2005 AMC 10B — Problems & Online Timed Exam

Timed

1:15:00

1.

A scout troop buys $1000$ candy bars at a price of five for $\$2.$ They sell all the candy bars at a price of two for $\$1.$ What was their profit, in dollars?

$100$

$200$

$300$

$400$

$500$

Answer: A

Concepts:money ratio and proportion

Difficulty rating: 880

Small Hint:

Find the total cost and the total revenue separately

Big Hint:

The bars come in $200$ groups of five and $500$ pairs

Solution:

The troop buys $1000 \div 5 = 200$ groups of five bars, costing $200 \cdot 2 = 400$ dollars.

They sell $1000 \div 2 = 500$ pairs of bars, earning $500 \cdot 1 = 500$ dollars.

The profit is $\$500 - \$400 = \$100.$

Thus, A is the correct answer.

2.

A positive number $x$ has the property that $x\%$ of $x$ is $4.$ What is $x?$

$2$

$4$

$10$

$20$

$40$

Answer: D

Concepts:percentage quadratic

Difficulty rating: 880

Small Hint:

$x\%$ means $\dfrac{x}{100}$

Big Hint:

Set up $\dfrac{x}{100}\cdot x=4$ and solve for $x$

Solution:

The statement translates to $\dfrac{x}{100}\cdot x = 4,$ so $x^2 = 400.$

Since $x$ is positive, $x = 20.$

Thus, D is the correct answer.

3.

A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{3}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

Answer: D

Concepts:fraction ratio and proportion

Difficulty rating: 870

Small Hint:

After day one, $\dfrac23$ of the paint remains

Big Hint:

On day two, $\dfrac13$ of that $\dfrac23$ is used

Solution:

After the first day, $1 - \dfrac13 = \dfrac23$ of the paint remains.

On the second day, $\dfrac13 \cdot \dfrac23 = \dfrac29$ of the original amount is used.

The fraction available on the third day is $\dfrac23 - \dfrac29 = \dfrac49.$

Thus, D is the correct answer.

4.

For real numbers $a$ and $b,$ define $a \diamond b = \sqrt{a^2 + b^2}.$ What is the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))?$

$0$

$\dfrac{17}{2}$

$13$

$13\sqrt{2}$

$26$

Answer: D

Concepts:custom operation radical

Difficulty rating: 1020

Small Hint:

Evaluate the two inner operations first

Big Hint:

Both $5\diamond12$ and $(-12)\diamond(-5)$ equal $13$

Solution:

Each inner expression evaluates to $5 \diamond 12 = \sqrt{5^2 + 12^2} = 13$ and similarly $(-12) \diamond (-5)$ $= \sqrt{144 + 25} = 13.$

Then $13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}.$

Thus, D is the correct answer.

5.

Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

$\dfrac15$

$\dfrac13$

$\dfrac25$

$\dfrac23$

$\dfrac45$

Answer: C

Concepts:fraction ratio and proportion money

Difficulty rating: 960

Small Hint:

Buying all the CDs costs three times as much as buying one third of them

Big Hint:

All the CDs cost $3\cdot\dfrac15$ of her money

Solution:

Buying all the CDs costs three times as much as buying one third of them, namely $3 \cdot \dfrac15 = \dfrac35$ of her money.

The fraction left over is $1 - \dfrac35 = \dfrac25.$

Thus, C is the correct answer.

6.

At the beginning of the school year, Lisa’s goal was to earn an A on at least $80\%$ of her $50$ quizzes for the year. She earned an A on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:percentage inequality

Difficulty rating: 960

Small Hint:

Find how many A’s she needs in total

Big Hint:

She needs $0.8\cdot50$ A’s, and $20$ quizzes remain

Solution:

Lisa needs an A on at least $0.8 \cdot 50 = 40$ quizzes.

She already has $22,$ so she needs $40 - 22 = 18$ A’s among the remaining $20$ quizzes.

That leaves at most $20 - 18 = 2$ quizzes with a grade lower than an A.

Thus, B is the correct answer.

7.

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

$\dfrac{\pi}{16}$

$\dfrac{\pi}{8}$

$\dfrac{3\pi}{16}$

$\dfrac{\pi}{4}$

$\dfrac{\pi}{2}$

Answer: B

Concepts:circle area square (geometry)area ratio

Difficulty rating: 1240

Small Hint:

Let the smaller circle have radius $r$ and work outward

Big Hint:

The larger circle’s radius is half the diagonal of the smaller square, $\sqrt2\,r$

Solution:

Let the smaller circle have radius $r,$ so its area is $\pi r^2.$

The smaller square, which circumscribes this circle, has side $2r,$ and its diagonal $2\sqrt2\,r$ is the diameter of the larger circle. So the larger circle has radius $\sqrt2\,r.$

The larger square circumscribes the larger circle, so it has side $2\sqrt2\,r$ and area $8r^2.$

The desired ratio is $\dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}.$

Thus, B is the correct answer.

