2018 AMC 10B Problems

The pan and each piece are rectangles; compare their areas

Divide the total area by the area of one $2 \times 2$ piece

The whole pan has area $20 \times 18 = 360$ square inches. Each piece is $2 \times 2 = 4$ square inches. So the number of pieces is $\frac{360}{4} = 90.$ Thus, A is the correct answer.

Distance = rate $\times$ time, and each $30$-minute leg is half an hour

Subtract the distance covered in the first hour from $96$ to get the last leg’s distance

Each leg is half an hour. In the first, Sam drove $60 \cdot \tfrac12 = 30$ miles; in the second, $65 \cdot \tfrac12 = 32.5$ miles. That’s $62.5$ miles so far. That leaves $96 - 62.5 = 33.5$ miles for the last half hour, which is a speed of $\frac{33.5}{\tfrac12} = 67$ mph. Therefore, the answer is D.

Neither multiplication nor the final addition depends on order, so only the way the four digits are split into two pairs matters

List the three ways to split $\{1,2,3,4\}$ into two pairs

Order inside a product doesn’t matter, and neither does the order we add the two products. So all that matters is how the four digits split into two pairs. There are three splits: $1\cdot2 + 3\cdot4 = 14,$ $1\cdot3 + 2\cdot4 = 11,$ and $1\cdot4 + 2\cdot3 = 10.$ That’s $3$ different values. Thus, B is the correct answer.

The three distinct face areas are the pairwise products $XY, XZ, YZ,$ equal to $24, 48, 72$

Multiplying all three gives $(XYZ)^2;$ take the square root, then divide by each face area

The three distinct face areas are the pairwise products $XY = 24,$ $XZ = 48,$ $YZ = 72$ in some order. Multiply all three: $(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,$ so $XYZ = 288.$ Now divide by each face area. We get $Z = \frac{288}{24} = 12,$ $Y = \frac{288}{48} = 6,$ and $X = \frac{288}{72} = 4,$ so $X + Y + Z = 22.$ Therefore, the answer is B.

Count the subsets with no prime and subtract from the total number of subsets

The non-prime elements are $4, 6, 8, 9;$ any subset of just those contains no prime

Count the complement. The set has $2^8 = 256$ subsets total. A subset avoids every prime exactly when it sticks to the non-primes $\{4, 6, 8, 9\},$ and there are $2^4 = 16$ of those. So $256 - 16 = 240$ subsets contain at least one prime. Thus, D is the correct answer.

A third draw is needed exactly when the first two chips still sum to at most $4$

Which pairs of distinct chips sum to $4$ or less? Count them among all ordered first-two-draw outcomes

We need a third draw exactly when the first two chips still sum to $4$ or less. The only such unordered pairs are $\{1,2\}$ and $\{1,3\}.$ Each can be drawn in either order, giving $4$ favorable ordered prefixes.

Imagine that a complete random ordering of all five chips is chosen in advance. Then all $5\cdot4=20$ ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is $\frac{4}{20}=\frac{1}{5}.$ Therefore, the answer is D.

If each small semicircle has radius $r,$ the large semicircle has radius $Nr$

Show $A : B = 1 : (N-1),$ then set this equal to $1 : 18$

Let each small semicircle have radius $r.$ The $N$ diameters cover the big diameter, so the large radius is $Nr.$ Then $A = N \cdot \tfrac12 \pi r^2,$ and the large semicircle has area $\tfrac12 \pi (Nr)^2,$ so the leftover region is $B = \tfrac12 \pi r^2(N^2 - N).$ This gives $A : B = N : N(N-1)$ $= 1 : (N-1).$ Set $N - 1 = 18,$ and $N = 19.$ Thus, D is the correct answer.

Count the toothpicks in an $n$-step staircase; the vertical ones number $(1 + 2 + \cdots + n) + n$

With equal horizontal and vertical counts the total is $n(n+3);$ solve $n(n+3) = 180$

In an $n$-step staircase the vertical toothpicks number $(1 + 2 + \cdots + n) + n$ $= \tfrac{n(n+1)}{2} + n,$ and there are just as many horizontal ones. That’s a total of $n(n+1) + 2n = n(n+3).$ Check: $n = 3$ gives $18,$ as it should. Now solve $n(n+3) = 180.$ This factors as $(n-12)(n+15) = 0,$ so $n = 12.$ Therefore, the answer is C.

