2010 AMC 10B — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of the following expression? $100(100-3)-(100 \cdot 100-3)$

$-20{,}000$

$-10{,}000$

$-297$

$-6$

$0$

Answer: C

Concepts:order of operations whole number operations

Difficulty rating: 560

Small Hint:

Expand both terms before subtracting

Big Hint:

Most of the $100\cdot100$ terms cancel

Solution:

We have that $\begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned}$ and $\begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{aligned}$

As such, their difference is: $9700 - 9997 = - 297.$

Thus, C is the correct answer.

2.

Makayla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$15$

$20$

$25$

$30$

$35$

Answer: C

Concepts:percentage unit conversion

Difficulty rating: 870

Small Hint:

Convert the meeting times to the same units

Big Hint:

The meetings last $45+90$ minutes out of $9$ hours

Solution:

Note that $45$ minutes is $\dfrac{45}{60} = \dfrac{3}{4}$ hours. The second meeting is then $2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$ hours. Both meetings take a total of $\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$ hours then. The percent of Makayla’s work day spent attending meetings is $100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%.$

Thus, C is the correct answer.

3.

A drawer contains red, green, blue, and white socks with at least $2$ of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?

$3$

$4$

$5$

$8$

$9$

Answer: C

Concepts:pigeonhole principle

Difficulty rating: 720

Small Hint:

Use the worst case before a pair appears

Big Hint:

You can draw one sock of each of the four colors first

Solution:

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are $4$ colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least $4 + 1 = 5$ socks to be drawn before a pair is guaranteed.

Thus, C is the correct answer.

4.

For a real number $x,$ define $\heartsuit(x)$ to be the average of $x$ and $x^2.$ What is the value of the following expression? $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$

$3$

$6$

$10$

$12$

$20$

Answer: C

Concepts:custom operation mean

Difficulty rating: 870

Small Hint:

Evaluate $\heartsuit(1),\heartsuit(2),\heartsuit(3)$ separately

Big Hint:

$\heartsuit(x)=\dfrac{x+x^2}{2}$

Solution:

We have $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$ and $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$

Then $1 + 3 + 6 = 10.$

Thus, C is the correct answer.

5.

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:date and time modular arithmetic

Difficulty rating: 960

Small Hint:

In $31$ days, three weekdays occur one extra time

Big Hint:

Monday and Wednesday must either both be extra days or both not be extra days

Solution:

Note that $31$ days leaves a remainder of $3$ when divided by $7.$

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

6.

A circle is centered at $O,$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ.$ What is the degree measure of $\angle CAB?$

$20$

$25$

$45$

$50$

$65$

Answer: B

Concepts:circle isosceles triangle angle chasing

Difficulty rating: 960

Small Hint:

Relate $\angle COB$ to $\angle AOC$

Big Hint:

$OA=OC$ , so $\triangle AOC$ is isosceles

Solution:

Consider the following diagram:

We have that $\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$

Since $\triangle AOC$ is isosceles, we have that $\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$

Since $\angle CAB = \angle CAO,$ we have that $\angle CAB = 25^{\circ}.$

Thus, B is the correct answer.

7.

A triangle has side lengths $10,$ $10,$ and $12.$ A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$16$

$24$

$28$

$32$

$36$

Answer: D

Concepts:triangle area Pythagorean Theorem perimeter

Difficulty rating: 1220

Small Hint:

Drop the altitude to the side of length $12$

Big Hint:

The altitude splits the triangle into two $6$ - $8$ - $10$ triangles

Solution:

To find the area of the triangle, we can drop the altitude to the side of length $12.$

Then we have a right triangle with one leg $12 \div 2 = 6$ and hypotenuse $10.$

The other leg has length $\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$

The area of the triangle is then $\dfrac{8 \cdot 12}{2} = 48.$

The length of the rectangle is then $48 \div 4 = 12.$ Its perimeter is $2(4 + 12) = 2 \cdot 16 = 32.$

Thus, D is the correct answer.

8.

A ticket to a school play costs $x$ dollars, where $x$ is a whole number. A group of $9$ th graders buys tickets costing a total of $\$48,$ and a group of $10$ th graders buys tickets costing a total of $\$64.$ How many values for $x$ are possible?

