG=BC=10, AG=20-10=10$ $AG=BG$ $GF=GF$ , by $HL$ $\triangle AGF \cong \triangle BGF$
So, $AF=BF=EF$
$\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}$ $\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$ $\angle AFB=\angle AFE$ $AF=AF$ $BF=EF$
$\triangle AFB \cong \triangle AFE$ by $SAS$
So, $AE=AB=\boxed{20}$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | All trigonometric functions in this solution are in degrees. We know \[\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)\] so \[\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)\] \[=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}\] \[=\frac{\sqrt{2}-\sqrt{6}}{4}\] Let $EC=x$ , then $BE=\sqrt{x^{2}+100}$ . By the definition of sine, \[\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}\] Squaring both sides, \[\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}\] Cross-multiplying, \[4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}\] Simplifying, \[\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}\] \[x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] Let $\frac{2-\sqrt{3}}{2+\sqrt{3}}=p$ . Notice that $\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1$ so $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$ $p$ is then \[\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}\] Recall that \[x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] which we now know is \[100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2}\] Therefore \[x=\frac{10}{2+\sqrt{3}}\] Rationalizing the denominator, \[\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\] Which by difference of squares reduces to \[20-10\sqrt{3}\] so $EC=20-10\sqrt{3}$ $ED$ is then $20-\left(20-10\sqrt{3}\right)=10\sqrt{3}$ and since we know $AD=10$ , by the Pythagorean theorem, $AE = 20$ . The answer is $\boxed{20}$ | E | 20 |
142179ef2260f4678ee5cd59e43de4f0 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | The ratio between the side lengths of a 15, 75, 90 triangle are $\sqrt{3}-1, \sqrt{3}+1, 2\sqrt{2}$ . Therefore, $\cos 75^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$ . Using triangular ratios, $\frac{\sqrt{3}+1}{10}=\frac{2\sqrt{2}}{BE}$ $BE=10\sqrt{2}(\sqrt{3}-1)$
From the law of cosines, $(AE)^2=(AB)^2+(BE)^2-2(AB)(BE)\cos 75^\circ$
So, $(AE)^2=20^2+(10\sqrt{2}(\sqrt{3}-1))^2-2(20)(10\sqrt{2}(\sqrt{3}-1))(\frac{\sqrt{3}-1}{2\sqrt{2}})$
At this stage, it is easier if we do not expand the numbers yet. Further simplifying, we have $(AE)^2=20^2+200(\sqrt{3}-1)^2-40(10\sqrt{2}(\sqrt{3}-1))(\frac{(\sqrt{3}-1)(\sqrt{2})}{4})$
Now, we have $(AE)^2=20^2+200(\sqrt{3}-1)^2-10(10\sqrt{2}(\sqrt{3}-1))(\sqrt{3}-1)(\sqrt{2})$
Combing like terms, $(AE)^2=20^2+200(\sqrt{3}-1)^2-(100(\sqrt{2})^2)(\sqrt{3}-1)^2)$
This equals, $(AE)^2=20^2+200(\sqrt{3}-1)^2-200(\sqrt{3}-1)^2$
The $200(\sqrt{3}-1)^2)$ cancel out each other. We are left with $(AE)^2=20^2$ . So, $AE=\boxed{20}$ .
~hwan | E | 20 |
a276ff37f4fb4f293550d46fa784b45c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_24 | A sequence of natural numbers is constructed by listing the first $4$ , then skipping one, listing the next $5$ , skipping $2$ , listing $6$ , skipping $3$ , and on the $n$ th iteration, listing $n+3$ and skipping $n$ . The sequence begins $1,2,3,4,6,7,8,9,10,13$ . What is the $500,\!000$ th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$ | If we list the rows by iterations, then we get
$1,2,3,4$
$6,7,8,9,10$
$13,14,15,16,17,18$ etc.