Replacing each die value $k$ by $7 - k$ pairs up equally likely outcomes

This replacement sends a total of $s$ to a total of $49 - s$

Replace each die’s value $k$ by $7 - k.$ This pairs up outcomes one-to-one and keeps their probabilities, and it sends a total of $s$ to $7 \cdot 7 - s = 49 - s.$ So the sums $s$ and $49 - s$ are equally likely. The partner of $10$ is $49 - 10 = 39.$ Thus, D is the correct answer.

Place $A$ at the origin with the edges along the axes; find the rectangle $BCHE$ and its area

Volume $= \tfrac13$ $\times (\text{base area})$ $\times (\text{distance from } M \text{ to base})$

Put $A$ at the origin with edges along the axes: $A = (0,0,0),$ $B = (3,0,0),$ $C = (3,1,0),$ $E = (0,0,2),$ $H = (0,1,2),$ $F = (3,0,2),$ $G = (3,1,2),$ so $M = (3, \tfrac12, 2).$ The base $BCHE$ is a rectangle with $BC = 1$ and $BE = \sqrt{3^2 + 2^2} = \sqrt{13},$ hence area $\sqrt{13}.$ Its plane is $2x + 3z = 6,$ and $M$ sits at distance $\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from it. The volume is $\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2.$ Therefore, the answer is E.

Test the expressions modulo $3;$ for a prime $p \ne 3,$ $p^2 \equiv 1 \pmod 3$

For the correct choice, also check $p = 3$ separately, then show it is a multiple of $3$ for all other primes

Look at $p^2 + 26.$ When $p = 3,$ it’s $35 = 5 \cdot 7.$ For any other prime, $p$ isn’t divisible by $3,$ so $p^2 \equiv 1 \pmod 3$ and $p^2 + 26 \equiv 1 + 2 \equiv 0 \pmod 3.$ Either way it’s a multiple of $3$ bigger than $3,$ hence composite. So it’s never prime. Thus, C is the correct answer.

The centroid of $\triangle ABC$ is $\tfrac13(A + B + C),$ and $A + B$ is twice the center

As $C$ moves on a circle of radius $12,$ the centroid moves on a circle of one third the radius

Put the center $O$ at the origin, so $A = (-12, 0)$ and $B = (12, 0),$ while $C$ runs over the circle of radius $12.$ Then $A + B = 0,$ so the centroid is $\tfrac13(A + B + C) = \tfrac13 C.$ As $C$ circles, $\tfrac13 C$ traces a circle of radius $\tfrac{12}{3} = 4$ (minus the two points where $C = A$ or $B$). Its area is $\pi \cdot 4^2 = 16\pi \approx 50.$ Therefore, the answer is C.

The $k$-th term is $10^{k+1} + 1,$ which is divisible by $101$ when $10^{k+1} \equiv -1 \pmod{101}$

$10^2 = 100 \equiv -1 \pmod{101},$ so $10^m \equiv -1$ exactly when $m \equiv 2 \pmod 4$

The $k$-th term is $10^{k+1} + 1,$ which $101$ divides iff $10^{k+1} \equiv -1 \pmod{101}.$ Notice $10^2 = 100 \equiv -1 \pmod{101}.$ So $10^m \equiv -1$ exactly when $m \equiv 2 \pmod 4,$ meaning $k + 1 \equiv 2,$ that is $k \equiv 1 \pmod 4.$ Among $k = 1, 2, \ldots, 2018,$ the values $1, 5, \ldots, 2017$ number $505.$ Thus, C is the correct answer.