8.

An $8$ -foot by $10$ -foot floor is tiled with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $\dfrac12$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

$80 - 20\pi$

$60 - 10\pi$

$80 - 10\pi$

$60 + 10\pi$

$80 + 10\pi$

Answer: A

Concepts:circle area area decomposition

Difficulty rating: 1170

Small Hint:

The four quarter circles on one tile combine into one full circle

Big Hint:

Each tile has shaded area $1-\pi\left(\dfrac12\right)^2,$ and there are $80$ tiles

Solution:

The four quarter circles on a tile together make one full circle of radius $\dfrac12,$ with area $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}.$

So each tile has shaded area $1 - \dfrac{\pi}{4}$ square feet.

There are $8 \cdot 10 = 80$ tiles, so the total shaded area is $80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi.$

Thus, A is the correct answer.

9.

One fair die has faces $1,$ $1,$ $2,$ $2,$ $3,$ $3$ and another has faces $4,$ $4,$ $5,$ $5,$ $6,$ $6.$ The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\dfrac13$

$\dfrac49$

$\dfrac12$

$\dfrac59$

$\dfrac23$

Answer: D

Concepts:dice (probability)parity independent events

Difficulty rating: 1240

Small Hint:

A sum is odd exactly when one die is even and the other is odd

Big Hint:

The first die is odd with probability $\dfrac23;$ the second is odd with probability $\dfrac13$

Solution:

The first die is odd (a $1$ or $3$ ) with probability $\dfrac23$ and even with probability $\dfrac13.$ The second die is odd (a $5$ ) with probability $\dfrac13$ and even with probability $\dfrac23.$

The sum is odd when the two parities differ: $\dfrac13 \cdot \dfrac13 + \dfrac23 \cdot \dfrac23 = \dfrac19 + \dfrac49 = \dfrac59.$

Thus, D is the correct answer.

10.

In $\triangle ABC,$ we have $AC = BC = 7$ and $AB = 2.$ Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD = 8.$ What is $BD?$

$3$

$2\sqrt{3}$

$4$

$5$

$4\sqrt{2}$

Answer: A

Concepts:isosceles triangle altitude Pythagorean Theorem

Difficulty rating: 1370

Small Hint:

Drop the altitude from $C$ to line $AB;$ its foot is the midpoint of $AB$

Big Hint:

The altitude has length $\sqrt{7^2-1^2},$ shared by right triangles $CHB$ and $CHD$

Solution:

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $AC = BC,$ $H$ is the midpoint of $AB,$ so $BH = 1$ and $CH^2 = 7^2 - 1^2 = 48.$

Applying the Pythagorean theorem in $\triangle CHD,$ where $HD = BH + BD = 1 + BD,$ gives $8^2 = 48 + (1 + BD)^2,$ so $(1 + BD)^2 = 16.$

Then $1 + BD = 4,$ so $BD = 3.$

Thus, A is the correct answer.

11.

The first term of a sequence is $2005.$ Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2005$ th term of the sequence?

$29$

$55$

$85$

$133$

$250$

Answer: E

Concepts:recursion digits pattern recognition

Difficulty rating: 1370

Small Hint:

Compute the first several terms until the values start repeating

Big Hint:

Once you find the repeating cycle, reduce the number of steps after the first term modulo its length

Solution:

The sequence begins $2005, 133, 55, 250, 133, \ldots,$ so after the first term it repeats the cycle $133, 55, 250$ of length $3.$

Terms $2,$ $3,$ and $4$ are the first, second, and third entries of this cycle. Because $2005 - 2 = 2003$ leaves remainder $2$ upon division by $3,$ the $2005$ th term matches the third entry, $250.$

Thus, E is the correct answer.

12.

