id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
3,001 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | 3 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | First we note that $x \in (-\infty,-4] \cup [4,\infty]$ . This will help us later with finding extraneous solutions.
Next, we have two cases: \[\text{IF } x\leq 4:\] \[\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3\] .
We note that $3$ is not in the range of possible $x$ 's and thus is... | 64 |
3,002 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | 1 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences... | 31 |
3,003 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | 3 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | Let $w - x = a$ $w - y = b$ $w - z = c$ . As above, we know that $c = 9$ . Thus, $a < b < c$ .
So, we have $w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44$ . This means $a + b + 9$ is a multiple of $4$ . Testing values of $a$ and $b$ , we find $(a, b, c) = (1, 6, 9), (3, 4, 9),$ and $(5, 6, 9)$ ... | 31 |
3,004 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | 5 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | Because we know that $w>x>y>z$ and that the positive differences are $1, 3, 4, 5, 6, 9$ , we can immediately come to the conclusion that $w-z= 9$ (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have $3$ equations, we have that $x+y+z+w=44$ (from t... | 31 |
3,005 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | 1 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$
To compute this, we use a clever application of the binomial theorem
$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$
In all of the other te... | 6 |
3,006 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | 2 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem , it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$
In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have \[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \eq... | 6 |
3,007 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | 3 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Notice that the hundreds digit of $2011^{2011}$ won't be affected by $2000$ . Essentially we could solve the problem by finding the hundreds digit of $11^{2011}$ . Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$... | 6 |
3,008 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | 5 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We know that the hundreds digit of $2011^{2011}$ is just the hundreds digit of $11^{2011} \equiv 011 \pmod{1000}$ . If we actually take a look at the powers of $11 \pmod{1000}$ , we notice a pattern:
\[11^{1} \equiv 011 \pmod{1000}\] \[11^{2} \equiv 121 \pmod{1000}\] \[11^{3} \equiv 331 \pmod{1000}\] \[11^{4} \equiv 64... | 6 |
3,009 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | 6 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We know that the hundreds digit of $2011^{11}$ is just the hundreds digit of $11^{11}$ (mod 1000). If we actually take a look at the powers of 11, we notice a pattern:
\[11^{1} = 11\]
\[11^{2} = 121\]
\[11^{3} = 1331\]
\[11^{4} = 14641\]
We notice that the hundreds digit just increases by 1 starting from 1(the hundreds... | 6 |
3,010 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_1 | 1 | Mary's top book shelf holds five books with the following widths, in centimeters: $6$ $\dfrac{1}{2}$ $1$ $2.5$ , and $10$ . What is the average book width, in centimeters?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | To find the average, we add up the widths $6$ $\dfrac{1}{2}$ $1$ $2.5$ , and $10$ , to get a total sum of $20$ . Since there are $5$ books, the average book width is $\frac{20}{5}=4$ The answer is $\boxed{4}$ | 4 |
3,011 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_3 | 1 | Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$ | Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\boxed{25}$ | 25 |
3,012 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_3 | 2 | Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$ | Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the p... | 25 |
3,013 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_5 | 1 | The area of a circle whose circumference is $24\pi$ is $k\pi$ . What is the value of $k$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$ | If the circumference of a circle is $24\pi$ , the radius would be $12$ . Since the area of a circle is $\pi r^2$ , the area is $144\pi$ . The answer is $\boxed{144}$ | 144 |
3,014 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_5 | 2 | The area of a circle whose circumference is $24\pi$ is $k\pi$ . What is the value of $k$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$ | By definition, $\pi$ is the ratio of the circumference to the diameter. Since the circumference is $24\pi$ , the diameter must be $24$ and the radius is $12$ . Therefore, by the area of circle formula $A=\pi r^{2}$ the area is $12^{2}\pi=144\pi$ and $k=144 \Longrightarrow \boxed{144}$ | 144 |
3,015 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_7 | 1 | Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
$\ma... | Crystal runs north one mile, then her next two moves can be broken up into four individual moves: for her northeast section, it forms a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . For her southeast section, it is also a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . Notice that the tw... | 3 |
