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2c7fb9677500f17cbd3a4d5fb1de56de | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_20 | Two circles lie outside regular hexagon $ABCDEF$ . The first is tangent to $\overline{AB}$ , and the second is tangent to $\overline{DE}$ . Both are tangent to lines $BC$ and $FA$ . What is the ratio of the area of the second circle to that of the first circle?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108$ | A good diagram is very helpful.
The first circle is in red, the second in blue.
With this diagram, we can see that the first circle is inscribed in equilateral triangle $GBA$ while the second circle is inscribed in $GKJ$ .
From this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas $\triangle GKJ$ to $\triangle GBA$
Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is $\left(\frac{GK}{GB}\right)^2$ .
From the diagram, we can see that this is $9^2=\boxed{81}$ | D | 81 |
2c7fb9677500f17cbd3a4d5fb1de56de | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_20 | Two circles lie outside regular hexagon $ABCDEF$ . The first is tangent to $\overline{AB}$ , and the second is tangent to $\overline{DE}$ . Both are tangent to lines $BC$ and $FA$ . What is the ratio of the area of the second circle to that of the first circle?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108$ | As above, we note that the first circle is inscribed in an equilateral triangle of sidelength 1 (if we assume, WLOG, that the regular hexagon has sidelength 1). The inradius of an equilateral triangle with sidelength 1 is equal to $\frac{\sqrt{3}}{6}$ . Therefore, the area of the first circle is $(\frac{\sqrt{3}}{6})^2 \cdot \pi =\frac{\pi}{12}$
Call the center of the second circle $O$ . Now we drop a perpendicular from $O$ to Circle O's point of tangency with $GK$ and draw another line connecting O to G. Note that because triangle $BGA$ is equilateral, $\angle BGA=60^{\circ}$ $OG$ bisects $\angle BGA$ , so we have a 30-60-90 triangle.
Call the radius of Circle O $r$ $OG=2r= \text{height of equilateral triangle} + \text{height of regular hexagon} + r$
The height of an equilateral triangle of sidelength 1 is $\frac{\sqrt{3}}{2}$ . The height of a regular hexagon of sidelength 1 is $\sqrt{3}$ . Therefore, $OG=\frac{\sqrt{3}}{2} + \sqrt{3} + r$
We can now set up the following equation:
$\frac{\sqrt{3}}{2} + \sqrt{3} + r=2r$
$\frac{\sqrt{3}}{2} + \sqrt{3}=r$
$\frac{3\sqrt{3}}{2}=r$
The area of Circle O equals $\pi r^2=\frac{27}{4} \pi$
Therefore, the ratio of the areas is $\frac{\frac{27}{4}}{\frac{1}{12}}=\boxed{81}$ | D | 81 |
f881171cdff412abec4e2f48fef0dd26 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?
$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934$ | We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.
Each candy has three choices; it can go in any of the three bags.
Since there are seven candies, that makes the total distribution $3^7=2187$
To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.
The number of distributions such that the red bag is empty is equal to $2^7$ , since it's equivalent to distributing the $7$ candies into $2$ bags.
We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also $2^7$
The case where both the red and the blue bags are empty (all $7$ candies are in the white bag) are included in both of the above calculations, and this case has only $1$ distribution.
The total overcount is $2^7+2^7-1=2^8-1$
The final answer will be $\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{1932}$ | C | 1932 |
f881171cdff412abec4e2f48fef0dd26 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?
$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934$ | We can use to our advantage the answer choices $\text{AMC}$ has given us, and eliminate the obvious wrong answer choices.
We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: $7\cdot 6=42$
Now we can look at the answer choices to find out which ones are divisible by $42$ , since the total number of combinations must be $42$ multiplied by some other number.
Since answers A, B, D, and E are not divisible by 3 (divisor of 42), the answer must be $\boxed{1932}$ | C | 1932 |
f881171cdff412abec4e2f48fef0dd26 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?
$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934$ | Let $r$ be the number of red bag candies. For ${1 \leq r \leq 6}$
So the number of candies left for the blue bag and the red bag is $7-r$ . Based on the problem, 1 candy must be fixed for the blue bag, which can be done $7-r$ ways. Now, before we continue, we need to realize that fixing a candy can lead to some over counting in cases where none in the white bag overlap. So we should try and find an alternative because we'll be over counting more than twice, and that will become extremely difficult to account for each case's over counting.
So, without fixing candies, we can put everything into options and work on the white bag, because once we figure out two bags, the remaining one is decided.
The options for the candies in the white bag are two: In the white bag or out of the white bag(by default, in the blue bag).
Now, for choosing the red bag, we have ${7 \choose r}$
Then, we have for the white bag: $2^{7-r}$ . Because we aren't fixing one for blue, the power is $7-r$ instead of $7-r-1$
We have: $({7 \choose r}) \cdot (2^{7-r})$ Adding them up for ${1 \leq r \leq 6}$ , we get 2058.
Now, for the invalid cases. Because we didn't fix candy for the blue bag, we need to subtract the cases where the blue bag remains empty. We can accomplish this pretty easily.
When $r = 6$ , how many ways can the remaining 1 candy $not$ be placed in the blue bag. This can be done ${7 \choose 1} = 7$ ways.
When $r = 5$ , how many ways can the remaining 2 candies $not$ be placed in the blue bag. This can be done ${7 \choose 2} = 21$ ways.
When $r = 4$ , how many ways can the remaining 3 candies $not$ be placed in the blue bag. This can be done ${7 \choose 3} = 35$ ways.
And because ${7 \choose 4} = {7 \choose 3}, {7 \choose 5} = {7 \choose 2}, and {7 \choose 6} = {7 \choose 1}$ , we just multiply by two.
Finally, we have: $2058 - 2 \cdot (7+21+35) = 2058 - 126 = \boxed{1932}$ | C | 1932 |
29876adbd9aa586bb7e1727913e40199 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_23 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.
\[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\]
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$
\[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\]
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$
\[12+18+12=\boxed{42}\] | D | 42 |
29876adbd9aa586bb7e1727913e40199 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_23 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$ | D | 42 |
d6fdd0589222bbda853c86f29fbc4315 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_24 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.
