2005 AMC 12B — 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:ratio and proportion money

Difficulty rating: 890

Small Hint:

Find the total cost and the total revenue separately

Big Hint:

There are $200$ groups of five bought and $500$ pairs sold

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 $500 - 400 = 100$ dollars.

Thus, the correct answer is A.

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: 980

Small Hint:

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

Big Hint:

The equation is $\dfrac{x}{100}\cdot x = 4$

Solution:

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

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

Thus, the correct answer is D.

3.

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?

$\dfrac{1}{5}$

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{2}{3}$

$\dfrac{4}{5}$

Answer: C

Concepts:fraction ratio and proportion

Difficulty rating: 1050

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 what one third of them cost, namely $3\cdot\dfrac15 = \dfrac35$ of her money.

She has $1 - \dfrac35 = \dfrac25$ of her money left.

Thus, the correct answer is C.

4.

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

Difficulty rating: 1050

Small Hint:

$80\%$ of $50$ quizzes is the number of A’s she needs

Big Hint:

She needs $40 - 22$ more A’s out of the remaining $20$ quizzes

Solution:

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

She has $22$ already, so she needs $40 - 22 = 18$ more of the remaining $20$ quizzes.

She can earn a lower grade on at most $20 - 18 = 2$ of them.

Thus, the correct answer is B.

5.

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

Difficulty rating: 1130

Small Hint:

The four quarter circles in 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 in a tile together form 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, the correct answer is A.

6.

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: 1350

Small Hint:

Drop the altitude from $C$ to line $AB$ ; it meets $AB$ at its midpoint $H$

Big Hint:

$CH^2 = 7^2 - 1^2,$ and $CD^2 = CH^2 + HD^2$ with $HD = 1 + BD$

Solution:

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $\triangle ABC$ is isosceles with $AC = BC,$ $H$ is the midpoint of $AB,$ so $AH = HB = 1.$

Then $CH^2 = 7^2 - 1^2 = 48.$ Applying the Pythagorean Theorem to $\triangle CHD$ with $HD = HB + BD = 1 + BD$ gives $8^2 = 48 + (1 + BD)^2,$ so $(1 + BD)^2 = 16.$

Therefore $1 + BD = 4,$ which means $BD = 3.$

Thus, the correct answer is A.

7.

What is the area enclosed by the graph of $|3x| + |4y| = 12?$

$6$

$12$

$16$

$24$

$25$

Answer: D

Concepts:absolute value rhombus

Difficulty rating: 1270

Small Hint:

Find the intercepts on the two axes

Big Hint:

The graph is a rhombus with diagonals along the axes

Solution:

Setting $y = 0$ gives $|3x| = 12,$ so $x = \pm 4.$ Setting $x = 0$ gives $|4y| = 12,$ so $y = \pm 3.$

The graph is a rhombus with vertices $(\pm 4, 0)$ and $(0, \pm 3),$ so its diagonals have lengths $8$ and $6.$

Its area is $\dfrac12 \cdot 8 \cdot 6 = 24.$

Thus, the correct answer is D.

8.

For how many values of $a$ is it true that the line $y = x + a$ passes through the vertex of the parabola $y = x^2 + a^2?$

$0$

$1$

$2$

$10$

infinitely many

Answer: C

Concepts:parabola quadratic

Difficulty rating: 1350

Small Hint:

The vertex of $y = x^2 + a^2$ is $(0, a^2)$

Big Hint:

Substitute the vertex into the line to get $a^2 = a$

Solution:

The vertex of the parabola $y = x^2 + a^2$ is $(0, a^2).$

The line $y = x + a$ passes through it exactly when $a^2 = 0 + a,$ that is $a^2 - a = 0.$

This gives $a = 0$ or $a = 1,$ so there are $2$ values.

Thus, the correct answer is C.

9.

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)

Difficulty rating: 1410

Small Hint:

The remaining percentage scored $95$ : $100 - 10 - 25 - 20 - 15$

Big Hint:

For the median, find where the cumulative percentage passes $50\%$

Solution:

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

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

Cumulatively, $10\%$ are below $80,$ $35\%$ are at or below $80,$ and $55\%$ are at or below $85.$ The middle scores fall at $85,$ so the median is $85.$

The difference is $86 - 85 = 1.$

Thus, the correct answer is B.

