AMC10 2018 B

The total area of cornbread is 20 · 18 = 360 in². Because each piece of cornbread has area 2·2 = 4 in², the pan contains 360 ÷ 4 = 90 pieces of cornbread.

Answer (D): Sam covered $\frac{1}{2}\cdot 60 = 30$ miles during the first 30 minutes and $\frac{1}{2}\cdot 65 = 32.5$ miles during the second 30 minutes, so he needed to cover $96 - 30 - 32.5 = 33.5$ miles during the last 30 minutes. Thus his average speed during the last 30 minutes was $\dfrac{33.5\ \text{miles}}{\frac{1}{2}\ \text{hour}} = 67\ \text{mph}.$

Answer (B): Both the multiplications and the addition can be performed in either order, so each possible value can be obtained by putting the 1 in the first position and one of the other three numbers in the second position. Therefore the only possible values are $(1 \times 2) + (3 \times 4) = 14,$ $(1 \times 3) + (2 \times 4) = 11,$ and $(1 \times 4) + (2 \times 3) = 10,$ so just 3 different values can be obtained.

Answer (B): Without loss of generality, assume that $X \le Y \le Z$. Then the geometric description of the problem can be translated into the system of equations, $XY=24$, $XZ=48$, and $YZ=72$. Dividing the second equation by the first yields $\frac{Z}{Y}=2$, so $Z=2Y$. Then $72=YZ=2Y^2$, so $Y^2=36$. Because $Y$ is positive, $Y=6$. It follows that $X=24 \div 6=4$ and $Z=72 \div 6=12$, so $X+Y+Z=22$.

Answer (D): The number of qualifying subsets equals the difference between the total number of subsets of $\{2,3,4,5,6,7,8,9\}$ and the number of such subsets containing no prime numbers, which is the number of subsets of $\{4,6,8,9\}$. A set with $n$ elements has $2^n$ subsets, so the requested number is $2^8-2^4=256-16=240$.

Answer (D): Three draws will be required if and only if the first two chips drawn have a sum of 4 or less. The draws $(1,2)$, $(2,1)$, $(1,3)$, and $(3,1)$ are the only draws meeting this condition. There are $5\cdot 4=20$ possible two-chip draws, so the requested probability is $\frac{4}{20}=\frac{1}{5}$. (Note that all 20 possible two-chip draws are considered in determining the denominator, even though some draws will end after the first chip is drawn.)

Answer (D): Suppose without loss of generality that each small semicircle has radius 1; then the large semicircle has radius $N$. The area of each small semicircle is $\frac{\pi}{2}$, and the area of the large semicircle is $N^2\cdot\frac{\pi}{2}$. The combined area $A$ of the $N$ small semicircles is $N\cdot\frac{\pi}{2}$, and the area $B$ inside the large semicircle but outside the small semicircles is $N^2\cdot\frac{\pi}{2}-N\cdot\frac{\pi}{2}=(N^2-N)\cdot\frac{\pi}{2}$. Thus the ratio $A:B$ of the areas is $N:(N^2-N)$, which is $1:(N-1)$. Because this ratio is given to be $1:18$, it follows that $N-1=18$ and $N=19$.

Answer (C): In the staircase with $n$ steps, the number of vertical toothpicks is $1+2+3+\cdots+n+n=\dfrac{n(n+1)}{2}+n.$ There are an equal number of horizontal toothpicks, for a total of $n(n+1)+2n$ toothpicks. Solving $n(n+1)+2n=180$ with $n>0$ yields $n=12$.

Answer (D): Without loss of generality, one can assume that the numbers on opposite faces of each die add up to 7. In other words, the 1 is opposite the 6, the 2 is opposite the 5, and the 3 is opposite the 4. (In fact, standard dice are numbered in this way.) The top faces give a sum of 10 if and only if the bottom faces give a sum of 7·7 − 10 = 39. By symmetry, the probability that the top faces give a sum of 39 is also p. The distribution of the outcomes of the dice rolls has the bell-shaped graph shown below, so no other outcome has the same probability as 10 and 39.

Answer (E): The volume of the rectangular pyramid with base $BCHE$ and apex $M$ equals the volume of the given rectangular parallelepiped, which is $6$, minus the combined volume of triangular prism $AEHDCB$, tetrahedron $BEFM$, and tetrahedron $CGHM$. Tetrahedra $BEFM$ and $CGHM$ each have three right angles at $F$ and $G$, respectively, and the edges of the tetrahedra emanating from $F$ and $G$ have lengths $2$, $3$, and $\frac{1}{2}$, so the volume of each of these tetrahedra is $\frac{1}{6}\cdot\left(2\cdot 3\cdot \frac{1}{2}\right)=\frac{1}{2}$. The volume of the triangular prism $AEHDCB$ is $3$ because it is half the volume of the rectangular parallelepiped. Therefore the requested volume is $6-3-\frac{1}{2}-\frac{1}{2}=2$.