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$\left(\dfrac{1}{12}\right)^{12}$

$\left(\dfrac{1}{6}\right)^{12}$

$2\left(\dfrac{1}{6}\right)^{11}$

$\dfrac52\left(\dfrac{1}{6}\right)^{11}$

$\left(\dfrac{1}{6}\right)^{10}$

Answer: E

Concepts:dice (probability)prime multiplication principle

Difficulty rating: 1480

Small Hint:

A product of positive integers is prime only when one factor is a prime and the rest are $1$

Big Hint:

Choose which die shows the prime, and which prime; the other $11$ dice must show $1$

Solution:

The product is prime exactly when one die shows a prime ( $2,$ $3,$ or $5$ ) and the other eleven all show $1.$

The probability that any single die is the prime one is $\dfrac36 = \dfrac12,$ and each of the other eleven shows $1$ with probability $\dfrac16.$ Accounting for which of the twelve dice is prime, the probability is $\begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10}. \end{aligned}$

Thus, E is the correct answer.

13.

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12?$

$501$

$668$

$835$

$1002$

$1169$

Answer: C

Concepts:multiple inclusion-exclusion counting integers in a range

Difficulty rating: 1370

Small Hint:

Count multiples of $3$ and of $4$ separately, noting that multiples of $12$ are the overlap

Big Hint:

From each of the $3$ -multiples and $4$ -multiples, remove the $12$ -multiples

Solution:

Between $1$ and $2005$ there are $668$ multiples of $3,$ $501$ multiples of $4,$ and $167$ multiples of $12.$

Every multiple of $12$ is both a multiple of $3$ and of $4,$ so removing them from each group gives $\begin{aligned} &(668 - 167) \\ &\quad {}+ (501 - 167) = 835 \end{aligned}$ numbers that are multiples of $3$ or $4$ but not $12.$

Thus, C is the correct answer.

14.

Equilateral $\triangle ABC$ has side length $2,$ $M$ is the midpoint of $\overline{AC},$ and $C$ is the midpoint of $\overline{BD}.$ What is the area of $\triangle CDM?$

$\dfrac{\sqrt{2}}{2}$

$\dfrac{3}{4}$

$\dfrac{\sqrt{3}}{2}$

$1$

$\sqrt{2}$

Answer: C

Concepts:equilateral triangle triangle area midpoint

Difficulty rating: 1370

Small Hint:

Take $\overline{CD}$ as the base; it has length $2$

Big Hint:

The height from $M$ to line $BD$ is half the height of $\triangle ABC$

Solution:

Take $\overline{CD}$ as the base. Since $C$ is the midpoint of $\overline{BD}$ and $BC = 2,$ we have $CD = 2.$

The height of $\triangle CDM$ is the distance from $M$ to line $BD.$ Because $M$ is the midpoint of $\overline{AC},$ this distance is half the height of $\triangle ABC,$ which is $\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}.$

The area is $\dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}.$

Thus, C is the correct answer.

15.

An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $\$20$ or more?

$\dfrac14$

$\dfrac25$

$\dfrac37$

$\dfrac12$

$\dfrac23$

Answer: D

Concepts:basic probability combinations casework

Difficulty rating: 1370

Small Hint:

There are $\binom82$ equally likely pairs of bills

Big Hint:

A sum of $\$20$ or more needs both tens, both twenties, or a twenty with any other bill

Solution:

There are $\binom82 = 28$ equally likely pairs.

A sum of at least $\$20$ comes from both twenties ( $1$ way), a twenty paired with any of the six smaller bills ( $2 \cdot 6 = 12$ ways), or both tens ( $1$ way).

The probability is $\dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12.$

Thus, D is the correct answer.

16.

The quadratic equation $x^2 + mx + n = 0$ has roots that are twice those of $x^2 + px + m = 0,$ and none of $m,$ $n,$ and $p$ is zero. What is the value of $\dfrac{n}{p}?$

$1$

$2$

$4$

$8$

$16$

Answer: D

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1480

Small Hint:

Let the roots of $x^2+px+m=0$ be $r_1$ and $r_2,$ then use Vieta’s formulas on both equations

Big Hint:

Express $m,$ $n,$ and $p$ in terms of $r_1+r_2$ and $r_1r_2$

Solution:

Let $r_1$ and $r_2$ be the roots of $x^2 + px + m = 0,$ so $m = r_1 r_2$ and $p = -(r_1 + r_2).$

The roots of $x^2 + mx + n = 0$ are $2r_1$ and $2r_2,$ so $n = 4r_1 r_2$ and $-m = 2(r_1 + r_2).$

Then $n = 4m$ and $m = -2(r_1 + r_2) = 2p,$ so $p = \dfrac{m}{2}.$ Therefore $\dfrac{n}{p} = \dfrac{4m}{\frac{m}{2}} = 8.$

Thus, D is the correct answer.

17.