3,016 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | 1 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ ... | Tony works $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the beginn... | 13 |
3,017 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | 2 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ ... | Let $x$ equal Tony's age at the end of the period. We know that his age changed during the time period (since $630$ does not evenly divide $50$ ). Thus, his age at the beginning of the time period is $x - 1$
Let $d$ be the number of days Tony worked while his age was $x$ . We know that his earnings every day equal his ... | 13 |
3,018 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | 3 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ ... | Since Tony worked for $50$ days, he has worked for $100$ hours. Let $k$ be his hourly wage. Then, we know that $100k = 630$
Dividing both sides by $50$ , we get that $2k$ (Daily wage, which is equal to his age) $\approx$ $12.6$ . Since we now know that his age was either 12 or 13 during the 6 month span, and we are ask... | 13 |
3,019 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_9 | 1 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qqu... | $x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$ | 24 |
3,020 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_10 | 1 | Marvin had a birthday on Tuesday, May 27 in the leap year $2008$ . In what year will his birthday next fall on a Saturday?
$\mathrm{(A)}\ 2011 \qquad \mathrm{(B)}\ 2012 \qquad \mathrm{(C)}\ 2013 \qquad \mathrm{(D)}\ 2015 \qquad \mathrm{(E)}\ 2017$ | $\boxed{2017}$ $2017$ | 17 |
3,021 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_11 | 1 | The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$ . What is $b - a$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$ | Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$
$a\le 2x+3\le b$
Subtract $3$ from all of the quantities:
$a-3\le 2x\le b-3$
Divide all of the quantities by $2$
$\frac{a-3}{2}\le x\le \frac{b-3}{2}$
Since we have the range of the solutions, we... | 20 |
3,022 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_14 | 1 | Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ ... | Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{A... | 90 |
3,023 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_15 | 1 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the... | 3 |
3,024 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_15 | 2 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means th... | 3 |
3,025 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_16 | 1 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), ... | 33 |
3,026 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | 1 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of ... | 7 |
3,027 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | 2 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included i... | 7 |
3,028 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | 3 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \bo... | 7 |
3,029 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_19 | 1 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqua... | It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $F... | 6 |
3,030 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | 2 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$
$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$
We do not care about $+bx$ in this case, because we are only looking for $a$ . We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$ We are trying to minimi... | 78 |
3,031 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | 3 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$ . The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$ . We want the smallest difference of the $3$ roots since by... | 78 |
3,032 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_22 | 1 | Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf... | To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circle's i... | 28 |
3,033 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_22 | 2 | Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf... | To make a triangle, where the $3$ points are arranged on a circle, you just need to choose $3$ points because no $3$ points are arranged in a straight line on a circle, meaning that to count the number of triangles we get $\binom{8}{3}=56$ . However, this is incorrect because of the restriction that no three chords int... | 28 |
3,034 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_25 | 1 | Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{arra... | We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$ , which is $x=2$
We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ f... | 3 |
3,035 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_25 | 2 | Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{arra... | Notice that to get the previous term, we must add the smallest square number, (let's call it $n^2$ ) such that the sum is less than $(n+1)^2$ . Otherwise, instead of subtracting $n^2$ from the previous term, we're subtracting a greater square number.
