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 |
|---|---|---|---|---|---|
4,801 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | 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 |
4,802 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | 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 |
4,803 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_20 | 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 |
4,804 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_20 | 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 |
4,805 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_21 | 1 | Let $p(x) = x^3 + ax^2 + bx + c$ , where $a$ $b$ , and $c$ are complex numbers. Suppose that
What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$ | Consider the graph of $x^4$ . It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$ . Clearly they are just being translated down some large amount. This will result in the $x$ -axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know ... | 8 |
4,806 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_25 | 1 | The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$ | Consider another sequence $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$ , and $0 \leq \theta_n < 180$
The given recurrence becomes
It follows that $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$ . Since $\theta_1 = 45, \theta_2 = 30$ , all terms in the sequence $\{\theta_1, \thet... | 0 |
4,807 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_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 |
4,808 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_2 | 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 |
4,809 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_3 | 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 |
4,810 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_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 |
4,811 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_5 | 1 | 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 = ... | 18 |
4,812 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_6 | 1 | 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 |
4,813 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_7 | 1 | 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\qqu... | 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 i... | 17 |
4,814 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | 1 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | Because the line $x + y = 7$ is parallel to $\overline {AB}$ , the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$ . In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed{6}$ | 6 |
4,815 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | 2 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$ . Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\frac 4{\sqrt 2} = 2\sqrt 2$ . Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$ | 6 |
4,816 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_10 | 1 | 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 1... | 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... | 12 |
4,817 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_12 | 1 | The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively. What is the first term?
$\mathrm{(A)}\ 60\qquad \mathrm{(B)}\ 75\qquad \mathrm{(C)}\ 120\qquad \mathrm{(D)}\ 225\qquad \mathrm{(E)}\ 315$ | Let the $n$ th term of the series be $ar^{n-1}$ . Because \[\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,\] it follows that $r = 2$ and the first term is $a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}$ | 315 |
4,818 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_13 | 1 | Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$ | Let $D$ be the foot of the altitude to $\overline BC$ . Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$ . Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ . The sum of the two possible values is $14 + 4 = \boxed{18}$ | 18 |
4,819 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | 1 | 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)--(... | 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 enti... | 23 |
4,820 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | 2 | 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)--(... | 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{... | 23 |
4,821 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | 3 | 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)--(... | 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 sq... | 23 |
4,822 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | 4 | 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)--(... | 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 $\fra... | 23 |
4,823 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_16 | 1 | Trapezoid $ABCD$ has $AD||BC$ $BD = 1$ $\angle DBA = 23^{\circ}$ , and $\angle BDC = 46^{\circ}$ . The ratio $BC: AD$ is $9: 5$ . What is $CD$
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$ | Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$ . Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$ . Because $\overline {AD} \parallel \overline {BC}$ , it follows ... | 45 |
4,824 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | 1 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$ | To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$ | 802 |
4,825 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | 2 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$ | We will now show a complete solution, with a proof that no other values are prime.
Consider the function $g(x) = x^2 - 360x + 400$ , then obviously $f(x) = g(x^2)$
The roots of $g$ are: \[x_{1,2} = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 = 180 \pm 80 \sqrt 5\]
We can then write $g(x) = (x - 180 - 80\sqrt 5)(x -... | 802 |
4,826 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 1 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\frac 12 \cdot 3 \cdot 200 = \boxed{300}$ edges. | 300 |
4,827 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 2 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$ . A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \boxed{3... | 300 |
4,828 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 3 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ . Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $... | 300 |
4,829 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 4 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.
[asy] pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F); [/asy]
But b... | 300 |
4,830 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 5 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \cdot 100 = 200$
$\frac{200}{4} = 50$ , so there are $50$ vertices.
The base of a pyramid has 4 edges, so each sliced vertex would add four... | 300 |
4,831 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | 6 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R... | I doubt anyone has the mental capacity to imagine $Q$ in its full complexity in their mind. So, we begin by examining a simpler structure: a cube.
Let $E_C$ denote the number of edges of a polyhedron $C$ . Let $V_C$ denote the number of vertices of $C$ . Let $B_C$ denote the number of edges that meet at each vertex.
