AMC10 2018 A

Answer (B): Computing inside to outside yields: $\left(\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1\right)=\left(\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1\right)$ $=\left(\left(\frac{7}{4}\right)^{-1}+1\right)$ $=\frac{11}{7}.$ Note: The successive denominators and numerators of numbers obtained from this pattern are the Lucas numbers.

Answer (A): Let $L$, $J$, and $A$ be the amounts of soda that Liliane, Jacqueline, and Alice have, respectively. The given information implies that $L=1.50J=\frac{3}{2}J$ and $A=1.25J=\frac{5}{4}J$, and hence $J=\frac{4}{5}A$. Then \[ L=\frac{3}{2}\cdot\frac{4}{5}A=\frac{6}{5}A=1.20A, \] so Liliane has $20\%$ more soda than Alice.

Answer (E): Converting \(10!\) seconds to days gives \[ \frac{10!}{60\cdot 60\cdot 24} =\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 120}{60\cdot 120\cdot 12} =\frac{9\cdot 8\cdot 7}{12} =42. \] Because 30 days after January 1 is January 31, 42 days after January 1 is February 12.

There are 4 choices for the periods in which the mathematics courses can be taken: periods 1, 3, 5; periods 1, 3, 6; periods 1, 4, 6; and periods 2, 4, 6. Each choice of periods allows 3! = 6 ways to order the 3 mathematics courses. Therefore there are 4 · 6 = 24 ways of arranging a schedule.

Because the statements of Alice, Bob, and Charlie are all incorrect, the actual distance d satisfies d < 6, d > 5, and d > 4. Hence the actual distance lies in the interval (5, 6).

Let N be the number of votes cast. Then 0.65N of them were like votes, and 0.35N of them were dislike votes. The current score for Sangho’s video is then 0.65N − 0.35N = 0.3N = 90. Thus N = 90 ÷ 0.3 = 300.

Answer (E): Because 4000 = $2^5 \cdot 5^3$, $4000 \cdot \left(\frac{2}{5}\right)^n = 2^{5+n} \cdot 5^{3-n}.$ This product will be an integer if and only if both of the factors $2^{5+n}$ and $5^{3-n}$ are integers, which happens if and only if both exponents are nonnegative. Therefore the given expression is an integer if and only if $5+n \ge 0$ and $3-n \ge 0$. The solutions are exactly the integers satisfying $-5 \le n \le 3$. There are $3-(-5)+1=9$ such values.

Answer (C): Let $n$ be the number of 5-cent coins Joe has, and let $x$ be the requested value—the number of 25-cent coins Joe has minus the number of 5-cent coins he has. Then Joe has $(n+3)$ 10-cent coins and $(n+x)$ 25-cent coins. The given information leads to the equations $$ n+(n+3)+(n+x)=23 $$ $$ 5n+10(n+3)+25(n+x)=320. $$ These equations simplify to $3n+x=20$ and $8n+5x=58$. Solving these equations simultaneously yields $n=6$ and $x=2$. Joe has 2 more 25-cent coins than 5-cent coins. Indeed, Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins.

Answer (E): The length of the base $\overline{DE}$ of $\triangle ADE$ is 4 times the length of the base of a small triangle, so the area of $\triangle ADE$ is $4^2 \cdot 1 = 16$. Therefore the area of $DBCE$ is the area of $\triangle ABC$ minus the area of $\triangle ADE$, which is $40 - 16 = 24$.

Answer (A): Let $a=\sqrt{49-x^2}-\sqrt{25-x^2}$ and $b=\sqrt{49-x^2}+\sqrt{25-x^2}$. Then $ab=(49-x^2)-(25-x^2)=24$, so $b=\dfrac{24}{a}=\dfrac{24}{3}=8$.

