2016 AMC 10B — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $\dfrac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}?$

$\ 1$

$\ 2$

$\ \dfrac{5}{2}$

$\ 10$

$\ 20$

Answer: D

Concepts:exponent substitution

Difficulty rating: 770

Small Hint:

Substitute $a=\frac{1}{2}$ after combining powers of $a$

Big Hint:

Dividing by $a$ turns $a^{-1}$ into $a^{-2}$

Solution:

The expression is equivalent to $2a^{-2}+\frac{a^{-2}}{2} = 2.5 (a^{-1})^2.$ Then $a^{-1}$ is equal to $\dfrac{1}{\frac 12} = 2,$ so our expression is equal to $2.5\cdot 2^2 = 10.$

Thus, the correct answer is D .

2.

If $n\heartsuit m=n^3m^2,$ what is $\frac{2\heartsuit 4}{4\heartsuit 2}?$

$\ \dfrac{1}{4}$

$\ \dfrac{1}{2}$

$\ 1$

$\ 2$

$\ 4$

Answer: B

Concepts:custom operation exponent

Difficulty rating: 870

Small Hint:

Compute the operation ratio symbolically first

Big Hint:

$\frac{n\heartsuit m}{m\heartsuit n}=\frac nm$

Solution:

For any nonzero $a,b,$ we have $\dfrac{a\heartsuit b}{b\heartsuit a} = \dfrac{a^3b^2}{b^3 a^2} = \dfrac ab .$ Using $a=2,b=4,$ we get that $\frac{ 2}{4} = \frac 12.$

Thus, the correct answer is B .

3.

Let $x=-2016.$ What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x?$

$\ -2016$

$\ 0$

$\ 2016$

$\ 4032$

$\ 6048$

Answer: D

Concepts:absolute value

Difficulty rating: 960

Small Hint:

For $x<0$ , $|x|=-x$

Big Hint:

Evaluate the absolute values from the inside outward

Solution:

Since $x=-2016<0,$ we have $|x|=2016.$ Thus $\begin{aligned} |x|-x&=2016-(-2016) \\ &=4032, \end{aligned}$ so the innermost absolute value is $4032.$ Therefore the full expression is $|4032-2016|+2016=4032.$

Thus, the correct answer is D .

4.

Zoey read $15$ books, one at a time. The first book took her $1$ day to read, the second book took her $2$ days to read, the third book took her $3$ days to read, and so on, with each book taking her $1$ more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day of the week did she finish her $15$ th book?

Sunday

Monday

Wednesday

Friday

Saturday

Answer: B

Concepts:triangular number modular arithmetic

Difficulty rating: 1020

Small Hint:

The total reading time is $1+2+\cdots+15$

Big Hint:

Compare the finish day of book $15$ with the finish day of book $1$ modulo $7$

Solution:

The number of days it takes to read $15$ books is $\dfrac{ 15\cdot 16}{2} =120.$ Therefore, it is $120-1=119$ days after the first book is read. This is a multiple of $7,$ so the $15$ th book was finished on the same day as the $1$ st book. Therefore, it was finished on a Monday.

Thus, the correct answer is B .

5.

The mean age of Amanda’s $4$ cousins is $8,$ and their median age is $5.$ What is the sum of the ages of Amanda’s youngest and oldest cousins?

$\ 13$

$\ 16$

$\ 19$

$\ 22$

$\ 25$

Answer: D

Concepts:mean median (data)

Difficulty rating: 900

Small Hint:

The total age is $4\cdot8$

Big Hint:

The two middle ages sum to $10$

Solution:

The total of the four cousin ages is $4\cdot8=32$ . Since the median is $5$ , the two middle ages have average $5$ , so their sum is $10$ .

Therefore the youngest and oldest ages sum to $32-10=22$ .

Thus, the correct answer is D.

6.

Laura added two three-digit positive integers. All six digits in these numbers are different. Laura’s sum is a three-digit number $S.$ What is the smallest possible value for the sum of the digits of $S?$

$\ 1$

$\ 4$

$\ 5$

$\ 15$

$\ 21$

Answer: B

Concepts:digits extremal argument

Difficulty rating: 960

Small Hint:

Show a digit sum below $4$ is impossible

Big Hint:

$157+243=400$ reaches digit sum $4$

Solution:

To minimize the sum of the addends, use the two smallest nonzero hundreds digits, $1$ and $2$ , and then the four smallest remaining digits, $0,3,4,5$ . Placing the smaller digits in the tens places shows that every possible sum is at least $104+235=339$ . On the other hand, any three-digit number with digit sum less than $4$ is at most $300$ . Therefore the digit sum of $S$ is at least $4$ .