so that the $500,000$ th number is the $506$ th number on the $997$ th row because $4+5+6+7......+999 = 499,494$ . The last number of the $996$ th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$ , (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \boxed{996,506}$ | A | 996,506 |
a276ff37f4fb4f293550d46fa784b45c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_24 | A sequence of natural numbers is constructed by listing the first $4$ , then skipping one, listing the next $5$ , skipping $2$ , listing $6$ , skipping $3$ , and on the $n$ th iteration, listing $n+3$ and skipping $n$ . The sequence begins $1,2,3,4,6,7,8,9,10,13$ . What is the $500,\!000$ th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$ | One may also note we simply need to add the number of skipped numbers to $500,000$ to get our answer. The number of skipped numbers is $\frac{996\cdot 997}{2}$ which has a units digit of $6$ . Looking at the answer choices, it becomes apparent that the answer is $\boxed{996,506}$ | A | 996,506 |
a276ff37f4fb4f293550d46fa784b45c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_24 | A sequence of natural numbers is constructed by listing the first $4$ , then skipping one, listing the next $5$ , skipping $2$ , listing $6$ , skipping $3$ , and on the $n$ th iteration, listing $n+3$ and skipping $n$ . The sequence begins $1,2,3,4,6,7,8,9,10,13$ . What is the $500,\!000$ th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$ | Let's start with natural numbers, with no skips in between.
$1,2,3,4,5,...,500,000$
All we need to do is count how many numbers are skipped, $n$ , and "push" (add on to) $500,000$ according to however many numbers are skipped.
Clearly, $\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}$ . This means that there are $999-3=996$ skipped number "blocks" in the sequence because we started counting from 4.
Therefore $n=\frac{996(997)}{2}=496,506$ , and the answer is $496,506+500000=\boxed{996,506}$ | A | 996,506 |
a276ff37f4fb4f293550d46fa784b45c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_24 | A sequence of natural numbers is constructed by listing the first $4$ , then skipping one, listing the next $5$ , skipping $2$ , listing $6$ , skipping $3$ , and on the $n$ th iteration, listing $n+3$ and skipping $n$ . The sequence begins $1,2,3,4,6,7,8,9,10,13$ . What is the $500,\!000$ th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$ | First, we group the numbers together in the following way: ${1, 2, 3, 4, (5)}; {6, 7, 8, 9, 10, (11, 12)}; ...$ We quickly realize that the number of terms in the curly braces follow a pattern: $5, 7, 9, 11, ... , n$ (where $n$ is the $n^\text{th}$ block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a $1, 3$ at the beginning, then the sum of all of the numbers before and including $n$ is $(n+2)^2$ . However, we have to subtract $1+3$ which results in $n^2+4n$ . The amount of numbers in the parenthesis are the $n^\text{th}$ triangular number or $\frac{n(n+1)}{2}$ . Next, we want to find the greatest $n$ , where $(n^2+4n) - \frac{n(n+1)}{2}<500000$ . Simplifying, we get $n^2+7n<1000000$ . We realize that $n=1000$ results in a number just $7000$ greater than our target. Next, we square $999$ $(1000 - 1)^2 = 1000000 - 2001$ . As we decrease $n$ by $1$ , we decrease the result of the equation by approximately $2000$ . In order to decrease by at least $7000$ , we have to decrease $4$ times leading to $n=1000-4=996$ . We plug it in to $n^2+4n$ getting $996\cdot1000=996000$ . This is the last number in the $996^\text{th}$ set. The number of terms used is $\frac{n\cdot(n+7)}{2}=\frac{996\cdot1003}{2}=498\cdot1003=498000+1494=499494$ . We need to add $500000-499494=506$ terms to get an answer of $\boxed{996,506}$ | A | 996,506 |
8dd58d71d622cdca7859918e42e278d8 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_25 | problem_id
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
Name: Text, dtype: object | The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$
There can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$
The first power of $2$ is between $5^n$ and $2 \cdot 5^n$
The second power of $2$ is between $2 \cdot 5^n$ and $4 \cdot 5^n$
The third power of $2$ is between $4 \cdot 5^n$ and $8 \cdot 5^n$ , meaning that it can be between $5^n$ and $5^{n+1}$ or not.