Answer (E): The only ways to achieve a sum of 10 by adding 7 unordered integers between 1 and 6 inclusive are (i) six 1s and one 4; (ii) five 1s, one 2, and one 3; or (iii) four 1s and three 2s. The number of ways to order the outcomes among the 7 dice are 7 in case (i), $7\cdot 6=42$ in case (ii), and $\binom{7}{3}=35$ in case (iii). There are $6^7$ possible outcomes. Therefore $n=7+42+35=84$.

Answer (C): The graph of the system is shown below. The graph of the first equation is a line with \(x\)-intercept \((3,0)\) and \(y\)-intercept \((0,1)\). To draw the graph of the second equation, consider the equation quadrant by quadrant. In the first quadrant \(x>0\) and \(y>0\), and thus the second equation is equivalent to \(|x-y|=1\), which in turn is equivalent to \(y=x\pm 1\). Its graph consists of the rays with endpoints \((0,1)\) and \((1,0)\), as shown. In the second quadrant \(x<0\) and \(y>0\). The corresponding graph is the reflection of the first quadrant graph across the \(y\)-axis. The rest of the graph can be sketched by further reflections of the first-quadrant graph across the coordinate axes, resulting in the figure shown. There are 3 intersection points: \((-3,2)\), \((0,1)\), and \(\left(\frac{3}{2},\frac{1}{2}\right)\), as shown.

Answer (D): The paper’s long edge $\overline{AB}$ is the hypotenuse of right triangle $ACB$, and the crease lies along the perpendicular bisector of $\overline{AB}$. Because $AC>BC$, the crease hits $\overline{AC}$ rather than $\overline{BC}$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the intersection of $\overline{AC}$ and the line through $D$ perpendicular to $\overline{AB}$. Then the crease in the paper is $\overline{DE}$. Because $\triangle ADE \sim \triangle ACB$, it follows that $\dfrac{DE}{AD}=\dfrac{CB}{AC}=\dfrac{3}{4}$. Thus $$ DE=AD\cdot \frac{CB}{AC}=\frac{5}{2}\cdot \frac{3}{4}=\frac{15}{8}. $$

Answer (A): Because the powers-of-3 terms greatly dominate the powers-of-2 terms, the given fraction should be close to $\frac{3^{100}}{3^{96}}=3^4=81$. Note that $(3^{100}+2^{100})-81(3^{96}+2^{96})=2^{100}-81\cdot 2^{96}=(16-81)\cdot 2^{96}<0$, so the given fraction is less than $81$. On the other hand $(3^{100}+2^{100})-80(3^{96}+2^{96})=3^{96}(81-80)-2^{96}(80-16)=3^{96}-2^{102}$. Because $3^2>2^3$, $3^{96}=(3^2)^{48}>(2^3)^{48}=2^{144}>2^{102}$, it follows that $(3^{100}+2^{100})-80(3^{96}+2^{96})>0$, and the given fraction is greater than $80$. Therefore the greatest integer less than or equal to the given fraction is $80$.

Answer (D): Let $C$ be the center of the larger circle, and let $D$ and $E$ be the centers of the two smaller circles, as shown. Points $C$, $D$, and $A$ are collinear because the radii are perpendicular to the common tangent at the point of tangency, and so are $C$, $E$, and $B$. These points form two isosceles triangles that share a vertex angle. Thus $\triangle CAB \sim \triangle CDE$, and therefore $\dfrac{AB}{DE}=\dfrac{CA}{CD}$, so $$ AB=\dfrac{DE\cdot CA}{CD}=\dfrac{(5+5)\cdot 13}{13-5}=\dfrac{65}{4}, $$ and the requested sum is $65+4=69$.

Answer (D): The area of $\triangle ABC$ is 210. Let $D$ be the foot of the altitude from $B$ to $AC$. By the Pythagorean Theorem, $AC=\sqrt{20^2+21^2}=29$, so $210=\frac12\cdot29\cdot BD$, and $BD=14\frac{14}{29}$. Two segments of every length from 15 through 19 can be constructed from $B$ to $AC$. In addition to these 10 segments and the 2 legs, there is a segment of length 20 from $B$ to a point on $AC$ near $C$, for a total of 13 segments with integer length.

