2016 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

What is the value of

$\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$

when $a=\tfrac12?$

$1$

$2$

$\dfrac52$

$10$

$20$

Concepts:exponent order of operations

Difficulty rating: 920

Small Hint:

$a^{-1}=\frac{1}{a},$ so $a^{-1}=2$ when $a=\tfrac12$

Big Hint:

The numerator is $2a^{-1}+\tfrac12 a^{-1}=\tfrac52 a^{-1};$ then divide by $a$

Solution:

With $a=\tfrac12,$ we have $a^{-1}=2.$ The numerator is $2\cdot2+\dfrac{2}{2}=4+1=5,$ and dividing by $a=\tfrac12$ gives $\dfrac{5}{\frac{1}{2}}=10.$

Thus, the correct answer is D.

2.

The harmonic mean of two numbers can be computed as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?

$2$

$45$

$504$

$1008$

$2015$

Concepts:harmonic mean fraction estimation

Difficulty rating: 1020

Small Hint:

Twice the product over the sum: $\dfrac{2\cdot1\cdot2016}{1+2016}$

Big Hint:

$\dfrac{4032}{2017}$ is just a hair under a whole number; round it

Solution:

The harmonic mean is $\dfrac{2\cdot1\cdot2016}{1+2016}=\dfrac{4032}{2017}.$ Since $\dfrac{2016}{2017}$ is very close to $1,$ this is just under $2,$ so the closest integer is $2.$

Thus, the correct answer is A.

3.

Let $x=-2016.$ What is the value of

$\Big|\;\big|\,|x|-x\,\big|-|x|\;\Big|-x?$

$-2016$

$0$

$2016$

$4032$

$6048$

Concepts:absolute value order of operations

Difficulty rating: 1130

Small Hint:

$|x|=2016,$ and the innermost part is $|x|-x=2016-(-2016)$

Big Hint:

Simplify outward: $|x|-x=4032,$ then $\big|4032\big|-|x|,$ then the outer absolute value, and finally $-x$

Solution:

Since $x=-2016,$ $|x|=2016.$ The innermost expression is $|x|-x=2016+2016=4032.$ Then $\big|4032\big|-|x|=4032-2016$ $=2016,$ and the outer absolute value leaves $2016.$ Finally subtracting $x$ gives $2016-(-2016)=4032.$

Thus, the correct answer is D.

4.

The ratio of the measures of two acute angles is $5:4,$ and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?

$75$

$90$

$135$

$150$

$270$

Concepts:ratio and proportion system of equations

Difficulty rating: 1200

Small Hint:

Let the angles be $\alpha\lt\beta$ with $\dfrac{\beta}{\alpha}=\dfrac54$

Big Hint:

The complements are $90-\alpha$ and $90-\beta;$ set $90-\alpha=2(90-\beta)$

Solution:

Let the angles be $\alpha\lt\beta$ with $\beta=\tfrac54\alpha.$ The larger complement belongs to the smaller angle, so $90-\alpha=2(90-\beta)=180-\tfrac52\alpha.$ This gives $\tfrac32\alpha=90,$ so $\alpha=60^\circ$ and $\beta=75^\circ.$ The sum is $135^\circ.$

Thus, the correct answer is C.

5.

The War of $1812$ started with a declaration of war on Thursday, June $18,$ $1812.$ The peace treaty to end the war was signed $919$ days later, on December $24,$ $1814.$ On what day of the week was the treaty signed?

Friday

Saturday

Sunday

Monday

Tuesday

Concepts:modular arithmetic date and time

Difficulty rating: 1200

Small Hint:

Days of the week repeat every $7$ days, so reduce $919$ modulo $7$

Big Hint:

$919=7\cdot131+2,$ so count $2$ days forward from Thursday

Solution:

Because $919=7\cdot131+2,$ the treaty was signed $131$ full weeks plus $2$ days after Thursday. Two days beyond Thursday is Saturday.

Thus, the correct answer is B.

6.