$1$

$2$

$3$

$4$

$5$

Answer: E

Concepts:greatest common divisor factor counting

Difficulty rating: 1020

Small Hint:

The ticket price must divide both totals

Big Hint:

Count the positive divisors of $\gcd(48,64)$

Solution:

Note that $x$ must divide both $48$ and $64,$ which means that it must divide their greatest common divisor.

The greatest common divisor of $48$ and $64$ is $16,$ which means $x$ divides $16.$

$16$ has $5$ factors, namely all the powers of $2$ up to $16.$

Thus, E is the correct answer.

9.

Lucky Larry’s teacher asked him to substitute numbers for $a,$ $b,$ $c,$ $d,$ and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a,$ $b,$ $c,$ and $d$ were $1,$ $2,$ $3,$ and $4,$ respectively. What number did Larry substitute for $e?$

$-5$

$-3$

$0$

$3$

$5$

Answer: D

Concepts:order of operations linear equation

Difficulty rating: 1280

Small Hint:

Compare Larry’s ignored-parentheses value to the correct value

Big Hint:

Correctly evaluated, the expression is $-2-e$

Solution:

Ignoring the parentheses, Larry would get $1 - 2 - 3 - 4 + e = e - 8.$

Evaluating with the parentheses, one would get $\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{aligned}$

Both of these values are the same, so $e - 8 = -2 - e$ $e = 3.$

Thus, D is the correct answer.

10.

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$18$

$21$

$24$

$27$

$30$

Answer: C

Concepts:distance rate and time linear equation

Difficulty rating: 1370

Small Hint:

Let $t$ be the rain time in minutes

Big Hint:

Use $20\cdot\dfrac{t}{60}+30\cdot\dfrac{40-t}{60}=16$

Solution:

Note that $40$ minutes is $\frac{2}{3}$ hours. Let Shelby drive $h$ hours in the sun.

Then she drives $\frac{2}{3} - h$ hours in the rain, which means she travels a total of $30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$ miles. We have this equals $16,$ so $10h + \dfrac{40}{3} = 16$ $h = \dfrac{4}{15}.$

Then Shelby drives $60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$ minutes in the rain.

Thus, C is the correct answer.

11.

A shopper plans to purchase an item that has a listed price greater than $\$100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\$30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\$100.$

Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x?$

$50$

$60$

$75$

$80$

$100$

Answer: A

Concepts:percentage inequality

Difficulty rating: 1540

Small Hint:

Write each coupon’s savings in terms of the listed price

Big Hint:

Coupon A must beat both $\$30$ and $0.25(p-100)$

Solution:

Let $p$ be the price of the item. Then coupon A saves $0.15p.$ Coupon B saves $\$ 30.$

Coupon C will save $0.25(p - 100) = 0.25p - 25.$

We must have that $0.15p \ge 30$ $p \geq 200$ and $0.15p \ge 0.25p - 25$ $250 \geq p.$

This shows that $x = 200$ and $y = 250.$ Therefore $y - x = 50.$

Thus, A is the correct answer.

12.

At the beginning of the school year, $50\%$ of all students in Mr. Wells’ math class answered “Yes” to the question “Do you love math”, and $50\%$ answered “No.” At the end of the school year, $70\%$ answered “Yes” and $30\%$ answered “No.” Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x?$

$0$

$20$

$40$

$60$

$80$

Answer: D

Concepts:percentage bounding to limit cases

Difficulty rating: 1420

Small Hint:

Think of $100$ students

Big Hint:

Minimize or maximize the number who switch answers

Solution:

To minimize $x,$ we want to have as many kids as possible maintain their answer.

We then need at least $70 - 50 = 20$ percent of the students to change their answer.

To maximize $x,$ we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need $70 - 50 = 20$ percent of the people who said yes to stay yes, which means only $50 - 20 = 30$ percent can switch.

This makes the maximum percent of students that can switch $50 + 30 = 80$ percent. The difference is then $80 - 20 = 60.$

Thus, D is the correct answer.

13.