Answer (C): If $p=3$, then $p^2+26=35=5\cdot 7$. If $p$ is a prime number other than 3, then $p=3k\pm 1$ for some positive integer $k$. In that case $$ p^2+26=(3k\pm 1)^2+26=9k^2\pm 6k+27=3(3k^2\pm 2k+9) $$ is a multiple of 3 and is not prime. The smallest counterexamples for the other choices are $5^2+16=41$, $7^2+24=73$, $5^2+46=71$, and $19^2+96=457$.

Answer (C): Let $O$ be the center of the circle. Triangle $ABC$ is a right triangle, and $O$ is the midpoint of the hypotenuse $AB$. Then $\overline{OC}$ is a radius, and it is also one of the medians of the triangle. The centroid is located one third of the way along the median from $O$ to $C$, so the centroid traces out a circle with center $O$ and radius $\frac{1}{3}\cdot 12=4$ (except for the two missing points corresponding to $C=A$ or $C=B$). The area of this smaller circle is then $\pi\cdot 4^2=16\pi \approx 16\cdot 3.14 \approx 50$.

Answer (C): The numbers in the given sequence are of the form $10^n+1$ for $n=2,3,\ldots,2019$. If $n$ is even, say $n=2k$ for some positive integer $k$, then $10^n+1=100^k+1\equiv(-1)^k+1\pmod{101}$. Thus $10^n+1$ is divisible by $101$ if and only if $k$ is odd, which means $n=2,6,10,\ldots,2018$. There are $\frac14(2018-2)+1=505$ such values. On the other hand, if $n$ is odd, say $n=2k+1$ for some positive integer $k$, then $$ 10^n+1=10\cdot10^{n-1}+1=10\cdot100^k+1\equiv10\cdot(-1)^k+1\pmod{101}, $$ which is congruent to $9$ or $11$, and $10^n+1$ is not divisible by $101$ in this case.

Answer (D): The list has $2018-10=2008$ entries that are not equal to the mode. Because the mode is unique, each of these 2008 entries can occur at most 9 times. There must be at least $\left\lceil\frac{2008}{9}\right\rceil=224$ distinct values in the list that are different from the mode, because if there were fewer than this many such values, then the size of the list would be at most $9\cdot\left(\left\lceil\frac{2008}{9}\right\rceil-1\right)+10=2017<2018$. (The ceiling function notation $\lceil x\rceil$ represents the least integer greater than or equal to $x$.) Therefore the least possible number of distinct values that can occur in the list is 225. One list satisfying the conditions of the problem contains 9 instances of each of the numbers 1 through 223, 10 instances of the number 224, and one instance of 225.

Answer (A): The figure shows that the distance $AO$ from a corner of the wrapping paper to the center is $$ \frac{w}{2}+h+\frac{w}{2}=w+h. $$ The side of the wrapping paper, $AB$ in the figure, is the hypotenuse of a $45-45-90^\circ$ right triangle, so its length is $\sqrt{2}\cdot AO=\sqrt{2}(w+h)$. Therefore the area of the wrapping paper is $$ \left(\sqrt{2}(w+h)\right)^2=2(w+h)^2. $$

Answer (E): Let $n$ be an integer. Because $n^3-n=(n-1)n(n+1)$, it follows that $n^3-n$ has at least one prime factor of $2$ and one prime factor of $3$ and therefore is divisible by $6$. Thus $n^3\equiv n\pmod 6$. Then $$ a_1^3+a_2^3+\cdots+a_{2018}^3\equiv a_1+a_2+\cdots+a_{2018}\equiv 2018^{2018}\pmod 6. $$ Because $2018\equiv 2\pmod 6$, the powers of $2018$ modulo $6$ are alternately $2,4,2,4,\ldots$, so $2018^{2018}\equiv 4\pmod 6$. Therefore the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$ is $4$.

Answer (B): Because $AP<4=\frac12 PQ$, it follows that $A$ is closer to $P$ than it is to $Q$ and that $A$ is between points $P$ and $B$. Because $AP=BQ$, $AH=BC$, and angles $APH$ and $BQC$ are right angles, $\triangle APH\cong\triangle BQC$. Thus $PH=QC$, and $PQCH$ is a rectangle. Because $CD=HG$, it follows that $HCDG$ is also a rectangle. Thus $GDRS$ is a rectangle and $DR=GS$, and it follows that $\triangle ERD\cong\triangle FSG$. Therefore segment $EF$ is centered in $RS$ just as congruent segment $AB$ is centered in $PQ$. Therefore $\triangle ERD\cong\triangle BQC$, and $CD$ is also centered in $QR$. Let $2x$ be the side length $AB=BC=CD=DE=EF=FG=GH=HA$ of the regular octagon; then $AP=BQ=4-x$ and $QC=RD=3-x$. Applying the Pythagorean Theorem to $\triangle BQC$ yields $(4-x)^2+(3-x)^2=(2x)^2$, which simplifies to $2x^2+14x-25=0$. Thus $x=\frac12\cdot(-7\pm3\sqrt{11})$, and because $x>0$, it follows that $2x=-7+3\sqrt{11}$. Hence $k+m+n=-7+3+11=7$.