Suppose not, and say $r = m/n$ where $m>n>1$ , and $\gcd(m,n)=1$ . Then $n, n^2, n^3$ must all divide $a$ so $a=n^3k$ for some integer $k$ . Then $S_R = n^3k + n^2mk + nm^2k + m^3k < 100$ and we see that even if $k=1$ and $n=2$ , we get $m < 4$ , which means that the only option for $r$ is $r=3/2$ . A quick check shows that even this doesn't work. Thus $r$ must be an integer.
Let $a, a+d, a+2d, a+3d$ be the quarterly scores for the Wildcats. Let $S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d$ . Let $S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)$ . Then $S_R<100$ implies that $r<5$ , so $r\in \{2, 3, 4\}$ . The Raiders win by one point, so \[a(1+r)(1+r^2) = 4a+6d+1.\]
Then the quarterly scores for the Raiders are $5, 10, 20, 40$ , and those for the Wildcats are $5, 14, 23, 32$ . Also $S_R = 75 = S_W + 1$ . The total number of points scored by the two teams in the first half is $5+10+5+14=\boxed{34}$ | E | 34 |
d6fdd0589222bbda853c86f29fbc4315 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_24 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$ . Now we can narrow our search for the values of $a,d$ , and $r$ . Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:
\[a(1+r+r^{2}+r^{3})=4a+6d+1\]
Now we can start trying out some values of $r$ . We try $r=2$ , which gives
\[15a=4a+6d+1\]
\[11a=6d+1\]
We need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\equiv 1 \pmod{6}$ . We see that this is $55$ , and therefore $a=5$ and $d=9$
So the Raiders' first two scores were $5$ and $10$ and the Wildcats' first two scores were $5$ and $14$
\[5+10+5+14=34 \longrightarrow \boxed{34}\] | E | 34 |
8c77a574f062843a158f36f3a1da1bc1 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_25 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$
Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$
Then, plugging in values of $2,4,6,8,$ we get
\[P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a\] \[P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a\] \[P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a\] \[P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a\]
$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$ .
Solving, we receive $315$ , so our answer is $\boxed{315}$ | B | 315 |
8c77a574f062843a158f36f3a1da1bc1 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_25 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | The evenly-spaced data suggests using discrete derivatives to tackle this problem. First, note that any polynomial of degree $n$
can also be written as
Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ . This latter form is convenient for calculating discrete derivatives of $P(x)$
The discrete derivative of a function $f(x)$ is the related function $\Delta f(x)$ defined as
With this definition, it's easy to see that for any positive integer $k$ we have
This in turn allows us to use successive discrete derivatives evaluated at $x=1$ to calculate all of the coefficients $b_i$ using
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
Thus we can read down the column for $x=1$ to find that $k! b_k = (-2)^k a$ for $k = 0, 1, \ldots, 7$ . Interestingly, even if we choose $P(x)$ to have degree greater than $7$ , the $8$ coefficients of lowest order always satisfy these conditions. Moreover, it's straightforward to show that the $P(x)$ of degree $7$ with $b_k$ satisfying these conditions will fit the data given in the problem statement. Thus, to find minimal necessary and sufficient conditions on the value of $a$ , we need only consider these $8$ equations. As a result, $P(x)$ with integer coefficients fitting the given data exists iff $k!$ divides $2^k a$ for $k = 0, 1, \ldots, 7$ . In other words, it's necessary and sufficient that
The last condition holds if $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ . Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{315}$ | B | 315 |
58ff06a5e5eedc52fdb1d28ea60853ea | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_1 | One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ( $128$ ounces) of soda?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$ | $10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\mathrm{\boxed{11}$ | E | 11 |
812b7c5628466484f2df77a740765f0b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_2 | Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55$ | As all five options are divisible by $5$ , we may not use any pennies. (This is because a penny is the only coin that is not divisible by $5$ , and if we used between $1$ and $4$ pennies, the sum would not be divisible by $5$ .)
Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is $4\cdot 5 = 20$ . Therefore the option that is not reachable is $\boxed{15}$ $\Rightarrow$ $(A)$ | null | 15 |
6256aaadc589c33014adb0886fe821c0 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_5 | What is the sum of the digits of the square of $\text 111111111$
$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$ | Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\boxed{81.}$ | E | 81. |
6256aaadc589c33014adb0886fe821c0 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_5 | What is the sum of the digits of the square of $\text 111111111$
$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$ | We note that
$1^2 = 1$
$11^2 = 121$
$111^2 = 12321$
and $1,111^2 = 1234321$
We can clearly see the pattern: If $X$ is $111\cdots111$ , with $n$ ones (and for the sake of simplicity, assume that $n<10$ ), then the sum of the digits of $X^2$ is
$1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1$
$=(1+2+3\cdots n)+(1+2+3+\cdots n-1)$
$=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}$
$=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.$
Aha! We know that $111,111,111$ has $9$ digits, so its digit sum is $9^2=\boxed{81}$ | E | 81 |
6256aaadc589c33014adb0886fe821c0 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_5 | What is the sum of the digits of the square of $\text 111111111$
$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$ | We see that $111^2$ can be written as $111(100+10+1)=11100+1110+111=12321$
We can apply this strategy to find $111,111,111^2$ , as seen below.
$111111111^2=111111111(100000000+10000000\cdots+10+1)$
$=11111111100000000+1111111110000000+\cdots+111111111$
$=12,345,678,987,654,321$
The digit sum is thus $1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{81}$ | E | 81 |
0b14472d35113590a637810028af0665 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_11 | One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216$ | Let the original cube have edge length $a$ . Then its volume is $a^3$ .
The new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ .
The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{125}$ | D | 125 |
86c4e81899ee034bbfbc00c4bf07591e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_12 | In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$
We get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$ | null | 13 |
c77d2566b88b53960240e7a48b0e0629 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_13 | Suppose that $P = 2^m$ and $Q = 3^n$ . Which of the following is equal to $12^{mn}$ for every pair of integers $(m,n)$
$\textbf{(A)}\ P^2Q \qquad \textbf{(B)}\ P^nQ^m \qquad \textbf{(C)}\ P^nQ^{2m} \qquad \textbf{(D)}\ P^{2m}Q^n \qquad \textbf{(E)}\ P^{2n}Q^m$ | We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{2}$ | E | 2 |
569e37555efc7af96b67e8a7e4f73300 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_14 | Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$ | Let the side length of the smaller square be $1$ , and let the smaller side of the rectangles be $y$ . Since the larger square's area is four times larger than the smaller square's, the larger square's side length is $2$ $2$ is equivalent to $2y+1$ , giving $y=1/2$ . Then, the longer side of the rectangles is $3/2$ $\frac{\frac{3}{2}}{\frac{1}{2}}=\boxed{3}$ | null | 3 |
73fa5c60cbb61fc87d98b9d1b271d7ad | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$ th triangular number is $\frac{n(n+1)}{2}$ . Putting this together for $F_{20}$ this gives:
$\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$ | null | 761 |
73fa5c60cbb61fc87d98b9d1b271d7ad | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle; draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); [/asy]
In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$ | null | 761 |
73fa5c60cbb61fc87d98b9d1b271d7ad | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761} | null | 761 |
73fa5c60cbb61fc87d98b9d1b271d7ad | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$ . The first finite differences go $4, 8, 12, 16, \dots$ . The second finite differences go $4, 4, 4, \dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we already know $d_1 = 1$ $d_2 = 3$ , and $d_3 = 13$ . Thus, \[a + b + c = d_1 = 1\] \[4a + 2b + c = d_2 = 3\] \[9a + 3b + c = d_3 = 13\] Solving this system for $a$ $b$ , and $c$ gives $a = 2$ $b = -2$ $c = 1$ . Finally, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{761}$ | E | 761 |
c7082847112cec41a0840c6b5ed7e99b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$ | From $|a-b|=2$ we get that $a=b\pm 2$
Similarly, $b=c\pm3$ and $c=d\pm4$
Substitution gives $a=d\pm 4\pm 3\pm 2$ . This gives $|a-d|=|\pm 4\pm 3\pm 2|$ . There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$
$4+3+2=9$
$4+3-2=5$
$4-3+2=3$
$-4+3+2=1$
$4-3-2=-1$
$-4+3-2=-3$
$-4-3+2=-5$
$-4-3-2=-9$
Therefore, the only possible values of $|a-d|$ are $9$ $5$ $3$ , and $1$ . Their sum is $\boxed{18}$ | D | 18 |
c7082847112cec41a0840c6b5ed7e99b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$ | If we add the same constant to all of $a$ $b$ $c$ , and $d$ , we will not change any of the differences. Hence we can assume that $a=0$
From $|a-b|=2$ we get that $|b|=2$ , hence $b\in\{-2,2\}$
If we multiply all four numbers by $-1$ , we will not change any of the differences. (This is due to the fact that we are calculating $|d|$ at the end ~Williamgolly) WLOG we can assume that $b=2$
From $|b-c|=3$ we get that $c\in\{-1,5\}$
From $|c-d|=4$ we get that $d\in\{-5,1,3,9\}$
Hence $|a-d|=|d|\in\{1,3,5,9\}$ , and the sum of possible values is $1+3+5+9 = \boxed{18}$ | D | 18 |
c7082847112cec41a0840c6b5ed7e99b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$ | Let \[\begin{cases} |a-b| = 2, \\ |b-c| = 3, \\ |c-d| = 4, \\ |d-a| = X. \\ \end{cases}\] Note that we have \[\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies\] \[\pm X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies X \pm 2 \pm 3 \pm 4 = 0,\] from which it follows that \[X = \pm 4 \pm 3 \pm 2.\]
Note that $X = |a-d|$ must be positive however, the only arrangements of $+$ and $-$ signs on the RHS which make $X$ positive are \[(4, 3, 2) \ \ \implies \ \ X = 9\] \[(4, 3, -2) \ \ \implies \ \ X = 5\] \[(4, -3, 2) \ \ \implies \ \ X = 3\] \[(-4, 3, 2) \ \ \implies \ \ X = 1.\] (There are no cases with $2$ or more negative as $4-3-2<0.$
Thus, the answer is \[1+3+5+9 = \boxed{18}.\] | D | 18 |
c7082847112cec41a0840c6b5ed7e99b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$ | Let $a=0$
Then $\begin{cases} b=2 \\ b=-2 \\ \end{cases}$
Thus $\begin{cases} c=5 \\ c=-1 \\ c=1 \\ c=-5 \\ \end{cases}$
Therefore $\begin{cases} d=9 \\ d=1 \\ d=3 \\ d=-5 \\ d=5 \\ d=-3 \\ d=-1 \\ d=-9 \\ \end{cases}$
So $|a-d|=1, 9, 3, 5$ and the sum is $\boxed{18}$ | D | 18 |
ec6fbbeafcd1fa9c7e96cf57c25d5903 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$ | Out of the soccer players, $40\%$ swim. As the soccer players are $60\%$ of the whole, the swimming soccer players are $0.4 \cdot 0.6 = 0.24 = 24\%$ of all children.
The non-swimming soccer players then form $60\% - 24\% = 36\%$ of all the children.
Out of all the children, $30\%$ swim. We know that $24\%$ of all the children swim and play soccer, hence $30\%-24\% = 6\%$ of all the children swim and don't play soccer.
Finally, we know that $70\%$ of all the children are non-swimmers. And as $36\%$ of all the children do not swim but play soccer, $70\% - 36\% = 34\%$ of all the children do not engage in any activity.
A quick summary of what we found out:
Now we can compute the answer. Out of all children, $70\%$ are non-swimmers, and again out of all children $36\%$ are non-swimmers that play soccer. Hence the percent of non-swimmers that play soccer is $\frac{36}{70} \approx 51\% \Rightarrow \boxed{51}$ | D | 51 |
ec6fbbeafcd1fa9c7e96cf57c25d5903 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$ | Let us set the total number of children as $100$ . So $60$ children play soccer, $30$ swim, and $0.4\times60=24$ play soccer and swim.
Thus, $60-24=36$ children only play soccer.
So our numerator is $36$
Our denominator is simply $100-\text{Swimmers}=100-30=70$
And so we get $\frac{36}{70}$ which is roughly $51.4\% \Rightarrow \boxed{51}$ | D | 51 |
ec6fbbeafcd1fa9c7e96cf57c25d5903 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$ | WLOG, let the total number of students be $100$ . Draw a venn diagram with 2 circles encompassing these 4 regions:
Non-soccer players, non-swimmers: 34 people
Soccer players, non-swimmers: 36 people
Soccer players, swimmers: 24 people
Non-soccer players, swimmers: 6 people.