What is the sum of all the solutions of the equation below? $x = |2x-|60-2x||$

$32$

$60$

$92$

$120$

$124$

Answer: C

Concepts:absolute value casework

Difficulty rating: 1660

Small Hint:

Split cases based on the sign of $60-2x$

Big Hint:

After resolving the inner absolute value, solve the resulting absolute-value equations

Solution:

We first take care of the outer absolute value. We have either $x = 2x - |60 - 2x|$ or $x = -2x + |60 - 2x|.$

These simplify to $x = |60 - 2x|$ and $3x = |60 - 2x|.$

We have $2$ cases for each equation. For the first one, we have $x = 60 - 2x$ and $x = 2x - 60.$

Solving both gives us $3x = 60$ $x = 20$ and $-x = -60$ $x = 60.$

For the other equation, we have $3x = 60 - 2x$ and $3x = 2x - 60.$

Again solving both, we have $5x = 60$ $x = 12$ and $x = -60.$ Note that $x$ cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us $20 + 60 + 12 = 92.$

Thus, C is the correct answer.

14.

The average of the numbers $1,$ $2,$ $3,$ $\cdots,$ $98,$ $99,$ and $x$ is $100x.$ What is $x?$

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

Answer: B

Concepts:mean arithmetic sequence linear equation

Difficulty rating: 1370

Small Hint:

Use the sum $1+2+\cdots+99$

Big Hint:

Set $\dfrac{99\cdot50+x}{100}=100x$

Solution:

Recall that the sum of the first $n$ integers is $\dfrac{n(n + 1)}{2}.$

Then, we have that $\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$ which simplifies to $99 \cdot 50 = (100^2 - 1)x$ $= 101 \cdot 99x,$ by difference of squares. Dividing gives us $x = \dfrac{50}{101}.$

Thus, B is the correct answer.

15.

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99.$ What is the maximum number of questions that Jesse could have answered correctly?

$25$

$27$

$29$

$31$

$33$

Answer: C

Concepts:linear equation inequality optimization

Difficulty rating: 1420

Small Hint:

Let $R$ be correct answers and $W$ wrong answers

Big Hint:

Use $4R-W=99$ and $R+W\le50$

Solution:

Let $x$ be the number of questions Jesse answered correctly and $y$ be the number he answered incorrectly.

Then $4x-y=99$ and $x+y\leq50.$

We have that $y = 4x - 99$ $5x - 99 \leq 50.$

Rearranging and simplifying tells us that $x \leq 29.8.$ Since $x$ is an integer, its maximum value is $29.$

Thus, C is the correct answer.

16.

A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$\dfrac{\pi}{3}-1$

$\dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3}$

$\dfrac{\pi}{18}$

$\dfrac{1}{4}$

$\dfrac{2\pi}{9}$

Answer: B

Concepts:circle area sector special right triangle area decomposition

Difficulty rating: 1790

Small Hint:

Find where the circle cuts one side of the square

Big Hint:

One side contributes a circular segment outside the square

Solution:

Drop the altitude from $O$ to $\overline{AB}$ at $X.$

Looking at the side lengths, we see that $\triangle OBX$ is a $30-60-90$ triangle.

This means that the area of sector $AOB$ is $\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$

We also have that the area of $\triangle AOB$ is $\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$

The area of the sector outside the square and inside the circle is then $\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$

There are $4$ of these regions, which gives us a total area of $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$

Thus, B is the correct answer.

17.

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea’s score was the median among all students, and hers was the highest score on her team. Andrea’s teammates Beth and Carla placed $37$ th and $64$ th, respectively. How many schools are in the city?

$22$

$23$

$24$

$25$

$26$

Answer: B

Concepts:median (data)inequality logical deduction

Difficulty rating: 1600

Small Hint:

If there are $n$ schools, there are $3n$ contestants

Big Hint:

Andrea’s median place is $\dfrac{3n+1}{2}$

Solution:

Let there be $n$ schools. Then there are $3n$ participants in the contest.

Because the median is one student’s score, $3n$ is odd, so $n$ is odd. Carla’s $64$ th place requires $3n\ge64,$ and hence the odd integer $n$ is at least $23.$

Andrea’s median place is $\dfrac{3n+1}{2}.$ Because she had the highest score on her team, she finished ahead of Beth, so $\dfrac{3n+1}{2}\lt37.$ This gives $3n\lt73,$ and therefore the odd integer $n$ is at most $23.$

Thus there are $23$ schools.

Thus, B is the correct answer.

18.