Answer (D): Let $X$, $Y$, and $Z$ denote the three different families in some order. Then the only possible arrangements are to have the second row be members of $XYZ$ and the third row be members of $ZXY$, or to have the second row be members of $XYZ$ and the third row be members of $YZX$. Note that these are not the same, because in the first case one sibling pair occupy the right-most seat in the second row and the left-most seat in the third row, whereas in the second case this does not happen. (Having members of $XYX$ in the second row does not work because then the third row must be members of $ZYZ$ to avoid consecutive members of $Z$; but in this case one of the $Y$ siblings would be seated directly in front of the other $Y$ sibling.) In each of these 2 cases there are $3! = 6$ ways to assign the families to the letters and $2^3 = 8$ ways to position the boy and girl within the seats assigned to the families. Therefore the total number of seating arrangements is $2 \cdot 6 \cdot 8 = 96$.

Answer (E): Let Chloe be $n$ years old today, so she is $n-1$ years older than Zoe. For integers $y \ge 0$, Chloe’s age will be a multiple of Zoe’s age $y$ years from now if and only if $$ \frac{n+y}{1+y}=1+\frac{n-1}{1+y} $$ is an integer, that is, $1+y$ is a divisor of $n-1$. Thus $n-1$ has exactly 9 positive integer divisors, so the prime factorization of $n-1$ has one of the two forms $p^2q^2$ or $p^8$. There are no two-digit integers of the form $p^8$, and the only one of the form $p^2q^2$ is $2^2 \cdot 3^2 = 36$. Therefore Chloe is 37 years old today, and Joey is 38. His age will be a multiple of Zoe’s age in $y$ years if and only if $1+y$ is a divisor of $38-1=37$. The nonnegative integer solutions for $y$ are 0 and 36, so the only other time Joey’s age will be a multiple of Zoe’s age will be when he is $38+36=74$ years old. The requested sum is $7+4=11$.

Answer (B): Applying the recursion for several steps leads to the conjecture that $$ f(n)= \begin{cases} n+2 & \text{if } n\equiv 0\pmod{6},\\ n & \text{if } n\equiv 1\pmod{6},\\ n-1 & \text{if } n\equiv 2\pmod{6},\\ n & \text{if } n\equiv 3\pmod{6},\\ n+2 & \text{if } n\equiv 4\pmod{6},\\ n+3 & \text{if } n\equiv 5\pmod{6}. \end{cases} $$ The conjecture can be verified using the strong form of mathematical induction with two base cases and six inductive steps. For example, if $n\equiv 2\pmod{6}$, then $n=6k+2$ for some nonnegative integer $k$ and $$ \begin{aligned} f(n) &= f(6k+2)\\ &= f(6k+1)-f(6k)+6k+2\\ &= (6k+1)-(6k+2)+6k+2\\ &= 6k+1\\ &= n-1. \end{aligned} $$ Therefore $f(2018)=f(6\cdot 336+2)=2018-1=2017.$

Answer (C): Let $d$ be the next divisor of $n$ after $323$. Then $\gcd(d,323)\ne 1$, because otherwise $n\ge 323d>323^2>100^2=10000$, contrary to the fact that $n$ is a $4$-digit number. Therefore $d-323\ge \gcd(d,323)>1$. The prime factorization of $323$ is $17\cdot 19$. Thus the next divisor of $n$ is at least $323+17=340=17\cdot 20$. Indeed, $340$ will be the next number in Mary’s list when $n=17\cdot 19\cdot 20=6460$.

Answer (C): The set of all possible ordered pairs $(x,y)$ occupies the unit square $0\le x\le 1,\ 0\le y\le 1$ in the Cartesian plane. The numbers $x$, $y$, and $1$ are the side lengths of a triangle if and only if $x+y>1$, which means that $(x,y)$ lies above the line $y=1-x$. By a generalization of the Pythagorean Theorem, the triangle is obtuse if and only if, in addition, $x^2+y^2<1^2$, which means that $(x,y)$ lies inside the circle of radius $1$ centered at the origin. Within the unit square, the region inside the circle of radius $1$ centered at the origin has area $\frac{\pi}{4}$, and the region below the line $y=1-x$ has area $\frac{1}{2}$. Therefore the ordered pairs that meet the required conditions occupy a region with area $\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4}$. The area of the unit square is $1$, so the required probability is also $\frac{\pi-2}{4}\approx \frac{1.14}{4}=0.285$, which is closest to $0.29$.

