AMC10 2013 B

Answer (C): Simplifying gives \[ \frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}=\frac{12}{9}-\frac{9}{12}=\frac{4}{3}-\frac{3}{4}=\frac{16-9}{12}=\frac{7}{12}. \]

The garden is 2 · 15 = 30 feet wide and 2 · 20 = 40 feet long. Hence Mr. Green expects 1/2 · 30 · 40 = 600 pounds of potatoes.

The difference between the high and low temperatures was 16 degrees, so the difference between each of these and the average temperature was 8 degrees. The low temperature was 8 degrees less than the average, so it was 3°−8°= −5°.

Answer (D): The number 201 is the 1st number counted when proceeding backwards from 201 to 3. In turn, 200 is the 2nd number, 199 is the 3rd number, and $x$ is the $(202-x)^{\text{th}}$ number. Therefore 53 is the $(202-53)^{\text{th}}$ number, which is the $149^{\text{th}}$ number.

Note that 2·a−a·b = (2−b)a. This expression is negative when b > 2. Hence the product is minimized when a and b are as large as possible. The minimum value is (2 −5) · 5 = −15.

Answer (C): The sum of all the ages is $55\cdot 33 + 33\cdot 11 = 33\cdot 66$, so the average of all the ages is $$ \frac{33\cdot 66}{55+33}=\frac{33\cdot 66}{88}=\frac{33\cdot 3}{4}=24.75. $$

Answer (B): The six points divide the circle into six arcs each measuring $60^\circ$. By the Inscribed Angle Theorem, the angles of the triangle can only be $30^\circ$, $60^\circ$, $90^\circ$, and $120^\circ$. Because the angles of the triangle are pairwise distinct the triangle must be a $30-60-90^\circ$ triangle. Therefore the hypotenuse of the triangle is the diameter of the circle, and the legs have lengths $1$ and $\sqrt{3}$. The area of the triangle is $\frac{1}{2}\cdot 1\cdot \sqrt{3}=\frac{\sqrt{3}}{2}$.

Answer (B): Let $D$ equal the distance traveled by each car. Then Ray's car uses $\frac{D}{40}$ gallons of gasoline and Tom's car uses $\frac{D}{10}$ gallons of gasoline. The cars combined miles per gallon of gasoline is $\frac{2D}{\left(\frac{D}{40}+\frac{D}{10}\right)}=16.$

Answer (D): Note that $27{,}000 = 2^3 \cdot 3^3 \cdot 5^3.$ The only three pairwise relatively prime positive integers greater than 1 with a product of 27,000 are 8, 27, and 125. The sum of these numbers is 160.

Let x denote the number of three-point shots attempted. Then the number of three-point shots made was 0.4x, resulting in 3(0.4x) = 1.2x points. The number of two-point shots attempted was 1.5x, and they were successful on 0.5(1.5x) = 0.75x of them resulting in 2(0.75x) = 1.5x points. The number of points scored was 1.2x + 1.5x = 54, so x = 20.

Answer (B): By completing the square the equation can be rewritten as follows: $x^2 + y^2 = 10x - 6y - 34,$ $x^2 - 10x + 25 + y^2 + 6y + 9 = 0,$ $(x - 5)^2 + (y + 3)^2 = 0.$ Therefore $x = 5$ and $y = -3$, so $x + y = 2$.

The five sides of the pentagon are congruent, and the five congruent diagonals are longer than the sides. Once one segment is selected, 4 of the 9 remaining segments have the same length as that segment. Therefore the requested probability is 4/9.

Answer (E): Note that Jo starts by saying 1 number, and this is followed by Blair saying 2 numbers, then Jo saying 3 numbers, and so on. After someone completes her turn after saying the number $n$, then $1+2+3+\cdots+n=\frac{1}{2}n(n+1)$ numbers have been said. If $n=9$ then 45 numbers have been said. Therefore there are $53-45=8$ more numbers that need to be said. The $53^{\mathrm{rd}}$ number said is 8.