10.

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:digits recursion pattern recognition

Difficulty rating: 1440

Small Hint:

Compute the first several terms and look for a repeating cycle

Big Hint:

Once a value repeats, the later terms repeat with the same period; use the term index modulo that period

Solution:

The sequence begins $2005, 133, 55, 250, 133, \ldots$ since $2^3 + 0^3 + 0^3 + 5^3 = 133,$ $1^3 + 3^3 + 3^3 = 55,$ $5^3 + 5^3 = 250,$ and $2^3 + 5^3 + 0^3 = 133.$

After the initial $2005,$ the terms cycle through $133, 55, 250$ with period $3.$

Term $n$ for $n \ge 2$ is the $((n-2)\bmod 3)$ th entry of $133, 55, 250.$ Since $2005 - 2 = 2003 \equiv 2 \pmod 3,$ the $2005$ th term is $250.$

Thus, the correct answer is E.

11.

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?

$\dfrac{1}{4}$

$\dfrac{2}{7}$

$\dfrac{3}{7}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: D

Concepts:basic probability combinations casework

Difficulty rating: 1500

Small Hint:

There are $\binom{8}{2} = 28$ equally likely pairs

Big Hint:

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

Solution:

There are $\binom{8}{2} = 28$ equally likely pairs of bills.

The sum is $\$20$ or more in these cases: both twenties ( $1$ way), one twenty with one of the six smaller bills ( $2 \cdot 6 = 12$ ways), or both tens ( $1$ way).

That is $1 + 12 + 1 = 14$ favorable pairs, so the probability is $\dfrac{14}{28} = \dfrac12.$

Thus, the correct answer is D.

12.

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: 1530

Small Hint:

Let $r_1, r_2$ be the roots of $x^2 + px + m = 0$ ; the other equation has roots $2r_1, 2r_2$

Big Hint:

Write $m, n, p$ in terms of $r_1 r_2$ and $r_1 + r_2$ using Vieta’s formulas

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 $m = 2p,$ which gives $p = \dfrac{m}{2},$ so $\dfrac{n}{p} = \dfrac{4m}{\tfrac{m}{2}} = 8.$

Thus, the correct answer is D.

13.

Suppose that $4^{x_1} = 5,$ $5^{x_2} = 6,$ $6^{x_3} = 7,$ $\ldots,$ $127^{x_{124}} = 128.$ What is $x_1 x_2 \cdots x_{124}?$

$2$

$\dfrac{5}{2}$

$3$

$\dfrac{7}{2}$

$4$

Answer: D

Concepts:logarithm telescoping

Difficulty rating: 1570

Small Hint:

Each equation gives $x_k = \log_{k+3}(k+4)$

Big Hint:

The product telescopes to $\log_4 128$

Solution:

From $4^{x_1} = 5$ we get $x_1 = \log_4 5,$ and in general $x_k = \log_{k+3}(k+4).$

The product telescopes: $\begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned}$

Since $128 = 2^7$ and $4 = 2^2,$ this equals $\dfrac{7\log 2}{2\log 2} = \dfrac72.$

Thus, the correct answer is D.

14.

A circle having center $(0, k),$ with $k \gt 6,$ is tangent to the lines $y = x,$ $y = -x$ and $y = 6.$ What is the radius of this circle?

$6\sqrt{2} - 6$

$6$

$6\sqrt{2}$

$12$

$6 + 6\sqrt{2}$

Answer: E

Concepts:circle tangent line coordinate geometry

Difficulty rating: 1630

Small Hint:

Tangency to $y = 6$ with center $(0,k)$ and $k \gt 6$ gives $r = k - 6$

Big Hint:

The distance from $(0,k)$ to the line $y = x$ is $\dfrac{k}{\sqrt2},$ and it also equals $r$

Solution:

Since the circle is tangent to $y = 6$ and its center $(0, k)$ is above that line, the radius is $r = k - 6.$

The distance from $(0, k)$ to the line $x - y = 0$ is $\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2},$ and this must also equal $r.$

Setting $\dfrac{k}{\sqrt2} = k - 6$ gives $k = \dfrac{6\sqrt2}{\sqrt2 - 1}$ $= 6\sqrt2\,(\sqrt2+1)$ $= 12 + 6\sqrt2.$

Then $r = k - 6 = 6 + 6\sqrt2.$

Thus, the correct answer is E.