If there are only $2$ power of $2$ s between every consecutive powers of $5$ up to $5^{867}$ , there would be $867\cdot 2 = 1734$ power of $2$ s. However, there are $2013$ powers of $2$ before $5^{867}$ , meaning the answer is $2013 - 1734 = \boxed{279}$ | B | 279 |
ffedbd76122b843da2874d8fddb18844 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_1 | Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
$\textbf {(A) } 33 \qquad \textbf {(B) } 35 \qquad \textbf {(C) } 37 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 41$ | If Leah has $1$ more nickel, she has $14$ total coins. Because she has the same number of nickels and pennies, she has $7$ nickels and $7$ pennies. This is after the nickel has been added, so we must subtract $1$ nickel to get $6$ nickels and $7$ pennies. Therefore, Leah has $6\cdot5+7=\boxed{37}$ cents. | C | 37 |
10924ff9283ca712f6f8323b93d1edfc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_2 | What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$
$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$ | We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{64}$ | E | 64 |
10924ff9283ca712f6f8323b93d1edfc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_2 | What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$
$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$ | We have \[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\] so our answer is $\boxed{64}$ | E | 64 |
a69deac110ef87469b3bd992620db455 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_5 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$ | We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle $2x$ , and the height $5x$ , we can see that the length is composed of $4$ widths and $5$ bars of length $2$ . This is equal to two heights of the small rectangles as well as $3$ bars of $2$ . Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$ . We quickly find that $x = 2$ . The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$ , or $\boxed{26}$ | A | 26 |
6e702acc52b2f3ea85bea0803ae40b33 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | We have that A is $x\%$ greater than B, so $A=\frac{100+x}{100}(B)$ . We solve for $x$ . We get
$\frac{A}{B}=\frac{100+x}{100}$
$100\frac{A}{B}=100+x$
$100\left(\frac{A}{B}-1\right)=x$
$100\left(\frac{A-B}{B}\right)=x$ $\boxed{100}$ | A | 100 |
6e702acc52b2f3ea85bea0803ae40b33 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | The question is basically asking the percentage increase from $B$ to $A$ . We know the formula for percentage increase is $\frac{\text{New-Original}}{\text{Original}}$ . We know the new is $A$ and the original is $B$ . We also must multiple by $100$ to get $x$ out of it's fractional/decimal form. Therefore, the answer is $100\left(\frac{A-B}{B}\right)$ or $\boxed{100}$ | A | 100 |
6e702acc52b2f3ea85bea0803ae40b33 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | Without loss of generality, let $A = 125$ and $B = 100,$ forcing $x$ to be $25$ . Plugging our values for $A$ and $B$ into these answer choices, we find that only $\boxed{100}$ returns $25$ | A | 100 |
8ff1fd67ffebeedbdd30639f7fc4ff9d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_8 | A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$ | Converting feet to yards and minutes to second, we see that the truck travels $\dfrac{b}{18}$ yards every $t$ seconds for $180$ seconds. We see that he does $\dfrac{180}{t}$ cycles of $\dfrac{b}{18}$ yards. Multiplying, we get $\dfrac{180b}{18t}$ , or $\dfrac{10b}{t}$ , or $\boxed{10}$ | E | 10 |
8ff1fd67ffebeedbdd30639f7fc4ff9d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_8 | A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$ | We set a proportion by letting the $x$ being the number of feet the truck travels in $3$ minutes.
$\frac{\frac{b}{6}}{t}=\frac{x}{180}$
$\frac{b}{6t}=\frac{x}{180}$
$\frac{180b}{6t}=x$
$x=\frac{30b}{t}$
Remember $x$ is the number of feet the truck travels, so we divide by $3$ to convert to yards.
$\frac{x}{3}=\frac{10b}{t}$ , which corresponds to $\boxed{10}$ | E | 10 |
a3f878a4d093629688057868b344bb4f | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_10 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}\]
$\textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9$ | Note from the addition of the last digits that $A+B=D\text{ or }A+B=D+10$ .
From the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$ .
Therefore $A+B=D$
Using the second or fourth column, this then implies that $C=0$ , so that $B+C=B$ and $C+D=D$ .