Answer (E): The equation $x\clubsuit y = y\clubsuit x$ is equivalent to $x^2y-xy^2 = y^2x-yx^2$. This equation is equivalent to gives $2xy(x-y)=0$. This equation will hold exactly if $x=0$, $y=0$, or $x=y$. The solution set consists of three lines: the $x$-axis, the $y$-axis, and the line $x=y$.

Answer (B): Let $s$ be the side length of the triangle and $h$ the side length of the hexagon. The hexagon can be subdivided into 6 equilateral triangles by drawing segments from the center of the hexagon to each vertex. Because the areas of the large triangle and hexagon are equal, the triangles in the hexagon each have area $\frac{1}{6}$ of the area of the large triangle. Thus $\frac{h}{s}=\sqrt{\frac{1}{6}}$ so $h=\frac{\sqrt{6}}{6}s.$ The perimeter of the triangle is $a=3s$ and the perimeter of the hexagon is $b=6h=\sqrt{6}s$, so $\frac{a}{b}=\frac{3s}{\sqrt{6}s}=\frac{\sqrt{6}}{2}.$

The ratio of PE:PD:DE is 3:4:5. Hence by the converse of the Pythagorean Theorem, $\triangle DPE$ is a right triangle. Therefore CE is perpendicular to AD, and the area of $\triangle AEDC$ is one-half the product of its diagonals. Because P is the centroid of $\triangle ABC$, it follows that CE = 3(PE) = 4.5 and AD = 3(PD) = 6. Therefore the area of $\triangle AEDC$ is 0.5(4.5)(6) = 13.5.

Answer (E): After Alex makes $m$ exchanges at the first booth and $n$ exchanges at the second booth, Alex has $75-(2m-n)$ red tokens, $75-(3n-m)$ blue tokens, and $m+n$ silver tokens. No more exchanges are possible when he has fewer than $2$ red tokens and fewer than $3$ blue tokens. Therefore no more exchanges are possible if and only if $2m-n\ge 74$ and $3n-m\ge 73$. Equality can be achieved when $(m,n)=(59,44)$, and Alex will have $59+44=103$ silver tokens. Note that the following exchanges produce 103 silver tokens: \[ \begin{array}{|l|c|c|c|} \hline & \text{Red Tokens} & \text{Blue Tokens} & \text{Silver Tokens} \\ \hline \text{Exchange 75 blue tokens} & 100 & 0 & 25 \\ \hline \text{Exchange 100 red tokens} & 0 & 50 & 75 \\ \hline \text{Exchange 48 blue tokens} & 16 & 2 & 91 \\ \hline \text{Exchange 16 red tokens} & 0 & 10 & 99 \\ \hline \text{Exchange 9 blue tokens} & 3 & 1 & 102 \\ \hline \text{Exchange 2 red tokens} & 1 & 2 & 103 \\ \hline \end{array} \]

Answer (D): First note that the only number between 2000 and 2013 that shares this property is 2002. Consider now the numbers in the range 1001 to 1999. There is exactly 1 number, 1001, that shares the property when the units digits is 1. There are exactly 2 numbers, 1102 and 1012, when the units digit is 2; exactly 3 numbers, 1203, 1113, and 1023, when the units digits is 3, and so on. Because the thousands digit is always 1, when the units digit is $n$, for $1 \le n \le 9$, the sum of the hundreds and tens digits must be $n-1$. There are exactly $n$ ways for this to occur. Hence there are exactly $$1+(1+2+\cdots+9)=1+\frac{9\cdot10}{2}=1+45=46$$ numbers that share this property.

Answer (D): Let the common difference in the arithmetic sequence be $d$, so that $a=b+d$ and $c=b-d$. Because the quadratic has exactly one root, $b^2-4ac=0$. Substitution gives $b^2=4(b+d)(b-d)$, and therefore $3b^2=4d^2$. Because $b\ge 0$ and $d\ge 0$, it follows that $\sqrt{3}b=2d$. Thus the real root is $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-b}{2a} = \frac{-b}{2(b+d)} = \frac{-b}{2\left(b+\frac{\sqrt{3}}{2}b\right)} = -2+\sqrt{3}. $$ Note that the quadratic equation $x^2+(4-2\sqrt{3})x+7-4\sqrt{3}$ satisfies the given conditions.