All three vertices of $\triangle ABC$ lie on the parabola defined by $y=x^2,$ with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64.$ What is the length of $BC?$

$4$

$6$

$8$

$10$

$16$

Concepts:parabola triangle area coordinate geometry

Difficulty rating: 1350

Small Hint:

Let a top vertex be $(x,x^2);$ by symmetry the other is $(-x,x^2)$

Big Hint:

The base is $BC=2x$ and the height is $x^2,$ so $\tfrac12(2x)(x^2)=64$

Solution:

Let the vertex in the first quadrant be $(x,x^2).$ By symmetry the base is $BC=2x$ and the height is $x^2,$ so $\tfrac12\cdot2x\cdot x^2=x^3=64.$ Thus $x=4$ and $BC=2x=8.$

Thus, the correct answer is C.

7.

Josh writes the numbers $1,$ $2,$ $3,$ $\ldots,$ $99,$ $100.$ He marks out $1,$ skips the next number $(2),$ marks out $3,$ and continues skipping and marking out the next number to the end of his list. Then he goes back to the start of his list, marks out the first remaining number $(2),$ skips the next number $(4),$ marks out $6,$ skips $8,$ marks out $10,$ and so on to the end. Josh continues in this manner until only one number remains. What is that number?

$13$

$32$

$56$

$64$

$96$

Concepts:power of 2 pattern recognition

Difficulty rating: 1440

Small Hint:

After the first pass only the multiples of $2$ remain; after the next, only the multiples of $4$

Big Hint:

Each pass keeps the multiples of the next power of $2;$ find the highest power of $2$ that is at most $100$

Solution:

The first pass removes the odd numbers, leaving the multiples of $2.$ The second pass removes $2,6,10,\ldots,$ leaving the multiples of $4.$ In general, after the $n$ th pass only the multiples of $2^n$ remain. The surviving number is the highest power of $2$ not exceeding $100,$ which is $2^6=64.$

Thus, the correct answer is D.

8.

A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?

$14.0$

$16.0$

$20.0$

$33.3$

$55.6$

Concepts:power scaling of length, area, and volume ratio and proportion

Difficulty rating: 1350

Small Hint:

The weight is proportional to the area, which scales with the square of the side length

Big Hint:

Multiply $12$ by $\left(\dfrac53\right)^2$

Solution:

Weight is proportional to area, and area scales with the square of the side length. The second side is $\tfrac53$ times the first, so its weight is $12\cdot\left(\dfrac53\right)^2=\dfrac{100}{3}\approx33.3$ ounces.

Thus, the correct answer is D.

9.

Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?

$256$

$336$

$384$

$448$

$512$

Concepts:fencepost counting system of equations rectangle perimeter

Difficulty rating: 1440

Small Hint:

Let the shorter side have $x$ posts and the longer side $2x$ posts; the four corner posts are shared

Big Hint:

Total posts $2x+2(2x)-4=20;$ a side with $k$ posts spans $(k-1)\cdot4$ yards

Solution:

Let the shorter side have $x$ posts, so the longer side has $2x.$ Counting all posts and subtracting the four corners counted twice, $2x+2(2x)-4=20,$ giving $x=4.$ The shorter side has $4$ posts, or $(4-1)\cdot4=12$ yards, and the longer side has $8$ posts, or $(8-1)\cdot4=28$ yards. The area is $12\cdot28=336.$

Thus, the correct answer is B.

10.

A quadrilateral has vertices $P(a,b),$ $Q(b,a),$ $R(-a,-b),$ and $S(-b,-a),$ where $a$ and $b$ are integers with $a\gt b\gt0.$ The area of $PQRS$ is $16.$ What is $a+b?$

$4$

$5$

$6$

$12$

$13$

Concepts:coordinate geometry rectangle difference of squares

Difficulty rating: 1500

Small Hint:

Check the slopes: $\overline{PQ}$ and $\overline{RS}$ have slope $-1,$ while $\overline{QR}$ and $\overline{PS}$ have slope $1,$ so $PQRS$ is a rectangle

