2017 AMC 10B Solutions | LIVE by Po-Shen Loh

1.

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11.$ Then she switched the digits of the result, obtaining a number between $71$ and $75,$ inclusive. What was Mary’s number?

$11$

$12$

$13$

$14$

$15$

Concepts:digits divisibility work backwards

Difficulty rating: 960

Small Hint:

Work backward from the possible reversed results

Big Hint:

After reversing and subtracting $11$ , check divisibility by $3$

Solution:

Work backward from the possible results. Reversing $71,72,73,74,75$ gives $17,27,37,47,57,$ respectively. Subtracting $11$ gives $6,16,26,36,46.$ Only $6$ and $36$ are divisible by $3,$ and only $36\div3=12$ is a two-digit number.

Thus, the correct answer is B .

2.

Sofia ran $5$ laps around the $400$ -meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

$5$ minutes and $35$ seconds

$6$ minutes and $40$ seconds

$7$ minutes and $5$ seconds

$7$ minutes and $25$ seconds

$8$ minutes and $10$ seconds

Concepts:distance rate and time unit conversion

Difficulty rating: 870

Small Hint:

Find the time for the first $100$ meters and the last $300$ meters of one lap

Big Hint:

Multiply one-lap time by $5$ and convert seconds to minutes

Solution:

She ran a total of $5\cdot 100=500$ meters at $4$ meters per second and $5\cdot 300=1500$ meters at $5$ meters per second.

Therefore, her time is $\frac{500}4 + \frac{1500}5 = 425$ seconds.

This is equal to a total of $7$ minutes and $5$ seconds.

Thus, the correct answer is C .

3.

Real numbers $x,$ $y,$ and $z$ satisfy the inequalities $0 < x < 1,$ $-1 < y < 0,$ and $1 < z < 2.$

Which of the following numbers is necessarily positive?

$y+x^2$

$y+xz$

$y+y^2$

$y+2y^2$

$y+z$

Concepts:inequality counterexample

Difficulty rating: 960

Small Hint:

The choice $y+z$ can be tested directly from the given bounds

Big Hint:

Use one small counterexample to reject each other expression

Solution:

Since $-1 < y$ and $1 < z,$ we can add the inequalities to see that $0 < y+z.$ This naturally proves choice E correct.

Furthermore, we can eliminate every other choice with the following values: $x=0.1,$ $y=-0.25,$ $z=1.25.$

Thus, the correct answer is E .

4.

Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2.$ What is the value of $\frac{x+3y}{3x-y}?$

$-3$

$-1$

$1$

$2$

$3$

Concepts:rational equation algebraic manipulation

Difficulty rating: 900

Small Hint:

Clear the denominator in the given equation

Big Hint:

The equation quickly forces a relation between $x$ and $y$

Solution:

Given that $\frac{3x+y}{x-3y}=-2,$ we can multiply by the denominator to get $3x+y = 6y-2x.$ Solving, we can see that $x=y.$

Therefore, $\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2.$

Thus, the correct answer is D .

5.

Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating $10$ pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?

$10$

$20$

$30$

$40$

$50$

Concepts:system of equations linear equation

Difficulty rating: 900

Small Hint:

Let the original cherry and blueberry counts be variables

Big Hint:

Use one equation before and one equation after eating the jelly beans

Solution:

Let the number of cherry jelly beans be $c$ and let the number of blueberry jelly beans be $b.$

Then, we know $b = 2c$ $b-10=3(c-10)$ from the first and second statements respectively.

Therefore, $2c-10 = 3c-30$ $c=20.$ This means that $b = 2\cdot 20=40.$

Thus, the correct answer is D .

6.

What is the largest number of solid $2$ -in $\times$ $2$ -in $\times$ $1$ -in blocks that can fit in a $3$ -in $\times$ $2$ -in $\times$ $3$ -in box?