Suppose that $4^a = 5,$ $5^b = 6,$ $6^c = 7,$ and $7^d = 8.$ What is $a \cdot b \cdot c \cdot d?$

$1$

$\dfrac32$

$2$

$\dfrac52$

$3$

Answer: B

Concepts:exponent telescoping

Difficulty rating: 1480

Small Hint:

Raise successive equations to powers so the exponents multiply

Big Hint:

$4^{abcd}$ telescopes down to $8;$ write $8$ as a power of $4$

Solution:

Chaining the equations, $\begin{aligned} 4^{abcd} &= \left(\left(\left(4^a\right)^b\right)^c\right)^d \\ &= \left(\left(5^b\right)^c\right)^d \\ &= \left(6^c\right)^d = 7^d = 8. \end{aligned}$

Since $8 = 4^{\frac{3}{2}},$ we conclude $a \cdot b \cdot c \cdot d = \dfrac32.$

Thus, B is the correct answer.

18.

All of David’s telephone numbers have the form $555\text{–}abc\text{–}defg,$ where $a,$ $b,$ $c,$ $d,$ $e,$ $f,$ and $g$ are distinct digits and in increasing order, and none is either $0$ or $1.$ How many different telephone numbers can David have?

$1$

$2$

$7$

$8$

$9$

Answer: D

Concepts:combinations bijection

Difficulty rating: 1510

Small Hint:

The seven digits come from $\{2,3,4,5,6,7,8,9\},$ and their order is forced

Big Hint:

Once the omitted digit is chosen, every digit’s position in the phone number is determined

Solution:

The seven digits are chosen from $\{2, 3, 4, 5, 6, 7, 8, 9\},$ and once chosen they must be written in increasing order, so only the choice of digits matters.

Choosing seven of these eight digits is the same as choosing the one digit to leave out, which can be done in $8$ ways.

Thus, D is the correct answer.

19.

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?

$0$

$1$

$2$

$4$

$5$

Answer: B

Concepts:mean median (data)percentage

Difficulty rating: 1420

Small Hint:

The remaining percentage scored $95;$ find it first

Big Hint:

For the median, find which score the middle student earned using the cumulative percentages

Solution:

The percentage scoring $95$ is $100 - 10 - 25 - 20 - 15 = 30.$

The mean is $\begin{aligned} &0.10(70) + 0.25(80) \\ &\quad {}+ 0.20(85) + 0.15(90) \\ &\quad {}+ 0.30(95) = 86. \end{aligned}$

Since $35\%$ scored below $85$ and $35\%$ scored above $85,$ the middle student scored $85,$ so the median is $85.$

The difference is $86 - 85 = 1.$

Thus, B is the correct answer.

20.

What is the average (mean) of all $5$ -digit numbers that can be formed by using each of the digits $1,$ $3,$ $5,$ $7,$ and $8$ exactly once?

$48000$

$49999.5$

$53332.8$

$55555$

$56432.8$

Answer: C

Concepts:permutations place value symmetry

Difficulty rating: 1540

Small Hint:

By symmetry, each digit appears equally often in each place

Big Hint:

The average digit in each place is $\dfrac{1+3+5+7+8}{5},$ then multiply by $11111$

Solution:

By symmetry, each of the five digits appears equally often in each place, so the average digit in every place is $\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8.$

The average number is therefore $\begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8. \end{gathered}$

Thus, C is the correct answer.

21.

Forty slips are placed into a hat, each bearing a number $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9,$ or $10,$ with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \ne a.$ What is the value of $\dfrac{q}{p}?$

$162$

$180$

$324$

$360$

$720$

Answer: A

Concepts:basic probability combinations sampling without replacement

Difficulty rating: 1660

Small Hint:

Both probabilities share the denominator $\binom{40}{4},$ so only the counts matter

Big Hint:

Count four-of-a-kind draws, then draws of the form two $a$ ’s and two $b$ ’s

Solution:

Both events draw from $\binom{40}{4}$ equally likely selections, so $\dfrac{q}{p}$ is the ratio of their favorable counts.

Exactly $10$ draws give four slips of the same number, one for each value.

For two $a$ ’s and two $b$ ’s, choose the two values in $\binom{10}{2}$ ways, then two of the four $a$ -slips and two of the four $b$ -slips: $\begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620. \end{aligned}$

Therefore $\dfrac{q}{p} = \dfrac{1620}{10} = 162.$

Thus, A is the correct answer.

22.