15.

The sum of four two-digit numbers is $221.$ None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:digits place value casework

Difficulty rating: 1660

Small Hint:

The eight distinct nonzero digits used have a total between $36$ and $44$

Big Hint:

If the units digits sum to $U$ and the tens digits to $T,$ then $10T + U = 221,$ so $U$ ends in $1$

Solution:

The eight digits are distinct and chosen from $1$ through $9,$ whose total is $45.$ So the eight used digits sum to between $45 - 9 = 36$ and $45 - 1 = 44.$

Let the four units digits sum to $U$ and the four tens digits sum to $T.$ Then $10T + U = 221,$ so $U$ ends in $1.$ Since $1+2+3+4 = 10$ $\le U \le$ $6+7+8+9 = 30,$ we have $U = 11$ or $U = 21.$

If $U = 11,$ then $10T = 210,$ so $T = 21$ and the eight digits sum to $32,$ which is below $36.$ So $U = 21,$ giving $T = 20$ and total $41.$

The missing digit is $45 - 41 = 4.$ For example, $13 + 25 + 86 + 97 = 221.$

Thus, the correct answer is D.

16.

Eight spheres of radius $1,$ one per octant, are each tangent to the coordinate planes. What is the radius of the smallest sphere, centered at the origin, that contains these eight spheres?

$\sqrt{2}$

$\sqrt{3}$

$1 + \sqrt{2}$

$1 + \sqrt{3}$

$3$

Answer: D

Concepts:3D geometry sphere distance formula

Difficulty rating: 1660

Small Hint:

A unit sphere tangent to all three coordinate planes in one octant has center $(\pm1, \pm1, \pm1)$

Big Hint:

Add the distance from the origin to a center and one more radius

Solution:

A sphere of radius $1$ tangent to the three coordinate planes in one octant has its center at a point like $(1, 1, 1),$ at distance $\sqrt{1^2 + 1^2 + 1^2} = \sqrt3$ from the origin.

The farthest point of that sphere from the origin is at distance $\sqrt3 + 1,$ so the containing sphere has radius $1 + \sqrt3.$

Thus, the correct answer is D.

17.

How many distinct four-tuples $(a, b, c, d)$ of rational numbers are there with $\begin{aligned} &a\log_{10} 2 + b\log_{10} 3 \\ &\quad {}+ c\log_{10} 5 + d\log_{10} 7 = 2005? \end{aligned}$

$0$

$1$

$17$

$2004$

infinitely many

Answer: B

Concepts:logarithm prime factorization

Difficulty rating: 1800

Small Hint:

Rewrite the left side as $\log_{10}\left(2^a 3^b 5^c 7^d\right)$

Big Hint:

Then $2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005}5^{2005}$ ; match exponents via unique factorization

Solution:

The equation is equivalent to $\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005,$ so $2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}.$

Clearing the denominators of $a, b, c, d$ with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: $a = 2005,$ $b = 0,$ $c = 2005,$ and $d = 0.$

So there is exactly $1$ such four-tuple.

Thus, the correct answer is B.

18.

Let $A(2, 2)$ and $B(7, 7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R?$

$25$

$39$

$51$

$60$

$80$

Answer: C

Concepts:coordinate geometry circle area decomposition

Difficulty rating: 1990

Small Hint:

Angle $A$ acute means $C$ is on the far side of the line through $A$ perpendicular to $AB$ ; similarly for $B$

Big Hint:

Angle $C$ acute means $C$ lies outside the circle with diameter $AB$ ; combine three regions

Solution:

Line $AB$ has slope $1.$ For $\angle A$ to be acute, $C$ must lie beyond the line through $A$ perpendicular to $AB;$ in the first quadrant that line runs between $P(4, 0)$ and $Q(0, 4).$ For $\angle B$ to be acute, $C$ must lie before the line through $B$ perpendicular to $AB,$ between $S(14, 0)$ and $T(0, 14).$