The example $157+243=400$ uses six distinct digits and has digit sum $4$ . Thus the smallest possible digit sum is $4$ .

Thus, the correct answer is B.

7.

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?

$\ 75$

$\ 90$

$\ 135$

$\ 150$

$\ 270$

Answer: C

Concepts:ratio and proportion linear equation

Difficulty rating: 1070

Small Hint:

Let the smaller acute angle be $x$

Big Hint:

The larger angle has the smaller complement

Solution:

Let the smaller angle be $x.$ Then, the larger angle is $\frac 54 x.$ Note that their sum is $\frac 94 x.$

Their complements are then $90-x$ and $90 - \frac 54x.$ Since $90-x$ is larger than $90-\frac 54 x,$ we know $\begin{aligned}90 -x &= 2\left(90 - \frac 54 x\right) \\90-x&= 180-\frac 52 x\\x&=60.\end{aligned}$

Therefore, the sum is $\dfrac 94 \cdot 60=135.$

Thus, the correct answer is C .

8.

What is the tens digit of $2015^{2016}-2017?$

$\ 0$

$\ 1$

$\ 3$

$\ 5$

$\ 8$

Answer: A

Concepts:modular arithmetic Chinese Remainder Theorem

Difficulty rating: 1420

Small Hint:

Work modulo $100$ to get the tens digit

Big Hint:

Use congruences modulo $4$ and $25$

Solution:

To find the tens digit, we first need to find the number modulo $100.$ First, we can find $2015^{2016} \mod 100.$ We can do this with the Chinese Remainder Theorem by first getting the number $\mod 4$ and then $\mod 25.$

The number $2015^{2016} \equiv 0 \mod 25$ since it is a multiple of $25.$

Then, observe that $2015^{2016} \equiv (-1)^{2016}$ $\equiv 1 \mod 4 .$

By the Chinese remainder theorem, we can get that our number is congruent to $25 \mod 100.$

Therefore, $2015^{2016}-2017$ $\equiv 25 - 17 \mod 100$ $\equiv 8 \mod 100 .$ This means the tens digit is $0.$

Thus, the correct answer is A .

9.

All three vertices of $\bigtriangleup ABC$ 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 $BC?$

$\ 4$

$\ 6$

$\ 8$

$\ 10$

$\ 16$

Answer: C

Concepts:parabola triangle area coordinate geometry

Difficulty rating: 1280

Small Hint:

Use symmetric points $(-t,t^2)$ and $(t,t^2)$

Big Hint:

The triangle area is $t^3$

Solution:

Let the points be $A=(0,0),$ $B=(x_1,y_1),$ $C=(x_2,y_2).$

Then, since $BC$ is parallel with the $x$ -axis, we know $y_1=y_2,$ which we will let be $y.$ Then, $x_1^2 = y=x_2^2,$ so $x_1^2=x_2^2 .$ This implies that either $x_1=-x_2$ or $x_1 = x_2.$ The second option cannot happen since that would set two points as the same, which would create an area of $0.$ As such, let $x= -x_1=x_2.$ Then $y=x^2$ also.

Then, the points are $A=(0,0),$ $B=(-x,x^2),$ $C=(x,x^2).$ With a base of $BC,$ the length is $2x$ and the height is $x^2.$ This would make the area $x^3 = 64.$ Therefore, $x=4,$ so $BC = 2\cdot 4 = 8.$

Thus, the correct answer is C .

10.

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?

$\ 14.0$

$\ 16.0$

$\ 20.0$

$\ 33.3$

$\ 55.6$

Answer: D

Difficulty rating: 960

Small Hint:

Weights scale like areas because thickness and density are fixed

Big Hint:

The side-length scale factor is $\frac{5}{3}$

Solution:

The surface area is increased by a factor of $\frac 53 ^2$ since the side lengths are increased by a factor of $\frac 53.$ Also since the thickness is constant, the volume is scaled up by that much. Then, the wood having the same density makes the weight increase by a factor of $\frac 53^2$ as well, so the new weight is $12 \cdot \dfrac {25}{9} = \dfrac{100}3.$ This is approximately $33.3.$

Thus, the correct answer is D .