Hence the answer is $\frac{36}{70}=\frac{18}{35}$ . We know this is a little bit larger than $\frac 12$ because $\frac{17.5}{35}=\frac 12$ $\boxed{51}$ | D | 51 |
bb63f04c90e117acf52abd3c739d1210 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_19 | Circle $A$ has radius $100$ . Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$ . The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?
$\mathrm{(A)}\ 4\ \qquad \mathrm{(B)}\ 8\ \qquad \mathrm{(C)}\ 9\ \qquad \mathrm{(D)}\ 50\ \qquad \mathrm{(E)}\ 90\ \qquad$ | The circumference of circle $A$ is $200\pi$ , and the circumference of circle $B$ with radius $r$ is $2r\pi$ . Since circle $B$ makes a complete revolution and ends up on the same point , the circumference of $A$ must be a multiple of the circumference of $B$ , therefore the quotient must be an integer.
Thus, $\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}$
Therefore $r$ must then be a factor of $100$ , excluding $100$ because the problem says that $r<100$ $100\: =\: 2^2\; \cdot \; 5^2$ . Therefore $100$ has $(2+1)\; \cdot \; (2+1)\;$ factors*. But you need to subtract $1$ from $9$ , in order to exclude $100$ . Therefore the answer is $\boxed{8}$ | null | 8 |
232f1c4b4752c433d376b543156a2d3e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$ | Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1\mathrm km/min = 60\mathrm km/h$ .) This solves to $v_A=45$ and $v_L=15$
As the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.
From this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$ , she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = \boxed{65}$ | null | 65 |
232f1c4b4752c433d376b543156a2d3e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$ | Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\frac{3}{4} \textbf{km/min}$ , and Lauren's $\frac{1}{4} \textbf{km/min}$ . Therefore, after 5 minutes, Andrea will have biked $\frac{3}{4} \cdot 5 = \frac{15}{4}km$
In all, Lauren will have to bike $20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}km$ . Because her speed is $\frac{1}{4} \textbf{km/min}$ , the time elapsed will be $\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{65}$ | D | 65 |
232f1c4b4752c433d376b543156a2d3e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$ | Since the distance between them decreases at a rate of $1$ kilometer per minute when they are both biking, their combined speed is $1$ kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of $\frac{3}{4}$ kilometers per minute and $\frac{1}{4}$ kilometers per minute, respectively.
After $5$ minutes, the distance between them will have decreased by $5$ kilometers, so they will be $20-5 = 15$ kilometers apart when Andrea stops. Then, Lauren will take $\frac{15}{\frac{1}{4}} = 15*4=60$ more minutes to reach Andrea.
They started to bike $5$ minutes before Andrea stopped, so the total time Lauren passed from the time they started biking to the time Lauren reached Andrea is $60+5=65$ minutes. Hence, the answer is $\boxed{65}$ . ~azc1027 | B | 65 |
e2ffc56c48214e519db9ad5fac445d77 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_23 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ . Diagonals $AC$ and $BD$ intersect at $E$ $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$ | The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\boxed{6}$ | A | 6 |
e2ffc56c48214e519db9ad5fac445d77 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_23 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ . Diagonals $AC$ and $BD$ intersect at $E$ $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$ | Using the fact that $[AED] = [BEC]$ and the fact that $\triangle AEB \sim \triangle EDC$ (which should be trivial given the two equal triangles) we have that
\[\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}\]
We know that $DC=EC,$ so we have
\[\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}\]
Thus
\[\frac{AE}{EC} = \frac{3}{4}\]
But $EC = 14 - AE$ so we have
\[\frac{AE}{14 - AE} = \frac{3}{4}\]
Simplifying gives $AE = \boxed{6}.$ | null | 6 |
e64062c65dfd2ba212a2550cff13f72f | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_24 | Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$ | We will try to use symmetry as much as possible.
Pick the first vertex $A$ , its choice clearly does not influence anything.
Pick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.
In the first case, there are $2$ faces that contain the edge $AB$ . In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$ , the entire triangle $ABC$ will be on a face. On the other hand, if $C$ is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good $C$ is $2/6 = 1/3$
In the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$ . Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$
In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$
Summing up all cases, the resulting probability is: \[\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{47}\] | null | 47 |
e64062c65dfd2ba212a2550cff13f72f | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_24 | Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$ | There are $\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
There are $\binom{4}{3}=4$ ways to choose three points from the vertices of a single face. Since there are six faces, $4 \times 6 = 24$
Thus, the probability of what we don't want is $\frac{24}{56} = \frac{3}{7}$ . Using complementary probability,
\[1- \frac 37 = \boxed{47}\] | null | 47 |
25001bffec3a1fdfd42805261c3a6fd2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$ | The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$
This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ $N(4) = \boxed{7}$ | null | 7 |
25001bffec3a1fdfd42805261c3a6fd2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$ | Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\boxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ $N(k)$ must be less than or equal to $7$ | B | 7 |
25001bffec3a1fdfd42805261c3a6fd2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$ | Similar to the other solutions, notice that $I_k$ can be written as $10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6$ . Factoring out $2^6$ we see that
$I_k = 2^6(2^{k-4}5^{k+2}+1)$
Notice that for $k < 4$ $2^{k-4}$ will not be an integer, and will "steal" some $2$ 's from the $2^6$ . We don't want this to happen, since we want to maximize the exponent of $2$ . We start by considering $k = 4$ . Then
$I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)$
$5^6$ is an odd number; more specifically, it ends in $25$ (all powers of $5$ after $5^1$ end in $25$ ). Therefore the value in the parentheses will be some large number that ends in $26$ . Considering the rules of divisibility, we find that $5^6+1$ is even, so it is divisible by $2$ . Now our exponent of $2$ is at $7$ . But the divisibility rule for $4$ is the last $2$ digits of the number must be divisible by $4$ . We know the last digits: $26$ , which is not divisible by $4$ . Therefore $5^6 + 1$ is divisible by $2$ , but not $4$ . Testing more values of $k$ , we find that for $k \ge 5$ , the last digit becomes $1$ , which means it is not even divisible by $2$ . Therefore the highest possible exponent of $2$ that we can reach is $7 \Rightarrow \boxed{7}$ | B | 7 |
25001bffec3a1fdfd42805261c3a6fd2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$ | Let $m=k+2$ $v_2(10^m+2^6)=6$ if $m>6$ and $v_2(10^m+2^6)=m$ if $m<6$ .