Positive integers $a,$ $b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}.$

What is the probability that $abc + ab + a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

Answer: E

Concepts:modular arithmetic basic probability casework

Difficulty rating: 1660

Small Hint:

Factor the expression as $a(bc+b+1)$

Big Hint:

If $a$ is not divisible by $3$ , work modulo $3$ with $b(c+1)$

Solution:

Note that $abc + ab + a = a(bc + b + 1).$ This means that if $a$ is divisible by $3,$ the whole expression is as well.

Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$

Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$

This means that $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$

For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability.

The total probability is then $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

Thus, E is the correct answer.

19.

A circle with center $O$ has area $156\pi.$ Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3},$ and point $O$ is outside $\triangle ABC.$ What is the side length of $\triangle ABC?$

$2\sqrt{3}$

$6$

$4\sqrt{3}$

$12$

$18$

Answer: B

Concepts:chord equilateral triangle Pythagorean Theorem

Difficulty rating: 1860

Small Hint:

Let $s$ be the side length and drop the altitude from $A$

Big Hint:

Use the circle radius $\sqrt{156}$ and the equilateral triangle altitude

Solution:

Consider the following diagram:

Using the formula for the area of a circle, we have that $BO = \sqrt{156}$ since it is a radius.

Extend $\overline{AO}$ to intersect $\overline{BC}$ at $X.$ Let $s$ be the side length of $\triangle ABC.$

Then we have that $BX = \dfrac{s}{2}$ and $AX = \dfrac{s\sqrt3}{2}.$

We have that $\triangle OXB$ is right, which means that we can apply the Pythagorean Theorem. This gives us $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

Simplifying, we get $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

Since $s$ is positive, we must have that $s = 6.$

Thus, B is the correct answer.

20.

Two circles lie outside regular hexagon $ABCDEF.$ The first is tangent to $\overline{AB},$ and the second is tangent to $\overline{DE}.$ Both are tangent to lines $BC$ and $FA.$ What is the ratio of the area of the second circle to that of the first circle?

$18$

$27$

$36$

$81$

$108$

Answer: D

Concepts:regular polygon tangent line special right triangle area ratio

Difficulty rating: 1960

Small Hint:

Compare the two circle radii

Big Hint:

The larger radius comes from a $30$ - $60$ - $90$ triangle

Solution:

Consider the following diagram:

Assume the regular hexagon has side length $1.$ The smaller circle is inscribed in an equilateral triangle of side length $1.$

The inradius of this equilateral triangle is $\dfrac{\sqrt3}{6}.$ The area of the circle is then $\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.$

Let $O$ be the center of the larger circle. Drop the perpendicular from $O$ to $\overline{GH}$ at $J.$ Draw $\overline{OG}.$

We have that $\triangle OJG$ is right. Since $\angle HGI = 60^{\circ},$ we also have that $\triangle OJG$ is a $30-60-90$ triangle.

Let $OJ = r.$ Then $OG = 2r.$ We also have that $OG$ is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then $OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.$

Substituting in $OG,$ we get $2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.$ Simplifying gives us $r = \dfrac{3\sqrt3}{2}.$

The area of the larger circle is then $\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.$

The desired ratio is then $\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.$

Thus, D is the correct answer.

21.

A palindrome between $1000$ and $10{,}000$ is chosen at random. What is the probability that it is divisible by $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

Answer: E

Concepts:palindrome divisibility basic probability

Difficulty rating: 1540

Small Hint:

Write a four-digit palindrome as $abba$

Big Hint:

The value is $1001a+110b$ , and $1001$ is divisible by $7$

Solution:

Note that we can express any $4$ digit number as $abcd.$ This can be expressed in long form as $10^3a + 10^2b + 10c + d.$ Since in a palindrome, we have that $a = d$ and $b = c.$ We can simplify this to get $1001a + 110b.$ Note that $1001$ is divisible by $7.$ This means that $110b$ must also be divisible by $7.$

The only way for this to happen is if $b$ is $0$ or $7$ since $110$ is not divisible by $7.$

There are $9$ options for $a$ and $2$ options for $b,$ for a total of $9 \cdot 2 = 18$ palindromes.

The total number of palindromes is $9 \cdot 10$ since there are $9$ options for the thousands digit and $10$ options for the hundreds digit.

The desired probability is then $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

Thus, E is the correct answer.

22.