Answer (C): If $1 \in S$, then $S$ can have only $1$ element, not $6$ elements. If $2$ is the least element of $S$, then $2, 3, 5, 7, 9$, and $11$ are available to be in $S$, but $3$ and $9$ cannot both be in $S$, so the largest possible size of $S$ is $5$. If $3$ is the least element, then $3, 4, 5, 7, 8, 10$, and $11$ are available, but at most one of $4$ and $8$ can be in $S$ and at most one of $5$ and $10$ can be in $S$, so again $S$ has size at most $5$. The set $S = \{4,6,7,9,10,11\}$ has the required property, so $4$ is the least possible element of $S$.

Answer (D): Let $S$ be the set of integers, both negative and non-negative, having the given form. Increasing the value of $a_i$ by 1 for $0 \le i \le 7$ creates a one-to-one correspondence between $S$ and the ternary (base 3) representation of the integers from 0 through $3^8-1$, so $S$ contains $3^8=6561$ elements. One of those is 0, and by symmetry, half of the others are positive, so $S$ contains $1+\frac{1}{2}\cdot(6561-1)=3281$ elements.

Answer (E): For $m \in \{11,13,15,17,19\}$, let $p(m)$ denote the probability that $m^n$ has units digit $1$, where $n$ is chosen at random from the set $S=\{1999,2000,2001,\ldots,2018\}$. Then the desired probability is equal to $\frac15\bigl(p(11)+p(13)+p(15)+p(17)+p(19)\bigr)$. Because any positive integral power of $11$ always has units digit $1$, $p(11)=1$, and because any positive integral power of $15$ always has units digit $5$, $p(15)=0$. Note that $S$ has $20$ elements, exactly $5$ of which are congruent to $j \bmod 4$ for each of $j=0,1,2,3$. The units digits of powers of $13$ and $17$ cycle in groups of $4$. More precisely, \[ (13^k \bmod 10)_{k=1999}^{2018}=(7,1,3,9,7,1,\ldots,3,9) \] and \[ (17^k \bmod 10)_{k=1999}^{2018}=(3,1,7,9,3,1,\ldots,7,9). \] Thus $p(13)=p(17)=\frac{5}{20}=\frac14$. Finally, note that the units digit of $19^k$ is $1$ or $9$, according to whether $k$ is even or odd, respectively. Thus $p(19)=\frac12$. Hence the requested probability is \[ \frac15\left(1+\frac14+0+\frac14+\frac12\right)=\frac25. \]

Answer (B): None of the squares that are marked with dots in the sample scanning code shown below can be mapped to any other marked square by reflections or non-identity rotations. Therefore these 10 squares can be arbitrarily colored black or white in a symmetric scanning code, with the exception of “all black” and “all white”. On the other hand, reflections or rotations will map these squares to all the other squares in the scanning code, so once these 10 colors are specified, the symmetric scanning code is completely determined. Thus there are $2^{10}-2=1022$ symmetric scanning codes.

Answer (E): Solving the second equation for $x^2$ gives $x^2=y+a$, and substituting into the first equation gives $y^2+y+(a-a^2)=0$. The polynomial in $y$ can be factored as $(y+(1-a))(y+a)$, so the solutions are $y=a-1$ and $y=-a$. (Alternatively, the solutions can be obtained using the quadratic formula.) The corresponding equations for $x$ are $x^2=2a-1$ and $x^2=0$. The second equation always has the solution $x=0$, corresponding to the point of tangency at the vertex of the parabola $y=x^2-a$. The first equation has 2 solutions if and only if $a>\frac12$, corresponding to the 2 symmetric intersection points of the parabola with the circle. Thus the two curves intersect at 3 points if and only if $a>\frac12$.

