### 2016 AMC 10B Exam Problems

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Used with permission of the Mathematical Association of America.

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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

###### Solution(s):

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

###### Solution(s):

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

###### Solution(s):

Observe that: $\begin{align*} &\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x\\ &=\Bigg\vert \Big\vert -x-x\Big\vert-|x|\Bigg\vert-x \\ &=\Bigg\vert\Big\vert -2x\Big\vert-|x|\Bigg\vert-x \end{align*}$ since $x < 0$ implies that $-x = |x|.$ Substituting values, we can see that $|4032-2016|-(-2016)$$= 2016 +2016$$= 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 the week did she finish her $15$th book?

Sunday

Monday

Wednesday

Friday

Saturday

###### Answer: B

###### Solution(s):

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

###### Solution(s):

The sum of everyone's age is $4\cdot 4 = 32.$ The median being $5$ means that the mean of the middle two cousins is $5.$ This makes the sum of the middle two cousins $2+5=10.$ Therefore, the sum of the oldest and youngest is $32-10=22.$

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

###### Solution(s):

As $300 \lt S \lt 1000,$ we know that the sum of the digits must be at least $4.$ This can be seen in $157+243=400.$

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

###### Solution(s):

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 $\begin{align*}90 -x &= 2\left(90 - \frac 54 x\right) \\90-x&= 180-\frac 52 x\\x&=60.\end{align*}$

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

###### Solution(s):

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

###### Solution(s):

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 of $5$ 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

###### Solution(s):

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 volume 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

###### Solution(s):

Let the number of posts on the longer side be $l$ and let the number of posts on the shorter side be $s.$ Then, $l+s+l+s-4 = 20$ since we take out the corners. Thus, $2l+2s = 6s = 24,$ making $s=4,l=8.$ Thus, on the shorter side, the corners are $3$ posts away and on the longers side, they are $7$ posts away. The shorter side is then $3\cdot 4=12,$ and the longer side is $7\cdot 4=28.$ Thus, the area is $12\cdot 28=336.$

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

###### Solution(s):

The product is odd if an 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 was 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

###### Solution(s):

Let the number of sets of twins, triplets, and quadruplets be $w,r,$ and $q$ respectively. We must find $4q$ which is $4$ times the number of sets of children. Then, we know $2w+3r+4q=1000$ as this is the number of total babies. Also, we have $r = 4q$ and $w=3r$ by the next statments. Thus, $w=24q.$ By substitution, we have $40q=1000,$ so $4q=100,$ which is our answer.

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

###### Solution(s):

The intersection of $x=5.1$ and the other two lines have an $x$-value of $5.1.$ The intersection of $y=\pi x$ and $y = -0.1$ has an $x$-value of $\dfrac{-0.1}\pi,$ which is greater than $-1.$

Thus, the $x$ value of some point on the square must be between $0$ and $5$ inclusive. Within this bound, $\pi x < -0.1,$ so $-0.1 < y < \pi x.$ This makes the largest points with the largest $y$-value given an $x$-value is $(0,0),$$(1,3),$$(2,6),$$(3,9),$$(4,12),$$(5,15).$

Now that we know the bounds of the triangle, we shall count the number of squares by looking at the top left corner of the square. We shall case on the size of the square.

If it is a $1 \times 1$ square, then the $y$ value must be greater than or equal to $1$ and the $x$ value must be less than or equal to $4.$ The number of possible values for this is $3+6+9+12=30.$

If it is a $2 \times 2$ square, then the $y$ value must be greater than or equal to $2$ and the $x$ value must be less than or equal to $3.$ The number of possible values for this is $2+5+8=15.$

If it is a $3 \times 3$ square, then the $y$ value must be greater than or equal to $3$ and the $x$ value must be less than or equal to $2.$ The number of possible values for this is $1+4=5.$

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

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

###### Solution(s):

We firstly claim that everything that is either in the center or a corner is odd. This is because every number is next to a consecutive number and therefore has the opposite parity as it.