Every value other than the mode can appear at most $9$ times, to keep the mode unique

With $d$ distinct values the list holds at most $10 + 9(d-1)$ entries; make this at least $2018$

The mode shows up $10$ times. To keep the number of distinct values small, let every other value repeat as much as the rules allow, which is $9$ times each (any more would tie the mode). With $d$ distinct values the list holds at most $10 + 9(d-1)$ entries. We need $10 + 9(d-1) \ge 2018,$ so $d - 1 \ge 223.1,$ giving $d \ge 225.$ Therefore, the answer is D.

If the sheet has side $s,$ its center-to-corner distance is $\dfrac{s}{\sqrt 2}$

Folding a corner to the top center covers, in a straight line, half the base, then the side of height $h,$ then the other half of the top

Let the sheet have side $s.$ The base sits as a square of side $w$ turned $45^\circ,$ so the center is $\tfrac{w}{2}$ from each base edge. A corner of the sheet lies $\tfrac{s}{\sqrt2}$ from the center. Folding that corner up to the top center traces a straight line: $\tfrac{w}{2}$ out to the base edge, then $h$ up the side, then $\tfrac{w}{2}$ across the top. So $\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h.$ Then $s = \sqrt2\,(w + h),$ and the area is $s^2 = 2(w + h)^2.$ Thus, A is the correct answer.

$n^3 \equiv n \pmod 6,$ because $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers

So the sum of cubes is congruent to $2018^{2018} \pmod 6;$ reduce that modulo $6$

For any integer $n,$ $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers, so it’s divisible by $6.$ That means $n^3 \equiv n \pmod 6.$ Summing, $\sum a_i^3 \equiv \sum a_i$ $= 2018^{2018} \pmod 6.$ Now $2018 \equiv 2 \pmod 6,$ and powers of $2$ mod $6$ alternate $2, 4, 2, 4, \ldots.$ The exponent $2018$ is even, so $2^{2018} \equiv 4 \pmod 6.$ The remainder is $4.$ Therefore, the answer is E.

The four cut corners are congruent right triangles with legs $x$ (on the sides of length $8$) and $y$ (on the sides of length $6$)

The equal sides give $8 - 2x = 6 - 2y = \sqrt{x^2 + y^2};$ the first equality gives $y = x - 1,$ then substitute and square

Let $s$ be the octagon’s side length and let $x=AP=BQ=\frac{8-s}{2}.$ The right triangles $\triangle APH$ and $\triangle BQC$ have the same hypotenuse $s$ and a leg of length $x,$ so they are congruent; write $PH=QC=y.$ Because $CD=HG=s$ and the vertical sides of the rectangle both have length $6,$ it follows that $DR=GS.$ The right triangles at $R$ and $S$ are then congruent, so $RE=SF.$ Since $RS=8$ and $EF=s,$ each of these equal lengths is $\frac{8-s}{2}=x.$ Thus all four cut corners have legs $x$ and $y.$

The equal octagon sides give $$8-2x=6-2y=\sqrt{x^2+y^2}.$$ The first equality gives $y=x-1.$ Substituting and squaring gives $2x^2-30x+63=0,$ so the root with $x<4$ is $x=\frac{15-3\sqrt{11}}{2}.$

The side length is $8-2x=-7+3\sqrt{11},$ so $k+m+n=-7+3+11=7.$ Thus, B is the correct answer.

Show that each row must contain exactly one child from each family

The third row’s families must differ from the second row’s in every column (a derangement), and each pair’s two children can be swapped between their seats

Suppose some family put both children in one row. They’d have to take the non-adjacent seats $1$ and $3,$ which forces the middle family’s two children into the same column. Not allowed. So each row holds exactly one child from each family. The second row is a permutation of the three families, $3! = 6$ ways. The third row needs a different family in every column, a derangement of the second row’s order, and there are $2$ of those. Finally, each pair can swap its two children between their seats, $2^3 = 8$ ways. The total is $6 \cdot 2 \cdot 8 = 96.$ Therefore, the answer is D.

Let Chloe be $n$ today. In $t$ years her age is a multiple of Zoe’s iff $1 + t$ divides $n - 1$

Exactly $9$ such birthdays means $n - 1$ has $9$ divisors; find the two-digit value, then repeat the divisor idea for Joey

Let Chloe be $n$ today; Zoe is $1.$ In $t$ years their age ratio is $$\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t},$$ which is an integer exactly when $1+t$ divides $n-1.$ Thus $n-1$ has exactly $9$ positive divisors.