However, if $m=6$ , then $v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)$ . By LTE, $v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3$ . Since $v_2(5^6-1)=3$ $v_2(5^6+1)$ must equal $1$ . So, the answer is $6+1=7 \Rightarrow \boxed{7}$ | B | 7 |
b5b5692fcc29a56f8f2797f0977341c9 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_1 | Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.
If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{2}$ bagels. | B | 2 |
16e7f155db0b3364edf768d1b477a542 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_2 | Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$ | Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
We can multiply both by $12$ , getting $\dfrac{4-3}{6-4} = \boxed{12}$ | null | 12 |
16e7f155db0b3364edf768d1b477a542 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_2 | Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$ | We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\cdot4=12$ .
Next, we multiply both the numerator and denominator by $12$ $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{12}$ .
-sosiaops | C | 12 |
8ac972cf92345277b9208138371cd0ad | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_3 | Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 25$ | Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$ | C | 15 |
beee1f89d8141011a57821601922498b | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_4 | A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?
$\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$ | Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{15}$ | C | 15 |
d6575969056bf0d03e78946dd6ec7822 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_5 | Twenty percent less than 60 is one-third more than what number?
$\mathrm{(A)}\ 16\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 32\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 48$ | Twenty percent less than 60 is $\frac 45 \cdot 60 = 48$ . One-third more than a number is $\frac 43n$ . Therefore $\frac 43n = 48$ and the number is $\boxed{36}$ | D | 36 |
ab7b312f949189ae3e607c2c5add90d8 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_6 | Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$ | The age of each person is a factor of $128 = 2^7$ . So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$ $32$ $8$ , or $2$ years old, respectively. Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$ | D | 18 |
49dd282ebb988acbca41ff09f1fc8ea7 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_7 | By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$ | The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct. So the answer is $\boxed{4}$ | C | 4 |
546ef2ca53c5b856e6bc7861ce60090d | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_8 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$ | Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$ | B | 17 |
546ef2ca53c5b856e6bc7861ce60090d | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_8 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$ | Without loss of generality, we can assume the price at the beginning of January was $$100$
When it rose by $20\%$ , it became $$120$ , when it fell by $20\%$ , it became $$96$ , and when it rose by $25\%$ , it became $$120$ again.
In order for the price at the end of April to be the same as it was at the beginning of January ( $$100$ ), the price must decrease by $$20$
20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{17}$ . ~azc1027 | B | 17 |
80865dd897fc6850c00cfa8f8120a2ac | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_9 | Segment $BD$ and $AE$ intersect at $C$ , as shown, $AB=BC=CD=CE$ , and $\angle A = \frac 52 \angle B$ . What is the degree measure of $\angle D$
[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), Ep=dir(35), D=dir(-35), B=dir(145); pair A=intersectionpoints(Circle(B,1),C--(-1*Ep))[0]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(A--Ep--D--B--cycle); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,N); label("$E$",Ep,E); label("$D$",D,E); [/asy]
$\text{(A) } 52.5 \qquad \text{(B) } 55 \qquad \text{(C) } 57.7 \qquad \text{(D) } 60 \qquad \text{(E) } 62.5$ | $\triangle ABC$ is isosceles, hence $\angle ACB = \angle CAB$
The sum of internal angles of $\triangle ABC$ can now be expressed as $\angle B + \frac 52 \angle B + \frac 52 \angle B = 6\angle B$ , hence $\angle B = 30^\circ$ , and each of the other two angles is $75^\circ$
Now we know that $\angle DCE = \angle ACB = 75^\circ$
Finally, $\triangle CDE$ is isosceles, hence each of the two remaining angles ( $\angle D$ and $\angle E$ ) is equal to $\frac{180^\circ - 75^\circ}2 = \frac{105^\circ}2 = \boxed{52.5}$ | null | 52.5 |
a38a2ce3d2b6a1245d801d7b7da95dbe | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_10 | A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x$
$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.3 \qquad \text{(E) } 2.4$ | The broken flagpole forms a right triangle with legs $1$ and $x$ , and hypotenuse $5-x$ . The Pythagorean theorem now states that $1^2 + x^2 = (5-x)^2$ , hence $10x = 24$ , and $x=\boxed{2.4}$ | null | 2.4 |
a38a2ce3d2b6a1245d801d7b7da95dbe | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_10 | A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x$
$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.3 \qquad \text{(E) } 2.4$ | Let $AB$ represent the flagpole in the diagram above. After the flagpole breaks at point $D$ , its tip lies at point $C$ . Since none of the flagpole is destroyed, we know that $DA=DC$ . Therefore, triangle $\triangle ADC$ is isosceles.
Draw the altitude $DE \perp AC$ . Since $\triangle ADC$ is isosceles, we know that $AE = EC$ . Also note that $\triangle AED \sim \triangle ABC$ . Therefore, \begin{align*} AD &= AE \times \frac{AD}{AE} \\ &= \frac{AC}{2} \times \frac{AC}{AB} \\ &= \frac{AC^2}{2 AB} \\ &= \frac{AB^2 + BC^2}{2 AB} \end{align*}
Since $AB = 5$ and $BC = 1$ , we have that $AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6$ , and thus $x = AB - AD = \boxed{2.4}$ | null | 2.4 |
576dfd7372cce3e5816c15c28b208630 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_11 | How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ $2$ $3$ $3$ $5$ $5$ $5$
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$ | A seven-digit palindrome is a number of the form $\overline{abcdcba}$ . Clearly, $d$ must be $5$ , as we have an odd number of fives. We are then left with $\{a,b,c\} = \{2,3,5\}$ . There are $3!$ permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. Each of the $\boxed{6}$ will give us one palindrome. | A | 6 |
576dfd7372cce3e5816c15c28b208630 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_11 | How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ $2$ $3$ $3$ $5$ $5$ $5$
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$ | Say we have a 2 first. Then, we have a 2 pinned as the last digit, so we have to fill in the remaining digits with only 3's and 5's. We have 2 options for the second digit then, and the rest is fixed. This means that we have $2$ ways for this case.