Answer (B): The prime factorization of 2013 is 3 · 11 · 61. There must be a factor of 61 in the numerator, so $a_1 \ge 61$. Since $a_1!$ will have a factor of 59 and 2013 does not, there must be a factor of 59 in the denominator, and $b_1 \ge 59$. Thus $a_1 + b_1 \ge 120$, and this minimum value can be achieved only if $a_1 = 61$ and $b_1 = 59$. Furthermore, this minimum value is attainable because $$ 2013=\frac{(61!)(11!)(3!)}{(59!)(10!)(5!)}. $$ Thus $|a_1-b_1|=a_1-b_1=61-59=2$.

Answer (C): Let the two sequences be $(a_n)$ and $(b_n)$, and assume without loss of generality that $a_1<b_1$. The definitions of the sequences imply that $a_7=5a_1+8a_2=5b_1+8b_2$, so $5(b_1-a_1)=8(a_2-b_2)$. Because 5 and 8 are relatively prime, 8 divides $b_1-a_1$ and 5 divides $a_2-b_2$. It follows that $a_1\le b_1-8\le b_2-8\le a_2-13$. The minimum value of $N$ results from choosing $a_1=0$, $b_1=b_2=8$, and $a_2=13$, in which case $N=104$.

Answer (C): The digit $j$ at $J$ contributes to all four sums, and each of the other digits contributes to exactly one sum. Therefore the sum of all four sums is $3j + (1 + 2 + 3 + \cdots + 9) = 45 + 3j$. Because all four sums are equal, this must be a multiple of $4$, so $j = 1, 5,$ or $9$. For each choice of $j$, pair up the remaining digits so that each pair has the same sum. For example, for $j = 1$ the pairs are $2$ and $9$, $3$ and $8$, $4$ and $7$, and $5$ and $6$. Then order the pairs so that they correspond to the vertex pairs $(A, E)$, $(B, F)$, $(C, G)$, $(D, H)$. This results in $2^4 \cdot 4!$ different combinations for each $j$. Thus the requirements can be met in $2^4 \cdot 4! \cdot 3 = 1152$ ways.

Answer (B): The Pythagorean Theorem applied to right triangles $ABD$ and $ACD$ gives $AB^2 - BD^2 = AD^2 = AC^2 - CD^2$; that is, $13^2 - BD^2 = 15^2 - (14 - BD)^2$, from which it follows that $BD = 5$, $CD = 9$, and $AD = 12$. Because triangles $AED$ and $ADC$ are similar, \[ \frac{AE}{12}=\frac{DE}{9}=\frac{12}{15}, \] implying that $ED=\frac{36}{5}$ and $AE=\frac{48}{5}$. Because $\angle AFB=\angle ADB=90^\circ$, it follows that $ABDF$ is cyclic. Thus $\angle ABD+\angle AFD=180^\circ$ from which $\angle ABD=\angle AFE$. Therefore right triangles $ABD$ and $AFE$ are similar. Hence \[ \frac{FE}{5}=\frac{\frac{48}{5}}{12}, \] from which it follows that $FE=4$. Consequently $DF=DE-FE=\frac{36}{5}-4=\frac{16}{5}$.