$3$

$4$

$5$

$6$

$7$

Concepts:volume rectangular prism bounding to limit cases

Difficulty rating: 1070

Small Hint:

Start with the volume upper bound

Big Hint:

Then check that four blocks really can be arranged

Solution:

The volume of the large solid object is $3\cdot 3\cdot 2 = 18$ and volume of the smaller object is $2\cdot 2\cdot 1=4.$ This means we can fit at most $4$ of the small objects.

We can make this happen by putting $3$ of the small objects in a $3 \times 2 \times 2$ rectangular prism, and then we have a $3 \times 2 \times 1$ space left where we can place one small object.

Thus, the correct answer is B .

7.

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend’s house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour.

In all, it took her $44$ minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$2.0$

$2.2$

$2.8$

$3.4$

$4.4$

Concepts:distance rate and time linear equation

Difficulty rating: 1220

Small Hint:

Let the walking distance be $d$ , so the biking distance is also $d$

Big Hint:

Add the biking time and walking time to get $44$ minutes

Solution:

Let $d$ be the distance Samia walked. She bicycled the same distance, so her total travel time gives $\dfrac{d}{17}+\dfrac{d}{5}=\dfrac{44}{60}.$ Solving yields $d=\dfrac{17}{6}=2.833\ldots,$ which rounds to $2.8$ kilometers.

Thus, the correct answer is C .

8.

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC.$ The altitude from $A$ meets the opposite side at $D(-1, 3).$ What are the coordinates of point $C?$

$(-8, 9)$

$(-4, 8)$

$(-4, 9)$

$(-2, 3)$

$(-1, 0)$

Concepts:coordinate geometry midpoint isosceles triangle

Difficulty rating: 1070

Small Hint:

In an isosceles triangle, the altitude from the vertex also bisects the base

Big Hint:

Use $D$ as the midpoint of $BC$

Solution:

Since $AB=AC,$ the altitude from $A$ also bisects the base $BC.$ Therefore, $D$ is the midpoint of $BC.$ If $C=(x,y),$ then we have $\frac{x+2}2 = -1,$ $\frac{y-3}2=3.$ As such, $C=(x,y)=(-4,9).$

Thus, the correct answer is C .

9.

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$\dfrac{1}{27}$

$\dfrac{1}{9}$

$\dfrac{2}{9}$

$\dfrac{7}{27}$

$\dfrac{1}{2}$

Concepts:binomial probability casework

Difficulty rating: 1140

Small Hint:

Count exactly three correct plus exactly two correct

Big Hint:

For exactly two correct, choose which question is missed

Solution:

The probability that a contestant gets all $3$ correct is $\left(\dfrac13\right)^3=\dfrac1{27}.$ The probability of getting exactly $2$ correct is $\binom32\left(\dfrac13\right)^2\left(\dfrac23\right)=\dfrac6{27}.$ The combined probability is $\dfrac6{27}+\dfrac1{27}=\dfrac7{27}.$

Thus, the correct answer is D .

10.

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5).$ What is $c?$

$-13$

$-8$

$2$

$8$

$13$

Concepts:coordinate geometry slope system of equations

Difficulty rating: 1370

Small Hint:

Rewrite both lines in slope-intercept form

Big Hint:

Perpendicular slopes and the point $(1,-5)$ determine the parameters

Solution:

The first equation can be rewritten as $y=\dfrac a2x-\dfrac c2,$ and the second as $y=-\dfrac2b x-\dfrac cb.$ Because the lines are perpendicular, their slopes multiply to $-1,$ so $a=b.$

Substituting $(x,y)=(1,-5)$ into the two original equations gives $a+10=c$ and $2-5a=-c.$ Adding these equations yields $12-4a=0,$ so $a=3$ and $c=13.$

Thus, the correct answer is E .

11.

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$10\%$

$12\%$

$20\%$

$25\%$

$33\frac{1}{3}\%$

Concepts:conditional probability percentage

Difficulty rating: 1370

Small Hint:

Separate actual preference from reported preference

Big Hint:

The denominator is everyone who says they dislike dancing

Solution:

Observe that of the $60\%$ of people that actually like dancing, only $80\%$ say they like dancing. This suggests that $48\%$ of the students say that they like dancing, and as such, $60\%-48\% = 12\%$ of the students who like dancing say they don’t like it.