Remember that $(x+1)^2 = x^2 + 2x + 1$ . Recall that to find the prev... | 3 |
3,036 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_1 | 1 | What is $100(100-3)-(100\cdot100-3)$
$\textbf{(A)}\ -20,000 \qquad \textbf{(B)}\ -10,000 \qquad \textbf{(C)}\ -297 \qquad \textbf{(D)}\ -6 \qquad \textbf{(E)}\ 0$ | $100(100-3)-(100\cdot{100}-3)=10000-300-10000+3=-300+3=\boxed{297}$ | 297 |
3,037 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_2 | 1 | Makarla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$ | The total number of minutes in her $9$ -hour work day is $9 \times 60 = 540.$ The total amount of time spend in meetings in minutes is $45 + 45 \times 2 = 135.$ The answer is then $\frac{135}{540}$ $= \boxed{25}$ | 25 |
3,038 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_3 | 1 | A drawer contains red, green, blue, and white socks with at least 2 of each color. What is
the minimum number of socks that must be pulled from the drawer to guarantee a matching
pair?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | After you draw $4$ socks, you can have one of each color, so (according to the pigeonhole principle ), if you pull $\boxed{5}$ then you will be guaranteed a matching pair. | 5 |
3,039 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_4 | 1 | For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$ | The average of two numbers, $a$ and $b$ , is defined as $\frac{a+b}{2}$ . Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$ . With that said, we need to find the sum when we plug, $1$ $2$ and $3$ into that equation. So:
\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + ... | 10 |
3,040 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_5 | 1 | A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | $31 \equiv 3 \pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\boxed{3}$ | 3 |
3,041 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_6 | 1 | A circle is centered at $O$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle... | 25 |
3,042 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_6 | 2 | A circle is centered at $O$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | Note that $\angle AOC = 180^\circ - 50^\circ = 130^\circ$ . Because triangle $AOC$ is isosceles, $\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25}$ | 25 |
3,043 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | 1 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - \left( \dfrac{12}{2} \right)^2} = \sqrt{64} = 8$
Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$
We have that the area of the rectangle is the same as the area of the triangle, namely $... | 32 |
3,044 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | 2 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | An alternative way to find the area of the triangle is by using Heron's formula, $A=\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$ . Thus the area equals $\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48.$ W... | 32 |
3,045 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | 3 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | Note that a triangle with side lengths $10,10,12$ is essentially $2$ “6,8,10” right triangles stuck together. Hence, the height is $8$ , and our area is $48$
So, the length of the rectangle is $\frac{48}{4}=12$ , and our perimeter $P=2(4+12)=\boxed{32}$ | 32 |
3,046 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_11 | 1 | A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 1... | Let the listed price be $(100 + p)$ , where $p > 0$
Coupon A saves us: $0.15(100+p) = (0.15p + 15)$
Coupon B saves us: $30$
Coupon C saves us: $0.25p$
Now, the condition is that A has to be greater than or equal to either B or C which gives us the following inequalities:
$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Righta... | 50 |
3,047 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_12 | 1 | At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end o... | Clearly, the minimum possible value would be $70 - 50 = 20\%$ . The maximum possible value would be $30 + 50 = 80\%$ . The difference is $80 - 20 = \boxed{60}$ | 60 |
3,048 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_13 | 1 | What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$ | We evaluate this in cases:
Case 1 $x<30$
When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ $x=|2x-60+2x|\implies x=|4x-60|$
Subcase 1 $30>x>15$
When $30>x>15$ we are going to h... | 92 |
3,049 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_13 | 2 | What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$ | From the equation $x = \left|2x-|60-2x|\right|$ , we have $x = 2x-|60-2x|$ , or $-x = 2x-|60-2x|$ . Therefore, $x=|60-2x|$ , or $3x=|60-2x|$ . From here we have four possible cases:
1. $x=60-2x$ ; this simplifies to $3x=60$ , so $x=20$
2. $-x=60-2x$ ; this simplifies to $x=60$
3. $3x=60-2x$ ; this simplifies to $5x=60$... | 92 |
3,050 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_15 | 1 | On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?