Answer (D): Because $\gcd(a,b)=24=2^3\cdot 3$ and $\gcd(b,c)=36=2^2\cdot 3^2$, it follows that $a$ is divisible by $2$ and $3$ but not by $3^2$. Similarly, because $\gcd(b,c)=2^2\cdot 3^2$ and $\gcd(c,d)=54=2\cdot 3^3$, it follows that $d$ is divisible by $2$ and $3$ but not by $2^2$. Therefore $\gcd(d,a)=2\cdot 3\cdot n$, where $n$ is a product of primes that do not include $2$ or $3$. Because $70<\gcd(d,a)<100$ and $n$ is an integer, it must be that $12\le n\le 16$, so $n=13$, and $13$ must also be a divisor of $a$. The conditions are satisfied if $a=2^3\cdot 3\cdot 13=312$, $b=2^3\cdot 3^2=72$, $c=2^2\cdot 3^3=108$, and $d=2\cdot 3^3\cdot 13=702$.

Answer (D): Let the triangle's vertices in the coordinate plane be (4,0), (0,3), and (0,0), with $[0,s]\times[0,s]$ representing the unplanted portion of the field. The equation of the hypotenuse is $3x+4y-12=0$, so the distance from $(s,s)$, the corner of $S$ closest to the hypotenuse, to this line is given by $$ \frac{|3s+4s-12|}{\sqrt{3^2+4^2}}. $$ Setting this equal to 2 and solving for $s$ gives $s=\frac{22}{7}$ and $s=\frac{2}{7}$, and the former is rejected because the square must lie within the triangle. The unplanted area is thus $\left(\frac{2}{7}\right)^2=\frac{4}{49}$, and the requested fraction is $$ 1-\frac{\frac{4}{49}}{\frac{1}{2}\cdot 4\cdot 3}=\frac{145}{147}. $$

Answer (D): Because $AB$ is $\frac{5}{6}$ of $AB+AC$, it follows from the Angle Bisector Theorem that $DF$ is $\frac{5}{6}$ of $DE$, and $BG$ is $\frac{5}{6}$ of $BC$. Because trapezoids $FDBG$ and $EDBC$ have the same height, the area of $FDBG$ is $\frac{5}{6}$ of the area of $EDBC$. Furthermore, the area of $\triangle ADE$ is $\frac{1}{4}$ of the area of $\triangle ABC$, so its area is $30$, and the area of trapezoid $EDBC$ is $120-30=90$. Therefore the area of quadrilateral $FDBG$ is $\frac{5}{6}\cdot 90=75$. Note: The figure (not drawn to scale) shows the situation in which $\angle ACB$ is acute. In this case $BC\approx 59.0$ and $\angle BAC\approx 151^\circ$. It is also possible for $\angle ACB$ to be obtuse, with $BC\approx 41.5$ and $\angle BAC\approx 29^\circ$. These values can be calculated using the Law of Cosines and the sine formula for area.

Answer (D): The equation $C_n-B_n=A_n^2$ is equivalent to $$ c\cdot\frac{10^{2n}-1}{9}-b\cdot\frac{10^n-1}{9} = a^2\left(\frac{10^n-1}{9}\right)^2. $$ Dividing by $10^n-1$ and clearing fractions yields $$ (9c-a^2)\cdot 10^n = 9b-9c-a^2. $$ As this must hold for two different values $n_1$ and $n_2$, there are two such equations, and subtracting them gives $$ (9c-a^2)\left(10^{n_1}-10^{n_2}\right)=0. $$ The second factor is non-zero, so $9c-a^2=0$ and thus $9b-9c-a^2=0$. From this it follows that $c=\left(\frac{a}{3}\right)^2$ and $b=2c$. Hence digit $a$ must be 3, 6, or 9, with corresponding values 1, 4, or 9 for $c$, and 2, 8, or 18 for $b$. The case $b=18$ is invalid, so there are just two triples of possible values for $a$, $b$, and $c$, namely $(3,2,1)$ and $(6,8,4)$. In fact, in these cases, $C_n-B_n=A_n^2$ for all positive integers $n$; for example, $4444-88=4356=66^2$. The second triple has the greater coordinate sum, $6+8+4=18$.