Say we have a 3 first. By symmetry, this is the same as the 2 cases, so we have $2$ ways.
Say we have a 5 first. We then have a 5 in the middle. We can either have a 2 second or a 3 second. So we have $2$ ways.
This means that our answer is $2+2+2=\boxed{6}$ | A | 6 |
0410b2dd9d9b679b593f07cc6287269d | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_12 | Distinct points $A$ $B$ $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$ | Consider the classical formula for triangle area: $\frac 12 \cdot b \cdot h$ .
Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.
Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are $1$ $2$ , and $3$ . Therefore there are only $\boxed{3}$ possible values for the area. | A | 3 |
0410b2dd9d9b679b593f07cc6287269d | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_12 | Distinct points $A$ $B$ $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$ | No matter what how we draw a triangle by selecting three non-linear points, its height will always remain the same. Therefore, we will only get different areas with different base-lengths. The possibilities are $1$ $2$ , and $3$ units for a total of $\boxed{3}$ | A | 3 |
1583aa620310ca0f50ce493fed9fd7c8 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_13 | As shown below, convex pentagon $ABCDE$ has sides $AB=3$ $BC=4$ $CD=6$ $DE=3$ , and $EA=7$ . The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$ -axis. The pentagon is then rolled clockwise to the right along the $x$ -axis. Which side will touch the point $x=2009$ on the $x$ -axis?
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,0), Ep=7*dir(105), B=3*dir(0); pair D=Ep+B; pair C=intersectionpoints(Circle(D,6),Circle(B,4))[1]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(B--C--D--Ep--A); draw((6,6)..(8,4)..(8,3),EndArrow(3)); xaxis("$x$",-8,14,EndArrow(3)); label("$E$",Ep,NW); label("$D$",D,NE); label("$C$",C,E); label("$B$",B,SE); label("$(0,0)=A$",A,SW); label("$3$",midpoint(A--B),N); label("$4$",midpoint(B--C),NW); label("$6$",midpoint(C--D),NE); label("$3$",midpoint(D--Ep),S); label("$7$",midpoint(Ep--A),W); [/asy]
$\text{(A) } \overline{AB} \qquad \text{(B) } \overline{BC} \qquad \text{(C) } \overline{CD} \qquad \text{(D) } \overline{DE} \qquad \text{(E) } \overline{EA}$ | The perimeter of the polygon is $3+4+6+3+7 = 23$ . Hence as we roll the polygon to the right, every $23$ units the side $\overline{AB}$ will be the bottom side.
We have $2009 = 23 \times 87 + 8$ . Thus at some point in time we will get the situation when $A=(2001,0)$ and $\overline{AB}$ is the bottom side. Obviously, at this moment $B=(2004,0)$
After that, the polygon rotates around $B$ until point $C$ hits the $x$ axis at $(2008,0)$
And finally, the polygon rotates around $C$ until point $D$ hits the $x$ axis at $(2014,0)$ .
At this point the side $\boxed{2009,0}$ | C | 2009,0 |
512a2e2581beac1c9c00c59124f3af41 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16 | Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$ | [asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C, dashed ); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); [/asy]
As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$ , hence $\angle OAC = \angle OCA = 30^\circ$ . Then $\angle AOC = 120^\circ$ , and from symmetry we have $\angle AOB = \angle COB = 60^\circ$ . Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$
We know that $DO = AO$ , as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$ , Hence $DO = \frac{BO}2$ , therefore $BD = BO - DO = \frac{BO}2$ , and $\frac{BD}{BO} = \boxed{12}$ | B | 12 |
512a2e2581beac1c9c00c59124f3af41 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16 | Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$ | [asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); label("$S$",Sp,(S+SSW)); [/asy]
As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$ . Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.
We then have $\angle SCD = \angle SAD = 30^\circ$ , hence $D$ is the incenter of $\triangle ABC$ , and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$ , and as $SD = SO$ , we have $2\cdot SD = SD + SO = OD$ , therefore $BD=OD$ , and as before we conclude that $\frac{BD}{BO} = \boxed{12}$ | null | 12 |
aa490c8ab49468b6cb75ec41c63d4f66 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | For $c\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.
Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of the entire triangle is $\frac{3(3-c)}2$ , therefore the area of the shaded part is $\frac{7-3c}{2}$
The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $c$ , we get $c=\boxed{23}$ | C | 23 |
aa490c8ab49468b6cb75ec41c63d4f66 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | The unit square is of area 1, so the five unit squares have area 5.
Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5.
It can then be inferred: $(3-c) * 3 = 7$
$3-c=\frac{7}{3}$ , so $3-\frac{7}{3}=c$
$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ $\boxed{23}$ | C | 23 |
aa490c8ab49468b6cb75ec41c63d4f66 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-c$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3c)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region.
If you add 3 unit squares to the top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion.
Since these areas are both equal, we set them equal to each other and solve for $c$ $(9-3c)/2 -1 = (9+3c)/2 -3$ $c$ is now solved to be $\frac{2}{3}$ $\boxed{23}$ | C | 23 |
aa490c8ab49468b6cb75ec41c63d4f66 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | We are looking for the area of the shaded region to be $\frac 52$ . We start by testing $(A) \frac 12$ . The area of the shaded region would be $\frac{(3-\frac 12)(3) }{2}-1=\frac {11}{4}$ when $c=\frac 12$ . This does not match our wanted answer.
We try $(B) \frac 35$ next. The area of the shaded region would be $\frac{(3-\frac 35)(3) }{2}-1=\frac {13}{5}$ when $c=\frac 35$ . This also does not match our desired answer.
We then try $(C) \frac 23$ . The area of the shaded region would be $\frac{(3-\frac 23)(3) }{2}-1=\frac {5}{2}$ when $c=\frac 23$ . Therefore, our answer is $\boxed{23}$ | C | 23 |
14b1f6481b0c8d91340b5b79cb1734ec | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_19 | A particular $12$ -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1$ , it mistakenly displays a $9$ . For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?