$\textbf{(A)}\ ... | Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ $4a-c=99$ . Also, $a+c\leq50 \Rightarrow c\leq50-a$ . W... | 29 |
3,051 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | 1 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot ... | 23 |
3,052 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | 2 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Als... | 23 |
3,053 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_19 | 1 | A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)... | The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$
Because $OA=4\sqrt{3} < \sqrt{156}, A$ must be in the interior of circle $O.$
Let $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equil... | 6 |
3,054 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_19 | 2 | A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)... | We can use the same diagram as Solution 1 and label the side length of $\triangle ABC$ as $s$ . Using congruent triangles, namely the two triangles $\triangle BOA$ and $\triangle COA$ , we get that $\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150$ . From this, we can use the Law of Cosines , to get... | 6 |
3,055 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_20 | 1 | 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{... | 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 ... | 81 |
3,056 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_20 | 2 | 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{... | 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... | 81 |
3,057 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | 1 | 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 ... | 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 calculat... | 932 |
3,058 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | 2 | 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 ... | 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 ... | 932 |
3,059 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | 3 | 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 ... | 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 co... | 932 |
3,060 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_23 | 1 | 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&\\... | 42 |
3,061 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_23 | 2 | 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 $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$ | 42 |
3,062 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_24 | 1 | 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 arit... | 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, ... | 34 |
3,063 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_24 | 2 | 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 arit... | 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 i... | 34 |
3,064 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_25 | 1 | 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 ... | 315 |
3,065 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_25 | 2 | 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 f... | 315 |
3,066 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_1 | 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 $\mathrm{\boxed{11}$ | 11 |
3,067 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_2 | 1 | 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 ... | 15 |
3,068 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_5 | 2 | 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
$1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1$
$=(1+2+3\cdots n)+(1+2+3+\cdots n-1)$... | 81 |
3,069 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_5 | 3 | 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+... | 81 |
3,070 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_11 | 1 | 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 \textb... | 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^... | 125 |
3,071 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_12 | 1 | 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 $BD=\boxed{13}$ | 13 |
3,072 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_13 | 1 | 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 $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{2}$ | 2 |
3,073 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_14 | 1 | 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)}\... | 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 $3/2$ $\f... | 3 |
3,074 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | 1 | 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... | 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 triang... | 761 |
3,075 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | 2 | 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... | 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,Fi... | 761 |
3,076 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | 3 | 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... | 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+1... | 761 |
3,077 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | 4 | 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... | 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_... | 761 |
3,078 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | 1 | 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, t... | 18 |
3,079 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | 2 | 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 calc... | 18 |
3,080 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | 3 | 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 = \p... | 18 |
3,081 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | 4 | 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 $\boxed{18}$ | 18 |
3,082 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | 1 | 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{(... | 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 c... | 51 |
3,083 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | 2 | 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{(... | 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\% \Ri... | 51 |
3,084 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | 3 | 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{(... | 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}{... | 51 |
3,085 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_19 | 1 | 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{(... | 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{20... | 8 |
3,086 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | 1 | 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 t... | 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 d... | 65 |
3,087 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | 2 | 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 t... | 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}k... | 65 |
3,088 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_23 | 1 | 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 ... | 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!), $\boxed{6}$ | 6 |
3,089 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_23 | 2 | 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 ... | 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}{... | 6 |
3,090 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_24 | 1 | 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 ... | 47 |
3,091 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_24 | 2 | 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.... | 47 |
3,092 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | 1 | 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)}\ 1... | 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}\... | 7 |
3,093 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | 2 | 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)}\ 1... | 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 $\box... | 7 |
3,094 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | 3 | 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)}\ 1... | 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 ... | 7 |
3,095 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | 4 | 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)}\ 1... | 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 $6+1=7 \Rightarrow \boxed{7}$ | 7 |
3,096 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_1 | 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 to... | 2 |
3,097 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_2 | 1 | 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 $\dfrac{4-3}{6-4} = \boxed{12}$ | 12 |
3,098 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_3 | 1 | 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 \math... | Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$ | 15 |
3,099 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_4 | 1 | 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 \... | 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... | 15 |
3,100 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_5 | 1 | 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 $\boxed{36}$ | 36 |
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