Note that all of the remaining equalities are now satisfied: $A+B=D, B+C=B,$ and $B+A=D$ . Since the digits must be distinct, the smallest possible value of $D$ is $1+2=3$ , and the largest possible value is $9$ . Thus we have that $3\le D\le9$ , so the number of possible values is $\boxed{7}$ | C | 7 |
ee6c916c6fc680739cc9c256e7b5b52d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$ | Let the original price be $x$ . Then, for option $1$ , the discounted price is $(1-.15)(1-.15)x = .7225x$ . For option $2$ , the discounted price is $(1-.1)(1-.1)(1-.1)x = .729x$ . Finally, for option $3$ , the discounted price is $(1-.25)(1-.05) = .7125x$ . Therefore, $n$ must be greater than $\max(x - .7225x, x-.729x, x-.7125x)$ . It follows $n/100$ must be greater than $.2875$ . We multiply this by $100$ to get the percent value, and then round up because $n$ is the smallest integer that provides a greater discount than $28.75$ , leaving us with the answer of $\boxed{29}$ | C | 29 |
ee6c916c6fc680739cc9c256e7b5b52d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$ | Assume the original price was $100$ dollars. Thus, after a discount of $n\%$ , the price will be $100-n$ dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: $100-n$ $\frac{289}{4}$ $\frac{9^3}{10}$ , and $\frac{15\cdot19}{4}$ . Simplify these to get $100-n$ $72$ (rounded down), $72.9$ , and $71$ (rounded down). To have the greatest discount, we need the least price, which is $71$ dollars. Now we get the original fractional value of this, and the discount of this is $100-\frac{285}{4}$ , and we round this down to $28$ . Now, it's pretty easy. The integer value greater than this is $\boxed{29}$ | C | 29 |
ee6c916c6fc680739cc9c256e7b5b52d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$ | Assume WLOG that the original price was $$100$ . Then, option 1 would cost $100 \cdot \frac{17}{20} \cdot \frac{17}{20} = $ 72.25$ . Option 2 would cost $100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = $72.90$ . Option 3 would cost $100 \cdot \frac{3}{4} \cdot \frac{19}{20} = $71.25$ . Therefore, the largest integer smaller than all of these three is $71$ , so $100-71= \boxed{29}$ | C | 29 |
e778b1b1c71f126e8e3fba2a7d37f637 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_14 | Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a\ge1$ and $a+b+c\le7$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2$
$\textbf {(A) } 26 \qquad \textbf {(B) } 27 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 37 \qquad \textbf {(E) } 41$ | We can set up an algebraic equation for this problem.
From what's given, we have that $100c+10b+a=55x+100a+10b+c$
This simplifies to be $0=55x+99a-99c\implies -55x=99a-99c$
Factoring, we get that $-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}$
Hence, notice that we want $a-c=-5$ so that $x=9$
The only pair that works for this problem that satisfies the original requirements is $(1,6)$
Hence, $a=1, b=0, c=6$
Checking, we have that $106+55(9)=601\implies 601=601$
Hence, the answer is $1^2+0^2+6^2=37\implies\boxed{37}$ | D | 37 |
f68e07dc92d254e38a85671844c306a8 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_18 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35$ | We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$
Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{35}$ | E | 35 |
f68e07dc92d254e38a85671844c306a8 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_18 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35$ | Note that $x_1 + \ldots + x_{11} = 110$ let $x_6 = 9$ so $x_1 + \ldots + x_5 + x_7 + \ldots + x_{11} = 101$ . To maximize the value of $x_i$ where $i$ ranges from $1$ to $11$ , we let any $7$ elements be $1,2,\ldots,7$ so $x_1 + x_2 + x_3 = 57$ . Now we have to let one of above $3$ values = $8$ hence $x_1 + x_2 = 49$ now let $x_1 = 35$ $x_2 = 14$ hence $\boxed{35}$ is the answer. | E | 35 |
3943c58cb1418f903e5861b3cf2fa0bf | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_19 | Two concentric circles have radii $1$ and $2$ . Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{1}{2}\qquad$ | Let the center of the two circles be $O$ . Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$ . We want to find the range of possible places for the second point, $A'$ , such that $AA'$ passes through the circle of radius $1$ . To do this, first draw the tangents from $A$ to the circle of radius $1$ . Let the intersection points of the tangents (when extended) with circle of radius $2$ be $B$ and $C$ . Let $H$ be the foot of the altitude from $O$ to $\overline{BC}$ . Then we have the following diagram.