Then, we know that $90\%$ of the $40\%$ of people who don’t like dancing say they don’t like it, which is $36\%$ of the total student population.

This means the total amount of people who say they don’t like dancing is $12\%+36\% = 48\%.$

We know then that the fraction of people who say they dislike dancing but actually like it is equal to: $\frac{12}{48} = \frac 14 = 25\% .$

Thus, the correct answer is D .

12.

Elmer’s new car gets $50\%$ better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car uses. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?

$20\%$

$26\tfrac23\%$

$27\tfrac79\%$

$33\tfrac13\%$

$41\tfrac23\%$

Concepts:rate percentage

Difficulty rating: 1280

Small Hint:

Compare cost per kilometer, not cost per liter

Big Hint:

New fuel uses fewer liters but each liter costs more

Solution:

Let the old car’s fuel efficiency be $M$ kilometers per liter and let gasoline cost $C$ dollars per liter. The old car therefore costs $\dfrac CM$ dollars per kilometer to fuel.

The new car gets $1.5M$ kilometers per liter and its fuel costs $1.2C$ dollars per liter, so its fuel cost per kilometer is $\dfrac{1.2C}{1.5M}=0.8\dfrac CM.$

The new fuel cost is $80\%$ of the old one, so Elmer saves $20\%.$

Thus, the correct answer is A .

13.

There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three.

There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?

$1$

$2$

$3$

$4$

$5$

Concepts:inclusion-exclusion system of equations

Difficulty rating: 1280

Small Hint:

Let $x,y,z$ count students taking exactly one, two, and three classes

Big Hint:

The total class enrollment counts these groups with weights $1,2,3$

Solution:

The number of classes taken total is $10+13+9=32.$

Let $x$ represent the number of people who take $1,$ let $y$ represent the number of people who take $2$ classes, and let $z$ represent the number of people who take $3$ classes.

Then, we know $x+2y+3z = 32.$

As such, the total number of people is $20,$ so $x+y+z = 20.$ This makes $y+2z=12.$

The number of people who take at least two classes is $9,$ so $y+z = 9.$

Therefore, $z=3,$ making that the answer.

Thus, the correct answer is C .

14.

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1?$

$\dfrac{1}{5}$

$\dfrac{2}{5}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

$1$

Concepts:Fermat’s Little Theorem modular exponentiation basic probability

Difficulty rating: 1370

Small Hint:

Modulo $5$ , only whether $N$ is divisible by $5$ matters

Big Hint:

Use Fermat’s little theorem or check residue classes modulo $5$

Solution:

By Fermat’s Little Theorem, $N^4\equiv1\pmod5$ whenever $N$ is not divisible by $5.$ Therefore, $N^{16}\equiv(N^4)^4\equiv1\pmod5.$

There are $\frac{2020}{5}=404$ multiples of $5,$ so there are $2020-404=1616$ allowable values of $N.$

A multiple of $5$ has $N^{16}\equiv0\pmod5,$ so no other values work. Thus the probability is $\dfrac{1616}{2020}=\dfrac45.$

Thus, the correct answer is D .

15.

Rectangle $ABCD$ has $AB=3$ and $BC=4.$ Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}.$ What is the area of $\triangle ADE?$

$1$

$\dfrac{42}{25}$

$\dfrac{28}{15}$

$2$

$\dfrac{54}{25}$

Difficulty rating: 1600

Small Hint:

Use similarity between $\triangle ABE$ and $\triangle ACB$

Big Hint:

Then compare $\triangle EAD$ to $\triangle CDA$ using their bases on $AC$

Solution:

The area of $\triangle CDA$ is $\dfrac{3\cdot4}{2}=6$ . Since $E$ lies on $AC$ , triangles $EAD$ and $CDA$ share the same altitude from $D$ , so $[EAD]=[CDA]\cdot \dfrac{AE}{AC}$ .