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$1930$

$1931$

$1932$

$1933$

$1934$

Answer: C

Concepts:complementary counting inclusion-exclusion multiplication principle

Difficulty rating: 1790

Small Hint:

Count all $3^7$ distributions first

Big Hint:

Subtract cases with the red or blue bag empty, then add back the overlap

Solution:

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is $3^7 = 2187.$

To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty.

For the case where the red bag is empty, each candy has $2$ options for the bag that goes into. There are then $2^7 = 128$ arrangements for this case. Similarly, there are $128$ arrangements for the case where the blue bag is empty.

There is an overlap of one case where both bags are empty. The final answer is then $2187 - (128 + 128 - 1) = 1932.$

Thus, C is the correct answer.

23.

The entries in a $3 \times 3$ array include all the digits from $1$ through $9,$ arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$18$

$24$

$36$

$42$

$60$

Answer: D

Concepts:arrangements with restrictions casework

Difficulty rating: 2030

Small Hint:

The center entry can only be $4,5,$ or $6$

Big Hint:

Case on the center and use the row-column increasing constraints

Solution:

Let $a_{ij}$ be the entry in row $i$ and column $j.$ The increasing conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}$ to be $4,$ $5,$ or $6.$

If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}.$ Choose which two of $5,$ $6,$ $7,$ and $8$ go in the bottom-left pair $\{a_{31},a_{32}\};$ the other two go in the top-right pair $\{a_{13},a_{23}\}.$ Each pair then has only one increasing order. There are $2\binom42=12$ arrays: two orders for $2,$ $3$ around the upper-left corner and $\binom42$ choices for the bottom-left pair. By reversing the digits and rotating the array, there are also $12$ arrays with center $6.$

If $a_{22}=5,$ choose the three entries in positions $a_{12},$ $a_{13},$ and $a_{23}.$ They can be any three of $\{2,3,4,6,7,8\}$ except $\{2,3,4\}$ or $\{6,7,8\};$ those two choices would put all three small or all three large entries on one side and violate a required comparison with the center. Every other choice uniquely determines the remaining entries and their increasing orders. This gives $\binom63-2=18$ arrays.

The total is $12+18+12=42.$

Thus, D is the correct answer.

24.

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?

$30$

$31$

$32$

$33$

$34$

Answer: E

Concepts:geometric sequence arithmetic sequence Diophantine Equation

Difficulty rating: 2180

Small Hint:

Let the tied first-quarter score be $a$

Big Hint:

Use the geometric and arithmetic total scores and the one-point final margin

Solution:

Let the first-quarter score for each team be $a.$ Let the Raiders have common ratio $r$ and let the Wildcats have common difference $d.$

Write $r=\frac{m}{n}$ in lowest terms. Since the Raiders’ four quarter scores are integers, $a=n^3A$ for some positive integer $A$ , and their total is $A(n^3+n^2m+nm^2+m^3)\le100$ . Thus the only possible ratios are $\dfrac{3}{2},2,3,4$ .

For $r=\dfrac{3}{2}$ or $r=4$ , the only possible Raiders totals give Wildcats totals that are not of the form $4a+6d$ . For $r=3$ , the equation $40a=4a+6d+1$ is impossible modulo $6$ .

For $r=2$ , the equation is $15a=4a+6d+1$ , so $11a=6d+1$ . The only positive solution with total at most $100$ is $a=5,d=9$ .

The first-half total is $5+10+5+14=34.$

Thus, E is the correct answer.

25.

Let $a \gt 0,$ and let $P(x)$ be a polynomial with integer coefficients such that $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a, \end{aligned}$ and $\begin{aligned} P(2)&=P(4)=P(6)\\ &=P(8)=-a. \end{aligned}$ What is the smallest possible value of $a?$

$105$

$315$

$945$

$7!$

$8!$

Answer: B

Concepts:polynomial divisibility least common multiple

Difficulty rating: 2350

Small Hint:

Use the roots of $P(x)-a$

Big Hint:

Plug $2,$ $4,$ $6,$ and $8$ into the factored form to force divisibility of $a$

Solution:

Because $1,$ $3,$ $5,$ and $7$ are roots of $P(x)-a$ , write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ , where $Q(x)$ has integer coefficients.

Substituting $2,$ $4,$ $6,$ and $8$ for $x$ gives $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ . Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$ .

This lower bound is attainable: take $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ and define $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ . This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.