11.

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?

$\ 256$

$\ 336$

$\ 384$

$\ 448$

$\ 512$

Answer: B

Concepts:system of equations perimeter area fencepost counting

Difficulty rating: 1280

Small Hint:

Let $s$ and $l$ be posts on the short and long sides, including corners

Big Hint:

Use $l=2s$ and $2l+2s-4=20$

Solution:

Let $l$ be the number of posts on a long side, including corners, and let $s$ be the number of posts on a short side, including corners. The total number of posts is $2l+2s-4=20,$ where the $-4$ avoids double-counting the corners.

Since a long side has twice as many posts as a short side, $l=2s$ . Thus $2(2s)+2s-4=20$ , so $s=4$ and $l=8$ .

There are $3$ equal gaps on a short side and $7$ equal gaps on a long side. With $4$ yards between neighboring posts, the side lengths are $12$ and $28$ , so the area is $12\cdot28=336$ .

Thus, the correct answer is B.

12.

Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?

$\ 0.2$

$\ 0.4$

$\ 0.5$

$\ 0.7$

$\ 0.8$

Answer: D

Concepts:basic probability complementary counting parity

Difficulty rating: 960

Small Hint:

Count the odd-product pairs instead

Big Hint:

There are $\binom32$ odd pairs out of $\binom52$ pairs

Solution:

The product is odd if and only if both numbers are odd. There are $\binom 32=3$ ways to do this out of a possible $\binom 52=10$ ways. This makes the probability of it being odd equal to $\frac 3{10} = 0.3.$ This means the probability it is even is $1-0.3=0.7.$

Thus, the correct answer is D .

13.

At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there were three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?

$\ 25$

$\ 40$

$\ 64$

$\ 100$

$\ 160$

Answer: D

Concepts:system of equations

Difficulty rating: 1140

Small Hint:

Let $w,r,q$ be the numbers of twin, triplet, and quadruplet sets

Big Hint:

Use $r=4q$ and $w=3r$

Solution:

Let $w,r,q$ be the numbers of sets of twins, triplets, and quadruplets. Then $2w+3r+4q=1000.$ The statement gives $r=4q$ and $w=3r=12q$ .

Substituting gives $2(12q)+3(4q)+4q$ $=40q=1000$ , so $q=25$ . The number of babies in sets of quadruplets is $4q=100$ .

Thus, the correct answer is D.

14.

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 $x=5.1?$

$\ 30$

$\ 41$

$\ 45$

$\ 50$

$\ 57$

Answer: D

Concepts:lattice point counting shapes in figures casework

Difficulty rating: 1970

Small Hint:

For each square size, count possible top-left lattice points

Big Hint:

Only side lengths $1,2,3$ can fit

Solution:

A square must lie above $y=-0.1$ , to the left of $x=5.1$ , and below $y=\pi x$ . Since $\pi$ is a little more than $3$ , the lattice heights available at $x=1,2,3,4$ are $3,6,9,12$ , and side lengths larger than $3$ cannot fit.

Count by side length using the top-left lattice point. For side length $1$ , there are $3+6+9+12=30$ choices. For side length $2$ , there are $2+5+8=15$ choices. For side length $3$ , there are $1+4=5$ choices.

The total is $30+15+5=50$ .

Thus, the correct answer is D.

15.

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 is the number in the center?

$\ 5$

$\ 6$

$\ 7$

$\ 8$

$\ 9$

Answer: C

Concepts:parity logical deduction

Difficulty rating: 1420

Small Hint:

Adjacent consecutive numbers have opposite parity

Big Hint:

The center and four corners must all have the same parity

Solution:

Color the array like a checkerboard, with the center and four corners one color and the four edge squares the other color. Consecutive numbers occupy adjacent squares, so the path $1,2,\ldots,9$ alternates colors. Therefore all five odd numbers occupy one color class and all four even numbers occupy the other.

The color class consisting of the center and four corners has five squares, so it contains the five odd numbers. Hence the sum of the corners and center is $1+3+5+7+9=25.$

Since the corners have a sum of $18,$ the center has a value of $25-18=7.$

Thus, the correct answer is C .

16.