$\mathrm{(A)}\ \frac 12\qquad \mathrm{(B)}\ \frac 58\qquad \mathrm{(C)}\ \frac 34\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac {9}{10}$ | The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$ . So the correct hour is displayed $\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$ , so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$ . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{12}$ | A | 12 |
26ccbca8be2d58fa910dbd0d5e635a0e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_21 | What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$ | The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed{4}$ | D | 4 |
26ccbca8be2d58fa910dbd0d5e635a0e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_21 | What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$ | We have $3^2 = 9 \equiv 1 \pmod 8$ . Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$ , and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8$
Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$ . There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs of $1+3$ , and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$ | null | 4 |
7af8c76494dcb75c6e629c6cacbfaa88 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$ | Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.
[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted ); label("$X$",(0,0),S); label("$Y$",(R,0),SE); label("$Z$",rotate(20)*(R,0),ENE); draw( arc( (0,0), (1.5,0), rotate(20)*(1.5,0) ) ); label("$20^\circ$", rotate(10)*(1.75,0), E ); [/asy]
Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$ . From $\triangle XYZ$ , the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$ , hence the degree measure of the larger interior angle is $180^\circ - 80^\circ = \boxed{100}$ | A | 100 |
7af8c76494dcb75c6e629c6cacbfaa88 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$ | A decagon can be formed from the trapezoids and the base. The sum of the decagon's angles is $180(10-2)=1440^\circ$ . Letting the larger angle in each trapezoid be $x$ , the two angles formed by the line each measures $(180-x)^\circ$ . There are $8$ congruent angles left. Each of those angles measures $(360-2x)^\circ$ . Putting it all together: $8(360-2x)+2(180-x)=1440\implies x=\boxed{100}$ | A | 100 |
7af8c76494dcb75c6e629c6cacbfaa88 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$ | If we reflect the arch across the line, we form an 18-gon. $\frac{180*(18-2)}{18} = 160^\circ$ so each interior angle of the 18-gon is $160^\circ$ . Let $x$ be the degree measure of the larger interior angle of a trapezoid. From the diagram, we see that $2x + 160 = 360$ , so $2x = 200$ and $x = 100$ , or $\boxed{100}$ | A | 100 |
fa9ba3dc8a60f714a79bf5426b37d604 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_6 | A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?
$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7$ | Since the three segments are all the same length, the triathlete's average speed is the harmonic mean of the three given rates. Therefore, the average speed is \[\frac{3}{\frac{1}{3}+\frac{1}{20}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{6}\] | D | 6 |
60ed795f22e22e4c9fb74057e36e6812 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_7 | The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$ | Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes
$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$
$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$
$= \boxed{9}$ | E | 9 |
28444d19c696b1c3f7d6aa9c5c29a63c | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_10 | Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4$ | Since the length ratio is $\frac{1}{\sqrt{2}}$ , then the area ratio is $\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \boxed{4}$ | E | 4 |
014fca40b286fd4a679de3886a652013 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_12 | In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?
$\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r$ | Let the number of red marbles be 100. You have $b = 80,$ and $g = 160.$
The answer is $\frac{100+80+160}{100} = \boxed{3.4}$ | null | 3.4 |
1ce44ec841e5865f4b9d7bd4d984621f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$ | Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$
We get the following equation for how much Han is ahead of Ian: $H+5I = 70.$
The expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$
This simplifies to: $2H+10+10I.$
However, this is just $2(H+5I)+10.$
Substitute, from the first equation, $H+5I$ as $70.$
Therefore, the answer is $140 + 10$ , which is $150$ , or $\boxed{150}$ | D | 150 |
1ce44ec841e5865f4b9d7bd4d984621f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$ | We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ $H_t$ $J_s$ $J_t$ . The problem gives us 5 equations
\begin{align} H_s&=I_s+5 \\ H_t&=I_t+1 \\ J_s&=I_s+10 \\ J_t&=I_t+2 \\ H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}
Substituting equations $(1)$ and $(2)$ into $(5)$ gives:
\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]
We are asked the difference between Jan's and Ian's distances, or
\[J_s J_t-I_s I_t=x,\]
Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:
\[(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow\]
\[2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x\]
Substituting $(*)$ into this equation gives:
\[2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x\]
Therefore, the answer is $150$ miles or $\boxed{150}$ | D | 150 |
1ce44ec841e5865f4b9d7bd4d984621f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$ | Let Ian drive $D$ miles, at a speed of $R$ , for some time $T$ (in hours). Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: \[D= RT\] \[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\] \[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\] Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: \[70 = 5T + R + 5 \Longrightarrow 5T + R = 65\] \[x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.\] Since $x = 150$ , our answer is $\boxed{150}$ | D | 150 |
1ce44ec841e5865f4b9d7bd4d984621f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$ | Let Ian drive $d$ miles, $t$ hours, and at speed $s$
Ian's Equation: $d=s \cdot t.$
Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.
Han's Equation: $d+70=(s+5) \cdot (t+1).$
Let Jan have driven $m$ miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.
Jan's Equation: $m=(s+10) \cdot (t+2).$
Let's group the equations together:
$(1) \phantom{a} d=s \cdot t$
$(2) \phantom{a} d+70=(s+5) \cdot (t+1)$
$(3) \phantom{a} m=(s+10) \cdot (t+2)$
Let's see what we want to find: We want to find $n$ . The equation is $m=d+n$ where $n$ is the number of more miles traveled by Jan than Ian.
Onto the calculating part.