[asy] scale(200); pair A,O,B,C,H; A = (0,1); O = (0,0); B = (-.866,-.5); C = (.866,-.5); H = (0, -.5); draw(A--C--cycle); draw(A--O--cycle); draw(O--C--cycle); draw(O--H,dashed+linewidth(.7)); draw(A--B--cycle); draw(B--C--cycle); draw(O--B--cycle); dot("$A$",A,N); dot("$O$",O,NW); dot("$B$",B,W); dot("$C$",C,E); dot("$H$",H,S); label("$2$",O--(-.7,-.385),N); label("$1$",O--H,E); draw(circle(O,.5)); draw(circle(O,1)); [/asy]
We want to find $\angle BOC$ , as the range of desired points $A'$ is the set of points on minor arc $\overarc{BC}$ . This is because $B$ and $C$ are part of the tangents, which "set the boundaries" for $A'$ . Since $OH = 1$ and $OB = 2$ as shown in the diagram, $\triangle OHB$ is a $30-60-90$ triangle with $\angle BOH = 60^\circ$ . Thus, $\angle BOC = 120^\circ$ , and the probability $A'$ lies on the minor arc $\overarc{BC}$ is thus $\dfrac{120}{360} = \boxed{13}$ | D | 13 |
efc19a38266226c7e25534b6ea66c0ea | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ | First, note that $50+1=51$ , which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$ . Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0<x^2-1$ . Solving this inequality, we find $1<x^2<50$ . There are exactly $12$ integers $x$ that satisfy this inequality, $\pm \{2,3,4,5,6,7\}$
Thus our answer is $\boxed{12}$ | C | 12 |
efc19a38266226c7e25534b6ea66c0ea | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ | Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$ $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$ . We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by $2$ . If we try $1$ , we get $1^4-51(1)^4+50 = -50+50 = 0$ , and $0$ therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set { ${2,3,4,5,6,7}$ }. That equates to $6$ numbers. Since each numbers' negative counterparts work, $6\cdot2=\boxed{12}$ | C | 12 |
efc19a38266226c7e25534b6ea66c0ea | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ | Let $x^{2}=u$ . Then the expression becomes $u^{2}-51u+50$ which can be factored as $\left(u-1\right)\left(u-50\right)$ . Since the expression is negative, one of $\left(u-1\right)$ and $\left(u-50\right)$ need to be negative. $u-1>u-50$ , so $u-1>0$ and $u-50<0$ , which yields the inequality $50>u>1$ . Remember, since $u=x^{2}$ where $x$ is an integer, this means that $u$ is a perfect square between $1$ and $50$ . There are $6$ values of $u$ that satisfy this constraint, namely $4$ $9$ $16$ $25$ $36$ , and $49$ . Solving each of these values for $x$ yields $12$ values (as $x$ can be negative or positive) $\Longrightarrow \boxed{12}$ .
~JH. L | C | 12 |
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,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",DD,N); label("$F$",CC,N); draw(C--CC); draw(D--DD); [/asy]
In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$ .
Denote $\overline{AE} = x$ and $\overline{DE} = h$ . In right triangle $AED$ , we have from the Pythagorean theorem: $x^2+h^2=100$ . Note that since $EF = DC$ , we have $BF = 33-DC-x = 12-x$ . Using the Pythagorean theorem in right triangle $BFC$ , we have $(12-x)^2 + h^2 = 196$
We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$
Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$ . Now, we can determine that $h^2 = 100-4 \implies h = 4\sqrt{6}$
[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",DD,SE); label("$F$",CC,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/2,N); label("$2$",(A+DD)/2,N); label("$10$",(CC+B)/2,N); label("$4\sqrt{6}$",(C+CC)/2,W); label("$4\sqrt{6}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]
The two diagonals are $\overline{AC}$ and $\overline{BD}$ . Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$ , we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$ . Since $\sqrt{96+529}<\sqrt{96+961}$ $25$ is the shorter length*, so the answer is $\boxed{25}$ | B | 25 |
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