Big Hint:

Its sides are $(a-b)\sqrt2$ and $(a+b)\sqrt2,$ so the area is $2(a^2-b^2)=16$

Solution:

The sides $\overline{PQ}$ and $\overline{RS}$ have slope $-1,$ and $\overline{QR}$ and $\overline{PS}$ have slope $1,$ so $PQRS$ is a rectangle with sides $(a-b)\sqrt2$ and $(a+b)\sqrt2.$ Its area is $2(a-b)(a+b)=2(a^2-b^2)$ $=16,$ so $a^2-b^2=8.$ The only perfect squares differing by $8$ are $9$ and $1,$ giving $a=3,$ $b=1,$ and $a+b=4.$

Thus, the correct answer is A.

11.

How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x,$ the line $y=-0.1,$ and the line $x=5.1?$

$30$

$41$

$45$

$50$

$57$

Concepts:counting shapes in figures lattice point casework

Difficulty rating: 1630

Small Hint:

The region sits below $y=\pi x;$ note $3\lt\pi\lt4,$ $6\lt2\pi\lt7,$ and $9\lt3\pi\lt10$

Big Hint:

In each vertical strip $k\le x\le k+1,$ count the $1\times1,$ $2\times2,$ and $3\times3$ squares that fit below the line, then add them up

Solution:

A square whose left edge is $x=k$ must fit below the lowest point of $y=\pi x$ over its width, namely $y=\pi k.$ For side length $1,$ the possible left edges $k=1,2,3,4$ contribute $3+6+9+12=30$ squares. For side length $2,$ the left edges $k=1,2,3$ contribute $2+5+8=15,$ and for side length $3,$ the left edges $k=1,2$ contribute $1+4=5.$ A square of side at least $4$ cannot fit, so the total is $30+15+5=50.$

Thus, the correct answer is D.

12.

All the numbers $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18.$ What number is in the center?

$5$

$6$

$7$

$8$

$9$

Concepts:parity logical deduction

Difficulty rating: 1590

Small Hint:

Color the grid like a checkerboard; consecutive numbers must lie on opposite colors, so they alternate

Big Hint:

The five same-colored cells (four corners plus center) must hold the five odd numbers, which sum to $25$

Solution:

Color the grid like a checkerboard so the four corners and the center share one color. Since consecutive numbers occupy adjacent (opposite colored) squares, the numbers alternate parity along the chain, so the five same-colored cells contain the five odd numbers $1,3,5,7,9,$ which sum to $25.$ The four corners add to $18,$ so the center is $25-18=7.$

Thus, the correct answer is C.

13.

Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice’s position and $60^\circ$ from Bob’s position. Which of the following is closest to the airplane’s altitude, in miles?

$3.5$

$4$

$4.5$

$5$

$5.5$

Concepts:special right triangle Pythagorean Theorem 3D geometry

Difficulty rating: 1630

Small Hint:

Let the plane be above point $D$ on the ground at height $h.$ Triangles $ACD$ and $BCD$ are $30$ - $60$ - $90$ right triangles

Big Hint:

$AD=\sqrt3\,h$ and $BD=\dfrac{h}{\sqrt3};$ then $AD^2+BD^2=10^2$

Solution:

Let the airplane be at $C,$ directly above point $D$ on the ground at altitude $h.$ Triangles $ACD$ and $BCD$ are $30$ - $60$ - $90$ right triangles, so $AD=\sqrt3\,h$ and $BD=\dfrac{h}{\sqrt3}.$ Since Alice looks north and Bob looks west, $\angle ADB=90^\circ,$ so $AD^2+BD^2=AB^2=100.$ Then $3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100,$ giving $h=\sqrt{30}\approx5.48,$ closest to $5.5.$

Thus, the correct answer is E.

14.