Therefore, all of the points that are the center or a corner are the same parity. The are $5$ such points, but only $4$ even numbers, so their parity isn't even. Thus, they are all odd. This means the sum of the corners plus the 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

###### Solution(s):

Let the first value of the series be $a,$ and let the ratio be $r.$ Thus, $\begin{align*}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{align*}$ 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

###### Solution(s):

Suppose that the pairs of opposite sides are labelled $(a_1,a_2),$$(b_1,b_2),$$(c_1,c_2).$ For the numbers on the same side as $a_1,$ the total is $a_1(b_1c_1+b_1c_2 + b_2c_2+b_2c_1)$$= a_1(b_1+b_2)(c_1+c_2).$ It is similar on the other side, so the total is: $(a_1+a_2)(b_1+b_2)(c_1+c_2).$ Then, $9 = \dfrac{(a_1+a_2)(b_1+b_2)(c_1+c_2)}3$$\geq ((a_1+a_2)(b_1+b_2)(c_1+c_2))^{\frac 13}$ by AM-GM, so our value is at most $9^3=729.$ Also, if each number $x$ is opposite to $9-x,$ we have the product as $729,$ which is a possible value. This makes it, by definition, the maximum.

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: B

###### Solution(s):

Let the smallest number be $x.$ Then, let the size of the sequence be $s.$ This makes the largest integer $x+s-1.$

Thus, the average of every number is $x + \dfrac{s-1}2.$ This makes the sum of all the numbers $s\left(x + \dfrac{s-1}2\right)=345.$ This is equivalent to $s(2x+s-1)=690,$ or equivalently: $2x+s-1=\dfrac{690}s$

As $x$ and $s$ are integers, we know that in order for the left hand side of the equation to be an integer, $s$ must be a factor of $690.$

Observing the $16$ factors of $690,$ observe that all factors strictly less than $30$ must work, as $\dfrac{345}{30}=11.5,$ suggesting that $11$ and $12$ are the middle numbers in the sequence. However, a sequence of $30$ integers with the middle numbers of $11$ and $12$ must have negative numbers at the start of the sequence. However, this isn't allowed!

As such, all sequences of length $0\lt s\lt 30$ work that are factors of $690.$ These are: $2,3,5,6,10,15,23$

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

###### Solution(s):

Observe that the value $AE = AB-EB$$= 5-1$$=4.$ Also, the value $FC = DC-DF$$= 5-2$$=3.$ Then, since $AE \mid \mid FC,$ we know $\triangle AEP \sim \triangle CFP$ by angle angle symmetry. Thus, $\dfrac{PF}{PE} = \dfrac{FC}{EA} = \dfrac 34 .$ This makes $\dfrac{PF}{EF} = \dfrac{3}{3+4} = \dfrac 37.$

Then, let $X$ be the extension of $AG$ to meet $CD.$ Since two sides are parallel, we have $\dfrac{AD}{DX} = \dfrac{BG}{AB}.$ Thus, $\dfrac{4}{DX} = \dfrac{3}{5}$ $DX = \dfrac{20}3.$ This makes $FX = \dfrac{20}3 - 2 = \dfrac{14}3 .$ Then, since $AE \mid \mid FX,$ we know $\triangle AEQ \sim \triangle XFQ$ by angle angle symmetry. Therefore, $\dfrac{QF}{QE} = \dfrac{FX}{EA} = \dfrac {\frac{14}{3}}4 = \dfrac 76.$ This makes $\dfrac{QF}{EF} = \dfrac{7}{7+6} = \dfrac 7{13}.$

As such, $\dfrac{QP}{EF} = \dfrac{QF}{EF} - \dfrac{PF}{EF}$$= \dfrac{7}{13} - \frac 37$$= \dfrac{10}{91}.$

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

###### Solution(s):

The line between the centers must be on the center of dialation as the center is dialated. Also, the dialation factor is $1.5$ since the radius goes from $2$ to $3.$ Let the center of dialation be $C.$ Then, $CA:CA' = 1:1.5.$ Thus, $CA:AA'= 1:0.5= 2:1.$

Therefore, we can go in the direction of the vector $A'A$ twice from $A.$ The vector means we go down $4$ and left $3.$ Doing this twice from $A$ is $(-4,-6).$

Then, $CO:CO' = 1:1.5$ since it dialates, meaning $CO:OO' = 1:0.5.$ This means the distance moved is the length of $CO$ divided by $2,$ which is equivalent to $\dfrac{\sqrt{52}}2 = \sqrt{13}.$

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

###### Solution(s):

$|x|^2+|y|^2 =x^2+y^2$$=|x|+|y|,$ so $(|x|-0.5)^2+(|y|-0.5)^2 = 0.5$ This means the graph is symmetric over all quandrants, so we could find the area of the region in just the first quadrant and multiply it by $4.$

Thus, we could look at the area of $(x-0.5)^2+(y-0.5)^2 = 0.5$ with $x,y \geq 0.$

If $x+y< 1,$ then we have a triangle with base and height of $1,$ making the area $0.5.$

Since the line $x+y=1$ goes through the center, the region $x+y \geq 1$ creates a semicircle of radius $\sqrt{0.5}.$ Thus, the area is $\dfrac{\pi r^2}2 = \dfrac{\pi} 4.$

This means the total area in each quadrant is $0.5 + \dfrac{\pi}{4}.$ Therefore, the total area is $4\left(0.5 + \dfrac{\pi}{4}\right) = 2 + \pi.$

Thus, the correct answer is **C**.