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 $S?$

$\ \dfrac{1+\sqrt{5}}{2}$

$\ 2$

$\ \sqrt{5}$

$\ 3$

$\ 4$

Answer: E

Concepts:geometric sequence completing the square optimization

Difficulty rating: 1540

Small Hint:

Write the first term as $a$ and common ratio as $r$

Big Hint:

The condition $ar=1$ makes $S=\frac{1}{r(1-r)}$

Solution:

Let the first value of the series be $a,$ and let the ratio be $r.$ Thus, $\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{aligned}$ This means we have to find $r$ that maximizes $r(1-r)= 0.25-(r-0.5)^2.$ This maximization will happen with $r=0.5.$

Therefore, $S = \dfrac{1}{0.5(0.5)}= 4.$

Thus, the correct answer is E .

17.

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?

$\ 312$

$\ 343$

$\ 625$

$\ 729$

$\ 1680$

Answer: D

Concepts:cube geometry optimization AM-GM Inequality

Difficulty rating: 1880

Small Hint:

Pair opposite faces of the cube

Big Hint:

Maximize $(a_1+a_2)(b_1+b_2)(c_1+c_2)$

Solution:

Pair opposite faces as $(a_1,a_2)$ , $(b_1,b_2)$ , and $(c_1,c_2)$ . Each vertex product uses one number from each pair, so the sum of all eight vertex products is $(a_1+a_2)(b_1+b_2)(c_1+c_2).$

The six face labels sum to $2+3+4+5+6+7=27$ , so the three opposite-pair sums have total $27$ . Their product is maximized when the sums are as equal as possible, namely $9,9,9$ , giving at most $9^3=729$ .

This maximum is attainable by pairing $2$ with $7$ , $3$ with $6$ , and $4$ with $5$ . Hence the greatest possible sum is $729$ .

Thus, the correct answer is D.

18.

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

$\ 1$

$\ 3$

$\ 5$

$\ 6$

$\ 7$

Answer: E

Concepts:arithmetic sequence factor counting

Difficulty rating: 1660

Small Hint:

If the length is $s$ , then $s(2x+s-1)=690$

Big Hint:

Check factor lengths that give positive integer $x$

Solution:

Suppose the sequence has length $s$ and first term $x$ . Then $345=s\left(x+\frac{s-1}{2}\right),$ or $s(2x+s-1)=690.$

Thus $s$ must be a divisor of $690$ , with $x=\frac{\frac{690}{s}-s+1}{2}$ a positive integer. Checking the possible divisor lengths gives $s=2,3,5,6,10,15,23.$ These are the only lengths that keep $x$ positive and integral.

Therefore there are $7$ representations.

Thus, the correct answer is E.

19.

Rectangle $ABCD$ has $AB=5$ and $BC=4.$ Point $E$ lies on $\overline{AB}$ so that $EB=1,$ point $G$ lies on $\overline{BC}$ so that $CG=1,$ and point $F$ lies on $\overline{CD}$ so that $DF=2.$ Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P,$ respectively. What is the value of $\dfrac{PQ}{EF}?$

$~\dfrac{\sqrt{3}}{16}$

$~\dfrac{\sqrt{2}}{13}$

$~\dfrac{9}{82}$

$~\dfrac{10}{91}$

$~\dfrac19$

Answer: D

Difficulty rating: 1970

Small Hint:

First find where $AC$ and $AG$ cut $EF$

Big Hint:

Use similar triangles to express $\frac{PF}{EF}$ and $\frac{QF}{EF}$

Solution:

We have $AE=AB-EB=4$ and $FC=DC-DF=3.$ Since $AE\parallel FC,$ the triangles $AEP$ and $CFP$ are similar. Therefore $\frac{PF}{PE}=\frac{FC}{AE}=\frac34,$ so $\frac{PF}{EF}=\frac37.$

Extend $AG$ to meet line $CD$ at $X.$ Because $BG=3,$ similarity gives $\frac{AD}{DX}=\frac{BG}{AB},\qquad \frac4{DX}=\frac35,$ and hence $DX=\frac{20}{3}$ . Thus $FX=DX-DF=\frac{14}{3}.$ Since $AE\parallel FX,$ triangles $AEQ$ and $XFQ$ are similar, so $\frac{QF}{QE}=\frac{FX}{AE}=\frac76.$ Consequently $\frac{QF}{EF}=\frac7{13}.$

It follows that $\begin{aligned} \frac{PQ}{EF}&=\frac{QF}{EF}-\frac{PF}{EF} \\ &=\frac7{13}-\frac37=\frac{10}{91}. \end{aligned}$

Thus, the correct answer is D .