The sum of an infinite geometric series is a positive number $S,$ and the second term in the series is $1.$ What is the smallest possible value of $S?$

$\dfrac{1+\sqrt5}{2}$

$2$

$\sqrt5$

$3$

$4$

Concepts:geometric sequence quadratic optimization

Difficulty rating: 1730

Small Hint:

With ratio $r$ and second term $1,$ the first term is $\frac{1}{r},$ so $S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}$

Big Hint:

$S$ is smallest when $r-r^2$ is largest; maximize the downward parabola $r-r^2$

Solution:

Let $r$ be the common ratio. Since the second term is $1,$ the first term is $\dfrac1r,$ so $S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}.$ Convergence requires $|r|\lt1,$ and $S\gt0$ then forces $0\lt r\lt1.$ Therefore $S$ is smallest when $r-r^2$ is largest. The parabola $r-r^2$ peaks at $r=\tfrac12,$ where it equals $\tfrac14,$ so the smallest value of $S$ is $\dfrac{1}{\frac{1}{4}}=4.$

Thus, the correct answer is E.

15.

All the numbers $2,$ $3,$ $4,$ $5,$ $6,$ $7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

$312$

$343$

$625$

$729$

$1680$

Concepts:factoring AM-GM Inequality

Difficulty rating: 1800

Small Hint:

Group the three pairs of opposite faces as $(a,b),(c,d),(e,f)$

Big Hint:

The sum of the eight vertex products factors as $(a+b)(c+d)(e+f);$ with a fixed total, a product is largest when the factors are equal

Solution:

Pair the opposite faces as $(a,b),(c,d),(e,f).$ Each vertex product uses one face from each pair, so the sum of all eight products factors as $(a+b)(c+d)(e+f).$ The three factors have fixed total $2+3+4+5+6+7=27,$ and a product with fixed sum is largest when the factors are equal, at $9$ each. This balance is achievable with $(2,7),(3,6),(4,5),$ giving $9\cdot9\cdot9=729.$

Thus, the correct answer is D.

16.

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$1$

$3$

$5$

$6$

$7$

Concepts:arithmetic sequence factor counting casework

Difficulty rating: 1800

Small Hint:

A run of $k$ consecutive integers equals $k$ times its median, and $345=3\cdot5\cdot23$

Big Hint:

For odd $k,$ the median is an integer factor; for even $k,$ the median is a half-integer. Count both while keeping all terms positive

Solution:

A sum of consecutive integers equals the count times the median. For an odd number of terms, the median is an integer divisor of $345,$ giving runs of $3$ (median $115$ ), $5$ (median $69$ ), $15$ (median $23$ ), and $23$ (median $15$ ) terms. For an even number of terms the median is a half-integer. The positive possibilities have lengths $2,6,10,$ with respective medians $172.5,57.5,34.5.$ Longer divisor-based runs would force a nonpositive first term. This gives $4+3=7$ ways.

Thus, the correct answer is E.

17.

In $\triangle ABC$ shown in the figure, $AB=7,$ $BC=8,$ $CA=9,$ and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB},$ respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P,$ respectively. What is $PQ?$

$1$

$\dfrac58\sqrt3$

$\dfrac45\sqrt2$

$\dfrac{8}{15}\sqrt5$

$\dfrac65$

Concepts:angle bisector theorem altitude Pythagorean Theorem

Difficulty rating: 1910

Small Hint:

Let $BH=x.$ Then $AH^2=7^2-x^2=9^2-(8-x)^2,$ which gives $x=2$ and $AH=\sqrt{45}$

Big Hint:

By the angle bisector theorem, $\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96$ and $\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72;$ then $PQ=AQ-AP$

Solution:

Let $x=BH.$ Then $CH=8-x,$ and from the two right triangles $AH^2=7^2-x^2=9^2-(8-x)^2.$ This gives $x=2$ and $AH=\sqrt{45}.$ By the angle bisector theorem in $\triangle ACH,$ $\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96,$ so $AP=\dfrac35 AH.$ Similarly in $\triangle ABH,$ $\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72,$ so $AQ=\dfrac79 AH.$ Then $\begin{aligned} PQ &= AQ-AP \\ &= \left(\dfrac79-\dfrac35\right)AH \\ &= \dfrac{8}{45}\sqrt{45} \\ &= \dfrac{8}{15}\sqrt5. \end{aligned}$

Thus, the correct answer is D.