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

###### Solution(s):

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

###### Solution(s):

Consider the following diagram:

The shape is symmetric, so we can find the ratio of the areas of $EFCD$ to the area of $ZFCW.$ For this, we can extend $FE$ and $CD$ until they meet each other, and let this point be $P.$

Also, the distance from $ED$ to $ZW$ is the same as the distance from $ZW$ to $YX,$ and the distance from $ZW$ to $YX$ is twice the distance as the distance from $ZW$ to $FC.$

Thus, the distance from $ED$ to $ZW$ is twice the distance as the distance from $ZW$ to $FC.$ Let the altitude from $P$ to $ED$ be $s.$ Then, if the altitude from $P$ to $ZW$ is $s+d,$ the altitude from $P$ to $CF$ is $s+1.5d$ because of the ratio. Then, since $\triangle FPC \sim \triangle PED,$$\dfrac{FP}{EP} = \dfrac{s+1.5d}{s}.$ Also, since $FE= ED = PE$ as $PED$ is equilateral, we have $2s = s+1.5d,$ and so $s=1.5d.$

Thus, the ratio between side lengths of $\triangle PED$ and $\triangle PZW$ is $\frac 53$ which makes the ratio between the area $\frac{25}9.$ Also the ratio between side lengths of $\triangle PED$ and $\triangle PFC$ is $2,$ which makes the ratio between the area $4.$

This makes the area of $EDCF$ $3$ times the area of $\triangle PED$ and the area of $EDWZ$ $\frac{16}{9}$ times the area of $\triangle PED$ Therefore, the area of $ZWCF$ is $\frac{11}{9}$ times the area of $\triangle PED.$

Thus, the ratio between $ZWCF$ and $EDCF$ is $\dfrac{\frac{11}9}{3} = \dfrac{11}{27} .$

Thus, the correct answer is **C**.

24.

How many four-digit 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

###### Solution(s):

We know $a \leq b \leq c$ by analyzing $ab < bc < cd.$ Also, $bc$ is the average of $ab$ and $cd$ so $\dfrac {10a+b+10c+d}2 = 10b+c.$ This means that $10(c-2b+a)=-b+2c-d,$ making the right hand side a multiple of $10.$ Thus, it must be $0$ or $10$ since it is digits that satisfy $a \leq b \leq c .$ Thus, we can case on that value.

**Case 1:** $-b+2c-d=10$$2b-a-c=1$

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 eqution. 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 eqution. 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 eqution. This makes four cases for $b=5,6,7,8.$ This case has $8$ 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. This case makes $9$ 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

###### Solution(s):

First, note that $x = \lfloor x \rfloor + \{x\}$ where $\{x\}$ is the fractional part of $x.$

Thus, $f(x)=\sum_{k=2}^{10}\lfloor k(\lfloor x \rfloor + \{x\}) \rfloor -k \lfloor x \rfloor$ $=\sum_{k=2}^{10}\lfloor k\{x\} \rfloor$ since $\lfloor kx \rfloor$ is an integer. This means $f(x)$ can only increase when the fractional part of $kx$ becomes greater than $\frac ik$ for some $i < k.$

Thus, we must count all possible fractions that are $\frac ik$ for some $k$ less than or equal to $10.$ Note that we could exclusively count $k \geq 6$ as all $k \leq 5$ would have every fraction equal to a fraction where the denominator is greater than $6.$

There are $k-1$ fractions for each $k,$ so we have $9+8+7+6+5=35$ fractions.

However, $\frac{4}{8},$$\frac 26,$$\frac 36 ,$$\frac 46$ are all duplicates from other greater $k,$ so we subtract $4.$

As such, we have $35-4=31$ possible values of $f.$ Then, we add one more for $f(x)=0.$ This makes the total count equal to $31+1=32.$

Thus, the correct answer is **A**.