A number with $9$ divisors has the form $p^8$ or $p^2q^2.$ The only two-digit possibility is $2^2\cdot3^2=36,$ so Chloe is $37$ and Joey is $38.$

Joey’s age $38+t$ is a multiple of Zoe’s age $1+t$ exactly when $1+t$ divides $37.$ The next time is $t=36,$ when Joey is $74.$ Its digit sum is $7+4=11.$ Thus, E is the correct answer.

Look for a particular solution of the form $f(n) = n + c;$ subtracting it leaves $g(n) = g(n-1) - g(n-2)$

That homogeneous recursion is periodic with period $6;$ reduce $2018$ modulo $6$

Notice $f(n) = n + 1$ solves the recurrence on its own, so write $f(n) = (n + 1) + g(n).$ Then $g$ satisfies the homogeneous version $g(n) = g(n-1) - g(n-2).$ With $g(1) = -1$ and $g(2) = -2,$ it cycles with period $6$: $-1, -2, -1, 1, 2, 1, \ldots.$ Since $2018 \equiv 2 \pmod 6,$ we get $g(2018) = -2,$ so $f(2018) = 2019 - 2 = 2017.$ Therefore, the answer is B.

$323 = 17 \cdot 19,$ and $n$ is an even $4$-digit multiple of $323$

For a candidate next divisor $d,$ $n$ must be a multiple of $\operatorname{lcm}(323, d);$ find the smallest $d > 323$ keeping that under $10000$

Let $d$ be the next divisor after $323.$ If $\gcd(d,323)=1,$ then $n$ is a multiple of $323d>323^2>9999,$ impossible for a four-digit number. Thus $\gcd(d,323)>1.$

Since $323=17\cdot19,$ this gcd is at least $17.$ It also divides $d-323,$ so $d-323\ge17$ and hence $d\ge340.$

This bound is attained: for $n=6460=2^2\cdot5\cdot17\cdot19,$ both $323$ and $340$ are divisors. The lower bound shows there is no divisor between them. Thus the smallest possible next divisor is $340,$ and C is the correct answer.

A triangle with sides $x, y, 1$ needs $x + y > 1;$ since $1$ is the longest side it is obtuse when $x^2 + y^2 < 1$

In the unit square this region is a quarter disk with the triangle below $x + y = 1$ removed

The three lengths $x, y, 1$ make a triangle iff $x + y > 1.$ Since $1$ is the longest side, that triangle is obtuse iff $x^2 + y^2 < 1.$ So in the unit square we want the region inside the quarter circle $x^2 + y^2 = 1$ but above the line $x + y = 1.$ That’s the quarter disk with the right triangle under the chord removed: $\tfrac{\pi}{4} - \tfrac12 \approx 0.285.$ The closest choice is $0.29.$ Therefore, the answer is C.

Use $ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b);$ let $x = \operatorname{lcm}(a,b)$ and $y = \gcd(a,b)$

The equation becomes $(x - 12)(y - 20) = 177 = 3 \cdot 59;$ also $y$ must divide $x$

Recall $ab=\gcd(a,b)\operatorname{lcm}(a,b).$ Let $x=\operatorname{lcm}(a,b)$ and $y=\gcd(a,b).$ The equation becomes $$(x-12)(y-20)=177=3\cdot59.$$

The positive factor pairs give $(x,y)=(13,197),$ $(189,21),$ $(15,79),$ and $(71,23).$ The negative factor pairs make either $x$ or $y$ negative, so they are impossible. Also $y$ must divide $x,$ and only $(x,y)=(189,21)$ passes.

Write $a=21u$ and $b=21v.$ Then $\gcd(u,v)=1$ and $uv=\frac{189}{21}=9,$ so $(u,v)=(1,9)$ or $(9,1).$ Hence the two ordered pairs are $(21,189)$ and $(189,21),$ and B is the correct answer.