18.

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$

$\pi+\sqrt2$

$\pi+2$

$\pi+2\sqrt2$

$2\pi+\sqrt2$

$2\pi+2\sqrt2$

Concepts:completing the square symmetry circle area

Difficulty rating: 1990

Small Hint:

The graph is symmetric about both axes; in the first quadrant the equation is $x^2+y^2=x+y$

Big Hint:

Complete the square to $\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2 =\tfrac12;$ in that quadrant the region is a right triangle plus a semicircle

Solution:

By symmetry, consider the first quadrant, where the equation is $x^2+y^2=x+y,$ or $\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2=\tfrac12.$ This is a circle centered at $\left(\tfrac12,\tfrac12\right)$ passing through $(1,0)$ and $(0,1);$ since the center is the midpoint of that chord, the enclosed first-quadrant region is the right triangle with legs to $(1,0)$ and $(0,1)$ (area $\tfrac12$ ) plus a semicircle of radius $\dfrac{\sqrt2}{2}$ (area $\dfrac\pi4$ ). Multiplying by $4$ for all quadrants gives $4\left(\tfrac12+\tfrac\pi4\right)=\pi+2.$

Thus, the correct answer is B.

19.

Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?

$\dfrac18$

$\dfrac17$

$\dfrac16$

$\dfrac14$

$\dfrac13$

Concepts:geometric distribution independent events

Difficulty rating: 1910

Small Hint:

A single player gets his first head on flip $n$ with probability $\left(\tfrac12\right)^n$

Big Hint:

All three stop on flip $n$ with probability $\left(\tfrac18\right)^n;$ sum this geometric series over $n\ge1$

Solution:

A player’s first head comes on flip $n$ with probability $\left(\tfrac12\right)^n.$ All three stopping on the same flip $n$ has probability $\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n.$ Summing over $n\ge1,$ $\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17.$

Thus, the correct answer is B.

20.

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A,B,C\}$ were there in which $A$ beat $B,$ $B$ beat $C,$ and $C$ beat $A?$

$385$

$665$

$945$

$1140$

$1330$

Concepts:graph theory complementary counting combinations

Difficulty rating: 2110

Small Hint:

Each team plays $20$ others, so there are $21$ teams and $\binom{21}{3}=1330$ triples in all

Big Hint:

A non-cyclic triple has exactly one team that beats the other two; count those and subtract from the total

Solution:

Since each team won $10$ and lost $10,$ there are $21$ teams and $\binom{21}{3}=1330$ triples. A triple is not cyclic exactly when one team beats both others. Choosing that team ( $21$ ways) and $2$ of the $10$ teams it beat gives $21\cdot\binom{10}{2}=21\cdot45=945$ non-cyclic triples. Thus the cyclic triples number $1330-945=385.$

Thus, the correct answer is A.

21.

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}.$ For $i=1,$ $2,$ $\ldots,$ let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD},$ and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}.$ What is

$\sum_{i=1}^{\infty}\text{Area of }\triangle DQ_iP_i?$

$\dfrac16$

$\dfrac14$

$\dfrac13$

$\dfrac12$

$1$

Difficulty rating: 2210

Small Hint:

Similar triangles give the recursion $DQ_{i+1}=\dfrac{DQ_i}{1+DQ_i},$ and $DQ_1=\tfrac12$ leads to $DQ_i=\dfrac{1}{i+1}$

Big Hint:

The area of $\triangle DQ_iP_i$ is $\tfrac12\cdot\dfrac{1}{i+1}$ $\cdot\dfrac{1}{i+2}$ $=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right);$ the sum telescopes

Solution:

Place $D=(0,0),$ $C=(1,0),$ $B=(1,1),$ $A=(0,1),$ and let $q_i=DQ_i.$ Intersecting line $AQ_i$ with $\overline{BD}$ (the line $y=x$ ) gives $P_i$ with both coordinates $\dfrac{q_i}{1+q_i},$ so $q_{i+1}=\dfrac{q_i}{1+q_i}.$ From $q_1=\tfrac12$ this yields $q_i=\dfrac{1}{i+1}.$ The base of $\triangle DQ_iP_i$ is $DQ_i=\dfrac{1}{i+1}$ and its height is the $y$ -coordinate of $P_i,$ which is $q_{i+1}=\dfrac{1}{i+2}.$ Then $\begin{gathered} \text{Area of }\triangle DQ_iP_i=\tfrac12\cdot\dfrac{1}{i+1} \\ \quad{}\cdot\dfrac{1}{i+2} \\ {}=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right). \end{gathered}$ Summing telescopes to $\tfrac12\cdot\tfrac12=\tfrac14.$

Thus, the correct answer is B.

22.

For a certain positive integer $n$ less than $1000,$ the decimal equivalent of $\dfrac1n$ is $0.\overline{abcdef},$ a repeating decimal of period $6,$ and the decimal equivalent of $\dfrac{1}{n+6}$ is $0.\overline{wxyz},$ a repeating decimal of period $4.$ In which interval does $n$ lie?

$[1,200]$

$[201,400]$

$[401,600]$

$[601,800]$

$[801,999]$

Concepts:repeating decimal multiplicative order divisibility

Difficulty rating: 2270

Small Hint:

Period $6$ means $10^6-1=3^3\cdot7\cdot11\cdot13\cdot37$ is divisible by $n;$ period $4$ means $10^4-1=3^2\cdot11\cdot101$ is divisible by $n+6$

Big Hint:

$n+6$ must divide $10^4-1$ but not $10^2-1,$ forcing $n+6=101k;$ test each $n=101k-6\lt1000$ to see whether $10^6-1$ is divisible by $n$

Solution:

Period $6$ requires $10^6-1=3^3\cdot7\cdot11\cdot13\cdot37$ to be divisible by $n.$ Period $4$ requires $10^4-1=3^2\cdot11\cdot101$ to be divisible by $n+6,$ while $10^2-1=3^2\cdot11$ is not divisible by $n+6$ (else the period would be $1$ or $2$ ). Hence $n+6$ is a multiple of $101.$ Since $n+6$ also divides $3^2\cdot11\cdot101$ and is less than $1006,$ the only possibilities are $n+6=101,303,909,$ giving $n=95,297,903.$ Only $297=3^3\cdot11$ divides $10^6-1,$ so $n=297.$

Finally, $10^6\equiv1\pmod{297},$ while $10^2\not\equiv1$ and $10^3\not\equiv1\pmod{297},$ so its period is exactly $6.$ Also $303$ divides $10^4-1$ but not $10^2-1,$ so the period of $\frac{1}{303}$ is exactly $4.$ Thus $n=297$ lies in $[201,400].$

Thus, the correct answer is B.

23.

What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+|z-1|\le1?$

$\dfrac16$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$1$

Concepts:3D geometry volume power scaling of length, area, and volume

Difficulty rating: 2270

Small Hint:

The set $|x|+|y|+|z|\le1$ is a regular octahedron with diagonals of length $2$ and volume $\tfrac43$

Big Hint:

The second inequality is that octahedron translated up by $1;$ their overlap is a similar octahedron with half the linear size

Solution:

The region $|x|+|y|+|z|\le1$ is a regular octahedron with vertices at $(\pm1,0,0),(0,\pm1,0),(0,0,\pm1),$ whose volume is $2\cdot\tfrac13\cdot(\sqrt2)^2\cdot1=\tfrac43.$ The second region is the same octahedron shifted up by $1.$ Their intersection is bounded by another regular octahedron with diagonals of length $1,$ half the linear dimensions of the first, so its volume is $\left(\tfrac12\right)^3\cdot\tfrac43=\tfrac16.$

Thus, the correct answer is A.