For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \cdots + n?$

$8$

$12$

$16$

$17$

$21$

Answer: C

Concepts:triangular number factorial divisibility prime

Difficulty rating: 1990

Small Hint:

Use $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ and simplify $\dfrac{n!}{\frac{n(n+1)}{2}}$

Big Hint:

The reduced fraction $\dfrac{2(n-1)!}{n+1}$ fails to be an integer exactly when $n+1$ is an odd prime

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ divisibility is equivalent to $\dfrac{n!}{\frac{n(n+1)}{2}} = \dfrac{2(n-1)!}{n+1}$ being an integer.

Put $N=n+1.$ If $N$ is composite and not a square, it has two distinct proper factors whose product is $N;$ both occur in $(N-2)!=(n-1)!.$ If $N=k^2$ with $k\ge3,$ the factors $k$ and $2k$ occur in that factorial, so it contains a multiple of $2N.$ The remaining composite case, $N=4,$ also divides $2(N-2)!=4.$ Thus the fraction is an integer whenever $N$ is composite. If $N=n+1$ is an odd prime, it divides neither $(n-1)!$ nor $2,$ so the fraction is not an integer. The even prime $N=2$ gives $n=1,$ which works.

The odd primes at most $25$ are $3,$ $5,$ $7,$ $11,$ $13,$ $17,$ $19,$ $23,$ giving $8$ failing values of $n.$ Hence $24 - 8 = 16$ values work.

Thus, C is the correct answer.

23.

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC},$ $E$ as the midpoint of $\overline{BC},$ and $F$ as the midpoint of $\overline{DA}.$ The area of $ABEF$ is twice the area of $FECD.$ What is $\dfrac{AB}{DC}?$

$2$

$3$

$5$

$6$

$8$

Answer: C

Concepts:trapezoid area ratio midpoint

Difficulty rating: 1630

Small Hint:

The midsegment $\overline{FE}$ has length $\dfrac{AB+DC}{2}$

Big Hint:

The two smaller trapezoids share a height, so their areas compare like the averages of their parallel sides

Solution:

Let $AB = a$ and $DC = c.$ The midsegment $\overline{FE}$ has length $\dfrac{a + c}{2},$ and $ABEF$ and $FECD$ have the same height.

Their areas are proportional to the averages of their parallel sides, so $\dfrac{a + \frac{a+c}{2}}{\frac{a+c}{2} + c} = \dfrac{3a + c}{a + 3c} = 2.$

Then $3a + c = 2a + 6c,$ so $a = 5c$ and $\dfrac{AB}{DC} = 5.$

Thus, C is the correct answer.

24.

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x.$ The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m.$ What is $x + y + m?$

$88$

$112$

$116$

$144$

$154$

Answer: E

Concepts:digits difference of squares perfect square

Difficulty rating: 1880

Small Hint:

Write $x=10a+b$ and $y=10b+a,$ then factor $x^2-y^2$

Big Hint:

$x^2-y^2=99(a+b)(a-b)$ must be a perfect square, which forces $a+b=11$

Solution:

Write $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $\begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}$

Since $99 = 9 \cdot 11,$ for $m^2$ to be a perfect square we need $(a+b)(a-b)$ to be divisible by $11.$ As $a + b \le 17,$ this forces $a + b = 11,$ and then $a - b$ must itself be a perfect square.

With $a - b \le 8,$ the only workable case is $a - b = 1,$ giving $(a, b) = (6, 5).$ Then $x = 65,$ $y = 56,$ and $m^2 = 99 \cdot 11 = 33^2,$ so $m = 33.$

Therefore $x + y + m$ $= 65 + 56 + 33$ $= 154.$

Thus, E is the correct answer.

25.

A subset $B$ of the set of integers from $1$ to $100,$ inclusive, has the property that no two elements of $B$ sum to $125.$ What is the maximum possible number of elements in $B?$

$50$

$51$

$62$

$65$

$68$

Answer: C

Concepts:subsets pairing and grouping extremal argument

Difficulty rating: 1720

Small Hint:

Which integers can pair with another to sum to $125?$

Big Hint:

Separate the integers with no possible partner from the disjoint pairs whose members sum to $125$

Solution:

The pairs summing to $125$ are $(25, 100), (26, 99), \ldots, (62, 63),$ which is $62 - 25 + 1 = 38$ pairs. From each pair, $B$ may contain at most one element.

The numbers $1$ through $24$ cannot pair with anything in range to sum to $125,$ so all $24$ of them may be included.

Thus $B$ has at most $38 + 24 = 62$ elements, and the set $\{1, 2, \ldots, 62\}$ achieves this.

Thus, C is the correct answer.