Answer (A): Let $n$ denote a nice number from the given set. An integer $m$ has exactly four divisors if and only if $m=p^3$ or $m=pq$, where $p$ and $q$ (with $p>q$) are prime numbers. In the former case, the sum of the four divisor is equal to $1+p+p^2+p^3$. Note that $1+11+11^2+11^3<2010\le n$ and $1+13+13^2+13^3>2019\ge n$. Therefore we must have $m=pq$ and $n=1+q+p+pq=(1+q)(1+p)$. Because $p$ is odd, $n$ must be an even number. If $q=2$, then $n$ must be divisible by $3$. In the given set only $2010=(1+2)(1+669)$ and $2016=(1+2)(1+671)$ satisfy these requirements. However neither $669$ nor $671$ are prime. If $q$ is odd, then $n$ must be divisible by $4$. In the given set, only $2012$ and $2016$ are divisible by $4$. None of the pairs of factors of $2012$, namely $1\cdot 2012$, $2\cdot 1006$, $4\cdot 503$, gives rise to primes $p$ and $q$. This leaves $2016=(1+3)(1+503)$, which is the only nice number in the given set. Remark: Note that $2016$ is nice in five ways. The other four ways are $(1+7)(1+251)$, $(1+11)(1+167)$, $(1+23)(1+83)$, and $(1+41)(1+47)$.

Answer (E): Expand the set of three-digit positive integers to include integers $N$, $0 \le N \le 99$, with leading zeros appended. Because $\operatorname{lcm}(5^2,6^2,10^2)=900$, such an integer $N$ meets the required condition if and only if $N+900$ does. Therefore $N$ can be considered to be chosen from the set of integers between $000$ and $899$, inclusive. Suppose that the last two digits in order of the base-5 representation of $N$ are $a_1$ and $a_0$. Similarly, suppose that the last two digits of the base-6 representation of $N$ are $b_1$ and $b_0$. By assumption, $2N\equiv a_0+b_0\pmod{10}$, but $N\equiv a_0\pmod{5}$ and so $a_0+b_0\equiv 2N\equiv 2a_0\pmod{10}.$ Thus $a_0\equiv b_0\pmod{10}$ and because $0\le a_0\le 4$ and $0\le b_0\le 5$, it follows that $a_0=b_0$. Because $N\equiv a_0\pmod{5}$, it follows that there is an integer $N_1$ such that $N=5N_1+a_0$. Also, $N\equiv a_0\pmod{6}$ implies that $5N_1+a_0\equiv a_0\pmod{6}$ and so $N_1\equiv 0\pmod{6}$. It follows that $N_1=6N_2$ for some integer $N_2$ and so $N=30N_2+a_0$. Similarly, $N\equiv 5a_1+a_0\pmod{25}$ implies that $30N_2+a_0\equiv 5a_1+a_0\pmod{25}$ and then $N_2\equiv 6N_2\equiv a_1\pmod{5}$. It follows that $N_2=5N_3+a_1$ for some integer $N_3$ and so $N=150N_3+30a_1+a_0$. Once more, $N\equiv 6b_1+a_0\pmod{36}$ implies that $6N_3-6a_1+a_0\equiv 150N_3+30a_1+a_0\equiv 6b_1+a_0\pmod{36}$ and then $N_3\equiv a_1+b_1\pmod{6}$. It follows that $N_3=6N_4+a_1+b_1$ for some integer $N_4$ and so $N=900N_4+180a_1+150b_1+a_0$. Finally, $2N\equiv 10(a_1+b_1)+2a_0\pmod{100}$ implies that $60a_1+2a_0\equiv 360a_1+300b_1+2a_0\equiv 10a_1+10b_1+2a_0\pmod{100}.$ Therefore $5a_1\equiv b_1\pmod{10}$, equivalently, $b_1\equiv 0\pmod{5}$ and $a_1\equiv b_1\pmod{2}$. Conversely, if $N=900N_4+180a_1+150b_1+a_0$, $a_0=b_0$, and $5a_1\equiv b_1\pmod{10}$, then $2N\equiv 60a_1+2a_0=10(a_1+5a_1)+a_0+b_0\equiv 10(a_1+b_1)+(a_0+b_0)\pmod{100}$. Because $0\le a_1\le 4$ and $0\le b_1\le 5$, it follows that there are exactly $5$ different pairs $(a_1,b_1)$, namely $(0,0)$, $(2,0)$, $(4,0)$, $(1,5)$, and $(3,5)$. Each of these can be combined with $5$ different values of $a_0$ ($0\le a_0\le 4$), to determine exactly $25$ different numbers $N$ with the required property.