Expanding the second equation, we get
$d+70=st+5t+s+5.$
Note that $st=d$ by the first equation, so substituting we get
$d+70=d+5t+s+5.$
Simplifying gets us
$(4) \phantom{a} s+5t=65.$
(Note for the above process: You could have substituted $d$ with $st$ but that would lead you to the same result since $d-d=st-st=0.$
Let's look at the third equation. Expanding, we get
$m=st+10t+2s+20.$
Since we want $n$ we want the equation $m=d+n$ . We write the expanded third equation into this form since $st=d.$
$(5) \phantom{a} m=d+(10t+2s+20)$
Let's take a closer look at the $n$ section of the equation:
$(6) \phantom{a} n=10t+2s+20$
This looks very similar to equation 4, if you multiply equation 4 by $2$ you get
$(7) \phantom{a} 10t+2s=130$
Plugging equation 7 into equation 6 we have
$n=(10t+2s)+20 \Rightarrow n=130+20$
Calculating gets us
$(8) \phantom{a} n=150.$
Substituting equation 8 into equation 5 gets us
$m=d+n \Rightarrow m=d+150.$
The $n$ term is $150$ which is what we want to find so the answer is $\boxed{150}.$ | D | 150 |
1ce44ec841e5865f4b9d7bd4d984621f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$ | Since Han drove for $1$ hour and drove $70$ miles more than Ian during that hour, we know that Ian's speed is $65$ miles per hour since Han drove $5$ mph faster than him. Now Jan went $10$ mph faster than Ian for $2$ hours, so we can tell that she drove $75 \cdot 2$ miles more than Ian, therefore the answer is $\boxed{150}.$ | D | 150 |
e952fe4882d1f1a8d4a71d2437514548 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23 | Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$ | First, choose the two letters to be repeated in each set. $\dbinom{5}{2}=10$ . Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted (Take for example $S_{1} = \{a,b,c,d \}$ and $S_{2} = \{a,b,e \}$ ). Notice how $S_{1}$ and $S_{2}$ are interchangeable. Division by two will fix this problem. Thus we have:
$\dfrac{10 \times 8}{2} = 40 \implies \boxed{40}$ | B | 40 |
e952fe4882d1f1a8d4a71d2437514548 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23 | Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$ | Another way of looking at this problem is to break it down into cases.
First, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are $\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets.
Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\cdot3=30$ ways to choose these sets.
This leads us to the answer:
$10+30=40 \implies \boxed{40}$ | B | 40 |
e952fe4882d1f1a8d4a71d2437514548 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23 | Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$ | There are $\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We now must place the 3 remaining elements into the subsets. Using stars and bars, we can notate this as: $I I I X \rightarrow \dbinom{4}{3}=4$ . Thus, $4*10=\boxed{40}$ | B | 40 |
39fc02e1caebc1612c0b90dbe5f2ade5 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_24 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | $k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$ | D | 6 |
39fc02e1caebc1612c0b90dbe5f2ade5 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_24 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6} | D | 6 |
39c3c929b03f132eb955d5aaae94534d | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_1 | A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$ | The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{6}$ | E | 6 |
43b2c056b96a9474e34431445277d923 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_2 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$ | B | 4 |
43b2c056b96a9474e34431445277d923 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_2 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | Notice that at baseline the diagonals sum to the same number ( $52$ ). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$ | B | 4 |
7ced1ddb9f1c670508ccf91a75e25200 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_4 | A semipro baseball league has teams with 21 players each. League rules state that a player must be paid at least $15,000 and that the total of all players' salaries for each team cannot exceed $700,000. What is the maximum possible salary, in dollars, for a single player?
$\mathrm{(A)}\ 270,000\qquad\mathrm{(B)}\ 385,000\qquad\mathrm{(C)}\ 400,000\qquad\mathrm{(D)}\ 430,000\qquad\mathrm{(E)}\ 700,000$ | The maximum salary for a single player occurs when the other 20 players receive the minimum salary. The total of all players' salaries is 700000. The answer is $700000-15000*20=400000\Rightarrow \boxed{400,000}$ | C | 400,000 |
b6c201bcb876693cbc8a61805837dc89 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_7 | An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$ | The area of the large triangle is $\frac{10^2\sqrt3}{4}$ , while the area of each small triangle is $\frac{1^2\sqrt3}{4}$ . Dividing these two quantities results in $100$ , therefore $\boxed{100}$ small triangles can fill the large one without overlap. | C | 100 |
b6c201bcb876693cbc8a61805837dc89 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_7 | An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$ | [asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]
The number of triangles is $1+3+\dots+19 = \boxed{100}$ | null | 100 |
454ad69a0a48c14359d39e22fdb27ea4 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8 | A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$ | The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$ . Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$ which solves to $r\leq 8 \frac13$ $r$ must be an integer, so there are $\boxed{9}$ possible values of $r$ , and each gives us one solution. | C | 9 |
454ad69a0a48c14359d39e22fdb27ea4 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8 | A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$ | Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$ . Since $50$ is an even number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ , and $16$ . There are $9$ numbers in total, so the answer is $\boxed{9}$ | C | 9 |
454ad69a0a48c14359d39e22fdb27ea4 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8 | A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$ | Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$ .
Using the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$ , meaning the complete solution will be, $r=50+\frac{2}{\gcd(2,3)}$ $k=50+2k$ $c=-50-\frac{3}{\gcd(2,3)}k=-50-3k$
The solution range for which both $r$ and $c$ are positive is $17$ $\leq k$ $\leq$ $25$ . There are $\boxed{9}$ possible values for $k$ | C | 9 |
d0cf7b9616bf4aa17eca65600e3f7d30 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_11 | Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$
and that $u_3=9$ and $u_6=128$ . What is $u_5$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$ | If we plug in $n=4$ , we get
By plugging in $n=3$ , we get
This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$ , therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{53}$ .
~ Mathkiddie | B | 53 |
69c0811df26c75e895b1b5a7ceb1161a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_12 | Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\mathrm{(A)}\ 2500\qquad\mathrm{(B)}\ 3000\qquad\mathrm{(C)}\ 3500\qquad\mathrm{(D)}\ 4000\qquad\mathrm{(E)}\ 4500$ | Every time the pedometer flips from $99999$ to
$00000$ Pete has walked $100000$ steps.
So, if the pedometer flipped $44$ times
Pete walked $44*100000+50000=4450000$ steps.
Dividing by $1800$ steps per mile gives $2472.\overline{2}$
This is closest to answer $\boxed{2500}$ | A | 2500 |
616ee74293cf497e3090950d3d87cb29 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$ | Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the $2008^{\text{th}}$ term of the sequence is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{4015}$ | B | 4015 |
616ee74293cf497e3090950d3d87cb29 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$ | Let $a_1, a_2, a_3, \cdots, a_n$ be the terms of the sequence. We know $\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n$ , so we must have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$ . The sum of consecutive odd numbers down to $1$ is a perfect square, if you don't believe me, try drawing squares with the sum, so $a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1$ , so the answer is $a_{2008} = 2(2007) + 1 = \boxed{4015}$ | B | 4015 |
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