Su... | 300 |
4,832 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | 1 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \tex... | Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 bloc... | 89 |
4,833 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | 2 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \tex... | Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers . Clearly, $\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem. | 89 |
4,834 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | 3 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \tex... | Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at... | 89 |
4,835 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | 1 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordi... | The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$
In our setting where $A=(0,0)$ $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$
In other words, we need to count the triples of integers $(k,s,t)$ where $k>... | 784 |
4,836 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | 2 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordi... | We know that $A$ is $(0, 0)$ . Since $B$ is on the line $y=x$ , let it be represented by the point $(b, b)$ . Similarly, let $D$ be $(d, kd)$ . Since this is a parallelogram, sides $\overline{AD}$ and $\overline{BC}$ are parallel. Therefore, the distance and relative position of $D$ to $A$ is equivalent to that of $C$ ... | 784 |
4,837 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_23 | 1 | A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$ . What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \fr... | We multiply $z$ and $(\frac{3}{4}+\frac{3}{4}i)$ to get \[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\] Since we want to find the probability that this number is in $S$ , we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives ... | 79 |
4,838 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | 3 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | Algebraically, the inverse function of a function should just cancel out, leaving $6x=x$ . However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return... | 4 |
4,839 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | 4 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | The most conventional domain and range for $\cos^{-1} x$ is $x \in [-1, 1]$ and $\cos^{-1} x \in [0, \pi]$ . For $\cos x$ $x \in [0, \pi]$ and $\cos x \in [-1, 1]$ . Since the domain of $\cos x$ is equal to the range of $\cos^{-1} x$ , and the range of $\cos x$ is equal to the domain of $\cos^{-1} x$ $\cos^{-1} (\cos x... | 4 |
4,840 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | 1 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8... | We need to find a reasonably easy way to count the squares.
First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$ , hence each square has exactly one vertex in each quadrant.
Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the... | 225 |
4,841 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | 2 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8... | Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the "mapping" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$
[asy] defaultpen(black+0.75bp+... | 225 |
4,842 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_11 | 1 | Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p)... | Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed f... | 164 |
4,843 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_15 | 1 | 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 unit... | 6 |
4,844 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_15 | 2 | 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 continui... | 6 |
4,845 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | 1 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | Let $A = \log(a)$ and $B = \log(b)$
The first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\text{th}$ term is $nB$
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$
Since the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $k$ th term i... | 112 |
4,846 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | 2 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | If $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are in arithmetic progression , then $a^3b^7$ $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression . Therefore,
\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]
Therefore, $a^3b^7=b^{13}$ $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^... | 112 |
4,847 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_18 | 1 | Triangle $ABC$ , with sides of length $5$ $6$ , and $7$ , has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin . What is the volume of tetrahedron $OABC$
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\... | Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem \begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2,... | 95 |
4,848 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | 1 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Let $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$ and $B = \left(1 + x + x^2 + \cdots + x^{27}\right)$ . We are expanding $A \cdot A \cdot B$
Since there are $15$ terms in $A$ , there are $15^2 = 225$ ways to choose one term from each $A$ . The product of the selected terms is $x^n$ for some integer $n$ between $0$ ... | 224 |
4,849 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | 2 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Let $P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}$ . Then the $x^{28}$ term from the product in question $\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})$ is
$1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1... | 224 |
4,850 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | 3 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | We expand $(1 + x + x^2 + x^3 + \cdots + x^{14})^2$ to $(1 + x + x^2 + x^3 + \cdots + x^{14}) * (1 + x + x^2 + x^3 + \cdots + x^{14})$ and use FOIL to multiply. It expands out to:
$1 + x + x^2 + x^3 + x^4 + \cdots + x^{14} +$
$\qquad x + x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} +$
$\qquad \qquad x^2 + x^3 + x^4 + \cd... | 224 |
4,851 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | 4 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Rewrite the product as $\frac{(x^{28} - 1)(x^{15} - 1)(x^{15} - 1)}{(x - 1)^3}$ . It is known that
\[\frac{1}{(1 - x)^n} = \binom{n - 1}{n - 1} + \binom{n}{n - 1}x + \binom{n + 1}{n - 1}x^2 + \binom{n + 2}{n - 1}x^3 + \cdots + \binom{n - 1 + k}{n - 1}x^k + \cdots .\]
Thus, our product becomes
\[-\left( x^{28} - 1 \righ... | 224 |
4,852 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_21 | 1 | A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$ . What is the number of heavy-tailed permutations?
$\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52$ | We use case work on the value of $a_3$
Case 1: $a_3 = 1$ . Since $a_1 + a_2 < a_4 + a_5$ $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$ . The values of $a_1$ and $a_2$ , as well as the values of $a_4$ and $a_5$ , are interchangeable, so this case produces a total of $2(2 \cdot 2) = 8$ solutions.