By the Pythagorean Theorem, $AC=5$ . Also $\triangle ABE\sim \triangle ACB$ , so $\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35$ , giving $AE=\dfrac95$ . Thus $\dfrac{AE}{AC}=\dfrac{9}{25}$ .

Therefore $[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}$ . Thus, E is the correct answer.

16.

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0?$

$469$

$471$

$475$

$478$

$481$

Concepts:digits complementary counting

Difficulty rating: 1480

Small Hint:

Count the complement: numbers with no digit $0$

Big Hint:

Split by one-, two-, three-, and four-digit numbers up to $2017$

Solution:

For numbers less than $100,$ we only have a $0$ if it is a multiple of $10,$ of which there are $9.$

For numbers between $100$ and $999$ inclusive, we will use complementary counting. There are $900$ total numbers in this range. Also, there are $9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each digit to not be $0.$ Thus, the total in this range is $171.$

For numbers between $1000$ and $1999$ inclusive, we will use complementary counting again. There are $1000$ total numbers in this range. Also, there are $1\cdot 9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each of the last $3$ digits to not be $0$ and the first digit must be $1.$ Thus, the total in this range is $271.$

There are $18$ numbers between $2000$ and $2017$ inclusive, each with a $0$ in the second digit from the left.

This makes the total $9+171+271+18=469.$

Thus, the correct answer is A .

17.

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3,$ $23578,$ and $987620$ are monotonous, but $88,$ $7434,$ and $23557$ are not. How many monotonous positive integers are there?

$1024$

$1524$

$1533$

$1536$

$2048$

Concepts:subsets digits bijection

Difficulty rating: 1660

Small Hint:

Increasing numbers are determined by their digit set

Big Hint:

Decreasing numbers may include $0$ , but the number $0$ itself is not positive

Solution:

The strictly increasing positive integers correspond to the nonempty subsets of $\{1,2,3,4,5,6,7,8,9\}$ , written in increasing order. There are $2^9-1=511$ of these.

The strictly decreasing positive integers correspond to subsets of $\{0,1,2,3,4,5,6,7,8,9\}$ , written in decreasing order, except for the empty set and $\{0\}$ . There are $2^{10}-2=1022$ of these.

The one-digit numbers $1$ through $9$ were counted in both groups, so the total is $511+1022-9=1524$ . Thus, B is the correct answer.

18.

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$6$

$8$

$9$

$12$

$15$

Concepts:combinations symmetry casework

Difficulty rating: 2010

Small Hint:

Use symmetry to reduce the green disk to two cases

Big Hint:

For each green position, count possible red-disk placements up to symmetry

Solution:

By symmetry, the green disk has two possible types of position: a corner or a side midpoint. Fix one representative of either type. There are $\binom52=10$ ways to choose the two red disks.

The reflection that fixes the green position fixes one of the other disks and exchanges the other four disks in two pairs. Exactly $2$ red-disk choices are unchanged by this reflection: choosing either exchanged pair. The other $8$ choices form $4$ mirror-image pairs. Hence there are $2+4=6$ paintings for each type of green position.

The two types therefore give $6+6=12$ paintings.

Thus, the correct answer is D .

19.

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB.$ Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC,$ and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA.$

What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC?$

$9:1$

$16:1$

$25:1$

$36:1$

$37:1$

Concepts:area ratio equilateral triangle triangle area

Difficulty rating: 1860

Small Hint:

Break the large triangle into the original triangle and six surrounding triangles

Big Hint:

Compare each surrounding triangle area to the original using base and height

Solution:

Let $X$ be the area of $\triangle ABC.$ Each of $\triangle BB'C,$ $\triangle CC'A,$ and $\triangle AA'B$ has a base three times as long as a side of $\triangle ABC$ and the same corresponding altitude. Each therefore has area $3X.$

Next, $\triangle AA'C'$ has three times the base and the same altitude as $\triangle ACC',$ whose area is $3X.$ Thus $\triangle AA'C'$ has area $9X.$ Similarly, $\triangle CC'B'$ and $\triangle BB'A'$ each have area $9X.$

These seven regions partition the large triangle, so $\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X.\end{aligned}$ The requested ratio is $37:1.$

Thus, the correct answer is E .