24.

There are exactly $77{,}000$ ordered quadruples $(a,b,c,d)$ such that $\gcd(a,b,c,d)=77$ and $\text{lcm}(a,b,c,d)=n.$ What is the smallest possible value of $n?$

$13{,}860$

$20{,}790$

$21{,}560$

$27{,}720$

$41{,}580$

Concepts:greatest common divisor least common multiple prime factorization

Difficulty rating: 2550

Small Hint:

Divide everything by $77.$ With $m=\frac{n}{77},$ you need reduced quadruples of gcd $1$ and lcm $m,$ analyzed one prime at a time

Big Hint:

For a prime with maximum exponent $M,$ the count of exponent quadruples is $2(6M^2+1).$ Use the factorization of $77000$ to determine how many primes divide $m$ and which values of $M$ can occur

Solution:

Writing each entry as $77$ times a reduced value, we need $\gcd=1$ and $\text{lcm}=m=\frac{n}{77}.$ For each prime $p$ dividing $m$ with maximum exponent $M,$ the number of valid exponent quadruples is $(M+1)^4-2M^4$ $+(M-1)^4$ $=2(6M^2+1).$ The total over all primes must equal $77{,}000=2^3\cdot5^3\cdot7\cdot11.$ Since $2(6M^2+1)$ equals $14,$ $50,$ and $110$ for $M=1,2,3,$ and $14\cdot50\cdot110=77{,}000,$ the exponents $1,2,3$ give one candidate factorization.

Every prime contributes exactly one factor of $2,$ so exactly three primes divide $m.$ Their odd factors $6M^2+1$ must divide $\frac{77{,}000}{2^3}=9625.$ Checking the divisors gives $M=1,2,3,8,$ with odd factors $7,25,55,385.$ The choice $M=8$ leaves only $25$ for the product of the other two odd factors, but each is at least $7,$ so this is impossible. Therefore the maximum exponents are exactly $1,2,3.$ To minimize $m=\frac{n}{77},$ assign the largest exponent to the smallest prime: $m=2^3\cdot3^2\cdot5=360,$ so $n=77\cdot360=27{,}720.$

Thus, the correct answer is D.

25.

The sequence $(a_n)$ is defined recursively by $a_0=1,$ $a_1=\sqrt[19]{2},$ and $a_n=a_{n-1}a_{n-2}^2$ for $n\ge2.$ What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?

$17$

$18$

$19$

$20$

$21$

Concepts:recursion modular arithmetic multiplicative order

Difficulty rating: 2650

Small Hint:

Write $a_n=2^{\frac{b_n}{19}}.$ Then $b_0=0,$ $b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$

Big Hint:

The product is an integer exactly when $b_1+b_2+\cdots+b_k$ is divisible by $19;$ solve $b_n=\tfrac13(2^n-(-1)^n)$ and use the order of $2$ modulo $19$

Solution:

Write $a_n=2^{\frac{b_n}{19}}.$ The recursion becomes $b_0=0,$ $b_1=1,$ $b_n=b_{n-1}+2b_{n-2},$ solved by $b_n=\tfrac13\bigl(2^n-(-1)^n\bigr).$ The product $a_1\cdots a_k$ is an integer exactly when $b_1+\cdots+b_k$ is divisible by $19.$ Summing the formula for $b_n$ gives $b_1+\cdots+b_k=\dfrac{2^{k+1}-1}{3}$ when $k$ is odd, and $b_1+\cdots+b_k=\dfrac{2^{k+1}-2}{3}$ when $k$ is even.

The order of $2$ modulo $19$ is $18$ because $2^9\equiv-1\pmod{19}$ and $2^6\not\equiv1\pmod{19}.$ For odd $k,$ divisibility therefore requires $k+1$ to be divisible by $18,$ first occurring at $k=17.$ For even $k,$ it requires $k$ to be divisible by $18,$ first occurring at $k=18.$ Hence the smallest positive $k$ is $17.$

Thus, the correct answer is A.