Answer (B): Recall that $a\cdot b=\gcd(a,b)\cdot\operatorname{lcm}(a,b)$. Let $x=\operatorname{lcm}(a,b)$ and $y=\gcd(a,b)$. The given equation is then $xy+63=20x+12y$, which can be rewritten as $$(x-12)(y-20)=240-63=177=3\cdot 59=1\cdot 177.$$ Because $x$ and $y$ are integers, one of the following must be true: - $x-12=1$ and $y-20=177$, - $x-12=177$ and $y-20=1$, - $x-12=3$ and $y-20=59$, - $x-12=59$ and $y-20=3$. Therefore $(x,y)$ must be $(13,197)$, $(189,21)$, $(15,79)$, or $(71,23)$. Because $x$ must be a multiple of $y$, only $(x,y)=(189,21)$ is possible. Therefore $\gcd(a,b)=21=7\cdot 3$, and $\operatorname{lcm}(a,b)=189=7\cdot 3^3$. Both $a$ and $b$ are divisible by $7$ but not by $7^2$; one of $a$ and $b$ is divisible by $3$ but not $3^2$, and the other is divisible by $3^3$ but not $3^4$; and neither is divisible by any other prime. Therefore one of them is $7\cdot 3=21$ and the other is $7\cdot 3^3=189$. There are 2 ordered pairs, $(a,b)=(21,189)$ and $(a,b)=(189,21)$.

Answer (C): Let $O$ be the center of the regular hexagon. Points $B, O, E$ are collinear and $BE=BO+OE=2$. Trapezoid $FABE$ is isosceles, and $XZ$ is its midline. Hence $XZ=\frac{3}{2}$ and analogously $XY=ZY=\frac{3}{2}$. Denote by $U_1$ the intersection of $\overline{AC}$ and $XZ$ and by $U_2$ the intersection of $\overline{AC}$ and $XY$. It is easy to see that $\triangle AXU_1$ and $\triangle U_2XU_1$ are congruent $30-60-90$ right triangles. By symmetry the area of the convex hexagon enclosed by the intersection of $\triangle ACE$ and $\triangle XYZ$, shaded in the figure, is equal to the area of $\triangle XYZ$ minus $3$ times the area of $\triangle U_2XU_1$. The hypotenuse of $\triangle U_2XU_1$ is $XU_2=AX=\frac{1}{2}$, so the area of $\triangle U_2XU_1$ is $$ \frac{1}{2}\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{2}\right)^2=\frac{1}{32}\sqrt{3}. $$ The area of the equilateral triangle $XYZ$ with side length $\frac{3}{2}$ is equal to $\frac{1}{4}\sqrt{3}\cdot\left(\frac{3}{2}\right)^2=\frac{9}{16}\sqrt{3}$. Hence the area of the shaded hexagon is $$ \frac{9}{16}\sqrt{3}-3\cdot\frac{1}{32}\sqrt{3} =3\sqrt{3}\left(\frac{3}{16}-\frac{1}{32}\right) =\frac{15}{32}\sqrt{3}. $$

Answer (C): Let $\{x\}=x-[x]$ denote the fractional part of $x$. Then $0\le \{x\}<1$. The given equation is equivalent to $x^2=10{,}000\{x\}$, that is, \[ \frac{x^2}{10{,}000}=\{x\}. \] Therefore if $x$ satisfies the equation, then \[ 0\le \frac{x^2}{10{,}000}<1. \] This implies that $x^2<10{,}000$, so $-100<x<100$. The figure shows a sketch of the graphs of \[ f(x)=\frac{x^2}{10{,}000}\qquad \text{and}\qquad g(x)=\{x\} \] for $-100<x<100$ on the same coordinate axes. The graph of $g$ consists of the 200 half-open line segments with slope 1 connecting the points $(k,0)$ and $(k+1,1)$ for $k=-100,-99,\ldots,98,99$. (The endpoints of these intervals that lie on the $x$-axis are part of the graph, but the endpoints with $y$-coordinate 1 are not.) It is clear that there is one intersection point for $x$ lying in each of the intervals $[-100,-99)$, $[-99,-98)$, $[-98,-97)$, $\ldots$, $[-1,0)$, $[0,1)$, $[1,2)$, $\ldots$, $[97,98)$, $[98,99)$ but no others. Thus the equation has 199 solutions.