20.

The number $21!$ $=51{,}090{,}942{,}171{,}709{,}440{,}000$ has over $60{,}000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

Concepts:factor counting prime factorization basic probability

Difficulty rating: 1540

Small Hint:

Count the exponent of $2$ in $21!$

Big Hint:

For each odd part of a divisor, only one exponent of $2$ gives an odd divisor

Solution:

The exponent of $2$ in $21!$ is $\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18. \end{aligned}$ Thus $21!=2^{18}d$ for some odd integer $d.$

For every divisor $x$ of $d,$ the divisors of $21!$ with odd part $x$ are $x,2x,\ldots,2^{18}x.$ Exactly one of these $19$ divisors is odd, so the probability is $\dfrac1{19}.$

Thus, the correct answer is B .

21.

In $\triangle ABC,$ $AB=6,$ $AC=8,$ $BC=10,$ and $D$ is the midpoint of $\overline{BC}.$ What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC?$

$\sqrt{5}$

$\dfrac{11}{4}$

$2\sqrt{2}$

$\dfrac{17}{6}$

$3$

Concepts:incircle, incenter, and inradius right triangle triangle area

Difficulty rating: 1720

Small Hint:

First recognize the $6,8,10$ triangle as right

Big Hint:

Use area equals inradius times semiperimeter for the two smaller triangles

Solution:

The triangle $ABC$ is a right triangle with a right angle at $A.$ This makes $D$ the circumcenter of the triangle since it is the midpoint of the hypotenuse.

Therefore, $AD = BD = DC = 5.$ Also, the area of $ABC$ is $\dfrac{6\cdot 8}2 = 24.$

The bases $BD$ and $DC$ are equal, and the two triangles share the same altitude from $A.$ Therefore, $\triangle ABD$ and $\triangle ACD$ each have area $12.$

Then, for each triangle, we have $A = rs$ where $A$ is the area, $r$ is the inradius, and $s$ is the semiperimeter. Equivalently, $12=\dfrac12rP,$ where $P$ is the perimeter, so $r=\dfrac{24}{P}.$ For $\triangle ABD,$ the inradius is $\dfrac{24}{5+5+6}=\dfrac32.$ For $\triangle ACD,$ it is $\dfrac{24}{5+5+8}=\dfrac43.$

Their sum is $\dfrac 32 + \dfrac 43 = \dfrac{17}6 .$

Thus, the correct answer is D .

22.

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3.$ Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD.$ Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E.$ What is the area of $\triangle ABC?$

$\dfrac{120}{37}$

$\dfrac{140}{39}$

$\dfrac{145}{39}$

$\dfrac{140}{37}$

$\dfrac{120}{31}$

Concepts:inscribed angle similarity area ratio

Difficulty rating: 1900

Small Hint:

Use the semicircle angle to get a right triangle

Big Hint:

Compare $\triangle ABC$ and $\triangle AED$ by similarity

Solution:

Since the radius is $2$ and $BD =3,$ we have $AD = 7.$ Since $ED = 5$ and the angle at $D$ is a right angle, the area of $ADE$ is $\dfrac{5\cdot 7}{2} = \dfrac{35}{2} .$

By the Pythagorean Theorem, $AE=\sqrt{5^2+7^2}=\sqrt{74}.$ Also, $\angle ACB$ is a right angle because $AB$ is a diameter. The triangles share the angle at $A,$ so $\triangle ABC\sim\triangle AED$ by angle-angle similarity.

Their corresponding hypotenuses are $AB=4$ and $AE=\sqrt{74},$ so their area ratio is $\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}.$ Therefore, $[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}.$

Thus, the correct answer is D .

23.

Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45?$

$1$

$4$

$9$

$18$

$44$

Concepts:modular arithmetic Chinese Remainder Theorem digits

Difficulty rating: 1660

Small Hint:

Find the remainder modulo $5$ and modulo $9$

Big Hint:

Combine them to get the remainder modulo $45$

Solution:

To find the remainder when divided by $45,$ we must find the remainder when divided by $5$ and $9.$ The remainder when divided by $5$ is the remainder when the units digit is divided by $5,$ making it $4.$

To find the remainder when divided by $9,$ we usually find the sum of the digits. However, each double digit number has the same remainder when divided by $9$ as its digit sum, so we can just sum the integers from $1$ to $44,$ because each integer is congruent to its own digit sum. This sum is $\dfrac{44\cdot45}{2}=990,$ which is a multiple of $9.$ Thus, $N$ is a multiple of $9.$

Since it is a multiple of $9$ and has a remainder of $4$ when divided by $5,$ the remainder when divided by $45$ is $9.$

Thus, the correct answer is C .

24.

The vertices of an equilateral triangle lie on the hyperbola $xy=1,$ and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$48$

$60$

$108$

$120$

$169$

Concepts:hyperbola equilateral triangle centroid symmetry

Difficulty rating: 2380

Small Hint:

Use symmetry of the hyperbola about $y=x$

Big Hint:

Relate the centroid-to-vertex distance to the circumradius of the equilateral triangle

Solution:

By symmetry, assume that the centroid is the hyperbola vertex $G=(1,1).$ At least two triangle vertices lie on the same branch of the hyperbola. They cannot both lie on the negative branch: if two of their $x$ -coordinates were negative, the third would exceed $3,$ while the sum of the three $y$ -coordinates would be less than $\frac{1}{3},$ contradicting that their centroid is $(1,1).$ Thus two vertices lie on the positive branch.

Write these vertices as $P=(a,\frac{1}{a})$ and $Q=(b,\frac{1}{b}),$ where $a,b>0.$ The centroid of an equilateral triangle is also its circumcenter, so $P$ and $Q$ are equidistant from $G.$ For $t>0,$ the squared distance from $(t,\frac{1}{t})$ to $G$ is $\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right).$ This is strictly increasing as $t+\frac{1}{t}$ increases from $2,$ so the distinct points must satisfy $b=\frac{1}{a}.$ Hence $P$ and $Q$ are reflections across $y=x.$

The third vertex lies on the perpendicular bisector $y=x.$ Its coordinates also satisfy $xy=1,$ so it is either $(1,1)$ or $(-1,-1).$ It cannot equal the centroid, so it is $(-1,-1).$ Therefore, the circumradius is the distance from $(1,1)$ to $(-1,-1),$ namely $2\sqrt2.$

Dividing the equilateral triangle into three triangles at its center gives its area as $3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3.$ The square of the area is $(6\sqrt3)^2=108.$

Thus, the correct answer is C .

25.

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100,$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95.$ What was her score on the sixth test?

$92$

$94$

$96$

$98$

$100$

Concepts:divisibility mean bounding to limit cases

Difficulty rating: 2300

Small Hint:

The final total score must be a multiple of $7$

Big Hint:

Use the seventh score to force the first six-score sum, then use divisibility by $5$

Solution:

Let $S$ be the sum of all seven scores. Since all seven averages were integers, $S$ is divisible by $7$ . Also the seven distinct scores are between $91$ and $100$ , so $91+92+\cdots+97$ $\le S\le$ $94+95+\cdots+100$ .

Thus $658\le S\le 679$ , and the possible multiples of $7$ are $658,665,672,679$ . Since the seventh score is $95$ , the first six scores sum to $S-95$ , which must be divisible by $6$ . This forces $S=665$ .

The first six scores sum to $570$ . The first five-score average was also an integer, so the sum of the first five scores is divisible by $5$ . Therefore the sixth score is divisible by $5$ . Since the seventh score is already $95$ and all scores are distinct, the sixth score is $100$ . Thus, E is the correct answer.