20.

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A'(5,6).$ What distance does the origin $O(0,0),$ move under this transformation?

$\ 0$

$\ 3$

$\ \sqrt{13}$

$\ 4$

$\ 5$

Answer: C

Concepts:homothety coordinate geometry vector

Difficulty rating: 1660

Small Hint:

The dilation scale factor is $\frac{3}{2}$

Big Hint:

The center of dilation lies on the line through the two circle centers

Solution:

The dilation scale factor is $\frac32$ , since the radius changes from $2$ to $3$ . The center of dilation $C$ lies on the line through $A(2,2)$ and $A'(5,6)$ .

The vector from $A$ to $A'$ is $(3,4)$ . If $C$ is the dilation center, then the scale factor gives $A'-A=\frac12(A-C).$ Thus $A-C=2(3,4)$ , so $C=(2,2)-2(3,4)=(-4,-6).$

Under a scale factor $\frac32$ dilation about $C$ , a point moves by half its distance from $C$ . Since $CO=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$ , the origin moves $\frac12\sqrt{52}=\sqrt{13}$ .

Thus, the correct answer is C.

21.

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

$\ \pi+\sqrt{2}$

$\ \pi+2$

$\ \pi+2\sqrt{2}$

$\ 2\pi+\sqrt{2}$

$\ 2\pi+2\sqrt{2}$

Answer: B

Concepts:absolute value circle area decomposition

Difficulty rating: 1860

Small Hint:

Work in one quadrant and multiply by $4$

Big Hint:

The first-quadrant region is a triangle plus a semicircle

Solution:

The equation is symmetric in all four quadrants. In the first quadrant it becomes $x^2+y^2=x+y,$ or $(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.$

In the first quadrant, the enclosed region is the triangle under $x+y=1$ , with area $\frac12$ , plus a semicircle of radius $\sqrt{\frac{1}{2}}$ , with area $\frac\pi4$ .

Multiplying by $4$ , the total area is $4\left(\frac12+\frac\pi4\right)=2+\pi.$

Thus, the correct answer is B.

22.

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 $A?$

$\ 385$

$\ 665$

$\ 945$

$\ 1140$

$\ 1330$

Answer: A

Concepts:graph theory complementary counting combinations

Difficulty rating: 1820

Small Hint:

Count all triples of teams, then subtract transitive triples

Big Hint:

Each team is the top team in $\binom{10}{2}$ transitive triples

Solution:

The total number of teams is $10+10+1=21.$ The total number of sets is therefore $\binom{21}{3} = 1330.$

Now, we must subtract the total number of sets such that there is no cycle. This only happens if one team beats the other two teams. There are $21$ choices for the team that beat the other two and $\binom{10}{2} = 45$ ways to choose the teams they beat. Thus, the total of non-cycles is $21\cdot 45=945.$ This means the total number of cycles is $1330-945=385.$

Thus, the correct answer is A .

23.

In regular hexagon $ABCDEF,$ points $W,$ $X,$ $Y,$ and $Z$ are chosen on sides $\overline{BC},$ $\overline{CD},$ $\overline{EF},$ and $\overline{FA}$ respectively, so lines $AB,$ $ZW,$ $YX,$ and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF?$

$\ \dfrac{1}{3}$

$\ \dfrac{10}{27}$

$\ \dfrac{11}{27}$

$\ \dfrac{4}{9}$

$\ \dfrac{13}{27}$

Answer: C

Concepts:regular polygon area ratio similarity

Difficulty rating: 2300

Small Hint:

Extend $\overline{FE}$ and $\overline{CD}$ until they meet

Big Hint:

Compare the areas of three similar triangles with vertex at the extension point

Solution:

Extend $\overline{FE}$ and $\overline{CD}$ until they meet at $P.$

Let $d$ be the distance between $ZW$ and $FC.$ Because the four given lines are equally spaced and $FC$ lies halfway between $ZW$ and $YX,$ the distance from $ED$ to $ZW$ is $2d.$ Let the altitude from $P$ to $ED$ be $h.$ The equilateral triangles $PED$ and $PFC$ have side lengths in the ratio $1:2,$ so their altitudes satisfy $h+3d=2h,$ giving $h=3d.$ Therefore the side-length ratio of $PZW$ to $PED$ is $\frac{h+2d}{h}=\frac53.$

Taking $[PED]=1,$ similarity gives $[PZW]=\frac{25}{9}$ and $[PFC]=4.$ Hence $\begin{aligned} [ZWCF]&=4-\frac{25}{9}=\frac{11}{9}, \\ [EDCF]&=4-1=3. \end{aligned}$ The desired hexagon consists of two congruent copies of $ZWCF,$ while the regular hexagon consists of two congruent copies of $EDCF.$ Thus the requested ratio is $\frac{\frac{11}{9}}{3}=\frac{11}{27}.$

Thus, the correct answer is C .