Case 2:... | 48 |
4,853 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_2 | 1 | $4\times 4$ 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&... | 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+... | 4 |
4,854 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_2 | 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&... | 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 $1+3=\boxed{4}$ | 4 |
4,855 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_5 | 1 | 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?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9... | The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{9}$ | 9 |
4,856 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_6 | 1 | 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 D... | Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{2500}$ | 500 |
4,857 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_12 | 1 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008$ th term of the sequence?
$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$ | Letting the sum of the sequence equal $a_1+a_2+\cdots+a_n$ yields the following two equations:
$\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008$ and
$\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007$
Therefore:
$a_1+a_2+\cdots+a_{2008}=2008^2$ and $a_1+a_2+\cdots+a_{2007}=2007^2$
Hence, by substitution, $a_{2008}=2008^2-2007^2=(200... | 15 |
4,858 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_18 | 1 | A pyramid has a square base $ABCD$ and vertex $E$ . The area of square $ABCD$ is $196$ , and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$ , respectively. What is the volume of the pyramid?
$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}... | Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$ . The side length of the base is $14$ . The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$ , respectively. Consider a side view of the pyramid from $\triangle BCE$ . We have a systems of equat... | 784 |
4,859 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | 1 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$ .
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$ .
We have $D(0)=200$
The tr... | 5 |
4,860 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | 2 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | The truck takes $20$ seconds to go from one pail to the next and then stops for $30$ seconds at the new pail. Thus it sets off from a pail every 50 sec. Let $t$ denote the time elapsed and write $t=50k + \Delta$ , where $\Delta \in [0,50)$ . In this time Michael has traveled $5t = 250k+5\Delta$ feet. What about the tr... | 5 |
4,861 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | 3 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | We make a chart by seconds in increments of ten.
$\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline 0&0&200\\ 10&50&300\\ 20&100&400\\ 30&150&400\\ 40&200&400\\ 50&250&400\\ 60&300&500\\ 70&350&600\\ 80&400&600\\ 90&450&600\\ 100&500&600\\ 110&550&700\\ 120&600&800\\ 130&650&800\\ 140&700... | 5 |
4,862 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | 4 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | This solution might be time consuming, but it is pretty rigorous. Also, throughout the solution refer to the graph in solution 1 to understand this one more. Lets first start off by defining the position function for Michael. We let $M(t) = 5t$ , where $t$ is the amount of seconds passed. Now, lets define the position ... | 5 |
4,863 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_23 | 1 | The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792$ . What is $n$
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$ | Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ . For any factor $2^a \times 5^b$ , there will ... | 11 |
4,864 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | 1 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qq... | Let $a_n=|A_{n-1}A_n|$ . We need to rewrite the recursion into something manageable. The two strange conditions, $B$ 's lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows: \[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\] which us... | 17 |
4,865 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | 2 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qq... | Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the $x$ axis and denote the resulting heights $h_n$ and $h_{n+1}$ . From 30-60-90 rules, the distance between the points where these altitudes meet the x-axis is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} = \fr... | 17 |
4,866 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | 3 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qq... | Note that $A_{1}$ is of the form $(2x,0)$ for some $x$ , and thus $B_{1}$ is of the form $(x, x \sqrt{3}).$ Then, we are told that $B_{1}$ lies on the graph of $y = \sqrt{x}$ , so \[(x \sqrt{3})^{2} = x.\] Solving for x, we get that $x = \frac{1}{3},$ and so $A_{1} = (2/3,0)$
Now, similarly to before, let $|A_{1}A_{2}|... | 17 |
4,867 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | 4 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qq... | We can iteratively calculate out the first few $A_i$ and $B_i$ . We know that $A_0 = (0,0)$ and the line through $A_0$ and $B_1$ needs to make a $60^{\circ}$ angle with the x-axis (because the triangle is equilateral). The equation of a line that makes an angle $\theta$ with the x-axis and passes through the origin has... | 17 |
4,868 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_4 | 1 | Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | \[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{7}\] | 7 |
4,869 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_9 | 1 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to hi... | Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r +... | 34 |
4,870 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_9 | 2 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to hi... | [asy] draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); [/asy] Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach ... | 34 |
4,871 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_11 | 1 | A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might be... | A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$ , so $S$ must be divisible by $37\ \mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequenc... | 37 |
4,872 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_12 | 1 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ cha... | 58 |
4,873 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_12 | 2 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.
Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$ ). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$ , because ea... | 58 |
4,874 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | 1 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}... | The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\frac{1}{5}$ , so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$ , ... | 10 |
4,875 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | 2 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}... | If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).$ The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \boxed{10}$ | 10 |
4,876 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | 1 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that \[n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.\] Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ ... | 4 |
4,877 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | 2 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$
Proof of claim:
We examine the positive integers mod $9$ . Here are the cases.
Case 1. $n \equiv 1 \pmod 9$ . Now, we examine $S(n)$ modulo $9$ .
Case 1.1. The tens digit of $n$ is different from the tens digit of the largest ... | 4 |
4,878 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | 3 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Let the number of digits of $n$ be $m$ . If $m = 5$ $n$ will already be greater than $2007$ . Notice that $S(n)$ is always at most $9m$ . Then if $m = 3$ $n$ will be at most $999$ $S(n)$ will be at most $27$ , and $S(S(n))$ will be even smaller than $27$ . Clearly we cannot reach a sum of $2007$ , unless $m = 4$ (i.e. ... | 4 |
4,879 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | 1 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$ . We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$
The spacy subsets of $S_{n + 1}$ can be divided into two groups:
Hence,
From this recursion , we find that
And so the answer is $\boxed{129}$ | 129 |
4,880 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | 2 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | ... | 129 |
4,881 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | 3 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most $4$ elements. Given any arrangment, we subract $2i-2$ from the $i-th$ element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size $k$ subsets of the s... | 129 |
4,882 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | 4 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Let us consider each size of subset individually. Since each integer in the subset must be at least $3$ away from any other integer in the subset, the largest spacy subset contains $4$ elements.
First, it is clear that there is $1$ spacy set with $0$ elements in it, the empty set. Next, there are $12$ spacy subsets wit... | 129 |
4,883 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_1 | 1 | Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mat... | There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \[2\cdot(12\cdot8+10\cdot8)-60=2\cdot176-60=292\] We have three bedrooms, so she must paint $292\cdot3=\boxed{876}$ square feet of walls. | 876 |
4,884 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_2 | 1 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 ... | The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$ | 24 |
4,885 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_2 | 2 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 ... | Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$ | 24 |
4,886 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_8 | 1 | Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found. | 5 |
4,887 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_10 | 1 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C... | If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20... | 8 |
4,888 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_10 | 2 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C... | Let $x$ be the number of people initially in the group and $g$ the number of girls. $\frac{2}{5}x = g$ , so $x = \frac{5}{2}g$ . Also, the problem states $\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \boxed{8}$ | 8 |
4,889 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_11 | 1 | The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?
$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$ | The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac... | 173 |
4,890 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_12 | 1 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B)... | We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior... | 93 |
4,891 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_12 | 2 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B)... | Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$ | 93 |
4,892 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_18 | 1 | Let $a$ $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qq... | Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds of t... | 16 |
4,893 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | 1 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qqu... | Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$ . Because $72= 4\cdot 18$ , we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$ , wh... | 16 |
4,894 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | 2 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qqu... | The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$ . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-... | 16 |
4,895 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | 3 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qqu... | Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$ ) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$ . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their ar... | 16 |
4,896 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_23 | 1 | How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$ | Let $a$ and $b$ be the two legs of the triangle, and $c$ be the hypotenuse.
By using $Area = \frac{r}{2} (a+b+c)$ , where $r$ is the in-radius, we get:
\[3(a+b+c) = \frac{r}{2} (a+b+c)\] \[r=6\]
In right triangle, $r = \frac{a+b-c}{2}$ \[a+b-c = 12\] \[c = a + b - 12\]
By the triangle's area we get:
\[\frac{ab}{2} = 6 ... | 6 |
4,897 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_23 | 2 | How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$ | Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is $\boxed{6}$ | 6 |
4,898 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24 | 1 | problem_id
c84575ab947ea92b2fa92f55386966ac Also refer to the 2007 AMC 10B #25 (same problem)
c84575ab947ea92b2fa92f55386966ac How many pairs of positive integers $(a,b)$ ar...
Name: Text, dtype: object | Rewrite the equation\[ $\frac{a}{b}+\frac{14b}{9a}=k$ \]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\[ $\frac{14b^2}{9a}=bk-a.$ \]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1, it follows ... | 4 |
4,899 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_1 | 1 | Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$ | The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=\boxed{31}$ dollars. | 31 |
4,900 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_3 | 1 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$ | 18 |
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