24.

How many four-digit positive integers $abcd,$ with $a \neq 0,$ have the property that the three two-digit integers $ab < bc < cd$ form an increasing arithmetic sequence?

One such number is $4692,$ where $a=4,$ $b=6,$ $c=9,$ and $d=2.$

$\ 9$

$\ 15$

$\ 16$

$\ 17$

$\ 20$

Answer: D

Concepts:arithmetic sequence digits casework

Difficulty rating: 2390

Small Hint:

The arithmetic-sequence condition is $2bc=ab+cd$

Big Hint:

The digit equation has two possible carry cases

Solution:

From $ab<bc<cd$ , we have $a\le b\le c.$ The arithmetic-sequence condition is $\begin{aligned} 2(10b+c)&=(10a+b) \\ &\quad +(10c+d), \end{aligned}$ which rearranges to $10(a-2b+c)=-b+2c-d.$ The right side is a multiple of $10.$ Because $b\le c$ and all four symbols are digits with $b\ge1,$ it lies between $-9$ and $17$ . Hence it is either $0$ or $10.$

Case $1:$ $\begin{aligned} -b+2c-d&=10, \\ a&=2b-c+1. \end{aligned}$

We can look at the possible values of $c.$

$c=6:$ $b+d=2, 2b-5=a.$ Thus, $b\leq 2$ from the first equation, but can’t work for the second equation.

$c=7:$ $b+d=4, 2b-6=a.$ Thus, $b\leq 4$ from the first equation, and $b > 3$ from the second equation. This makes one case for $b=4.$

$c=8:$ $b+d=6, 2b-7=a.$ Thus, $b\leq 6$ from the first equation, and $b > 3$ from the second equation. This makes three cases for $b=4,5,6.$

$c=9:$ $b+d=8, 2b-8=a.$ Thus, $b\leq 8$ from the first equation, and $b > 4$ from the second equation. This makes four cases for $b=5,6,7,8.$ Altogether this case gives $\begin{gathered} 2470,1482,3581,5680, \\ 2593,4692,6791,8890, \end{gathered}$ for $8$ solutions.

Case $2:$ $-b+2c-d=0,$ $2b-a-c=0,$ which means the digits are an arithmetic sequence.

If the difference is $1,$ then $1 \leq a \leq 6$ makes $6$ solutions.

If the difference is $2,$ then $1 \leq a \leq 3$ makes $3$ solutions. A difference of at least $3$ would force $d=a+3r>9.$ This case therefore gives $9$ solutions.

In total, the number of solutions is $8+9 = 17.$

Thus, the correct answer is D .

25.

Let $f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor),$ where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r.$ How many distinct values does $f(x)$ assume for $x \ge 0?$

$\ 32$

$\ 36$

$\ 45$

$\ 46$

$\ \text{infinitely many}$

Answer: A

Concepts:floor and ceiling functions Euler’s Totient Function

Difficulty rating: 2440

Small Hint:

Only the fractional part of $x$ matters

Big Hint:

Count distinct breakpoints $\frac{i}{k}$ for $2\le k\le10$

Solution:

Write $x=\lfloor x\rfloor+t$ , where $0\le t\lt1$ . Then $\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor,$ so $f(x)$ depends only on the fractional part $t$ .

The value of $f$ changes only when $t$ crosses a fraction $\frac{i}{k}$ , where $2\le k\le10$ and $1\le i\lt k$ . The number of distinct such fractions in $(0,1)$ is $\begin{aligned} &\varphi(2)+\varphi(3) \\ &\quad {}+\cdots+\varphi(10) \\ &=1+2+2+4+2 \\ &\quad {}+6+4+6+4 \\ &=31. \end{aligned}$

Including the initial value before the first breakpoint, $f$ assumes $31+1=32$ distinct values.

Thus, the correct answer is A.