For $\angle C$ to be acute, $C$ must lie outside the circle $U$ with diameter $AB,$ whose radius is $\dfrac{AB}{2} = \dfrac{5\sqrt2}{2}.$

The circle lies entirely inside this strip and in the first quadrant. Thus the region is the large right triangle $OST$ minus the small right triangle $OPQ$ and the full circle $U:$ $\begin{aligned} &\dfrac12 \cdot 14^2 - \dfrac12 \cdot 4^2 \\ &\quad {}- \pi\left(\dfrac{5\sqrt2}{2}\right)^2 \\ &= 98 - 8 - \dfrac{25\pi}{2} \\ &= 90 - \dfrac{25\pi}{2} \approx 51. \end{aligned}$

Thus, the correct answer is C.

19.

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: 1840

Small Hint:

Write $x = 10a + b$ and $y = 10b + a$ ; then $x^2 - y^2 = 99(a^2 - b^2)$

Big Hint:

For $99(a-b)(a+b)$ to be a perfect square, $a+b$ must be $11$

Solution:

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

Since $99 = 9 \cdot 11,$ for this to be a perfect square we need $(a+b)(a-b)$ to be a multiple of $11.$ As $a + b \le 17$ and $a - b \le 8,$ the only multiple of $11$ available is $a + b = 11.$

Then $x^2 - y^2 = 9 \cdot 11^2 (a - b),$ which is a perfect square exactly when $a - b$ is a perfect square. Because $a+b=11$ is odd, $a-b$ is odd; and because $1 \le a-b \le 8,$ its only possible square value is $1.$ Hence $(a, b) = (6, 5).$

So $x = 65,$ $y = 56,$ and $m = \sqrt{65^2 - 56^2}$ $= \sqrt{1089} = 33.$ Thus $x + y + m$ $= 65 + 56 + 33 = 154.$

Thus, the correct answer is E.

20.

Let $a,$ $b,$ $c,$ $d,$ $e,$ $f,$ $g$ and $h$ be distinct elements in the set $\{-7, -5, -3, -2, 2, 4, 6, 13\}.$ What is the minimum possible value of $\begin{aligned} &(a + b + c + d)^2 \\ &\quad {}+ (e + f + g + h)^2? \end{aligned}$

$30$

$32$

$34$

$40$

$50$

Answer: C

Concepts:optimization completing the square

Difficulty rating: 1910

Small Hint:

The whole set sums to $8,$ so if one group sums to $x,$ the other sums to $8 - x$

Big Hint:

$x^2 + (8-x)^2 = 2(x-4)^2 + 32$ ; check whether $x = 4$ is actually attainable

Solution:

The elements sum to $8.$ If $a + b + c + d = x,$ then $e + f + g + h = 8 - x,$ so $x^2 + (8 - x)^2 = 2(x - 4)^2 + 32.$

This is minimized when $x = 4,$ giving $32.$ But $13$ must lie in one group, and no three of the remaining elements add with $13$ to make $4$ (that would need three of them to sum to $-9$ ). With $-7,$ the other two would need to sum to $-2,$ which no available pair does; without $-7,$ replacing any term in $-5-3-2=-10$ raises the sum past $-9.$ So $x = 4$ is unattainable and $(x - 4)^2 \ge 1.$

The minimum is $2(1) + 32 = 34,$ achieved for instance by $\{-7, -5, 2, 13\}$ (sum $3$ ) and $\{-3, -2, 4, 6\}$ (sum $5$ ).

Thus, the correct answer is C.

21.

A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n?$

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:factor counting prime factorization

Difficulty rating: 1990

Small Hint:

Write $n = 7^k Q$ with $Q$ not divisible by $7$ ; let $d$ be the number of divisors of $Q$

Big Hint:

Then $n$ has $(k+1)d$ divisors and $7n$ has $(k+2)d$ divisors

Solution:

Write $n = 7^k Q$ where $Q$ is not divisible by $7,$ and let $d$ be the number of divisors of $Q.$ Then $n$ has $(k + 1)d = 60$ divisors and $7n = 7^{k+1}Q$ has $(k + 2)d = 80$ divisors.

Dividing, $\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43,$ so $3(k + 2) = 4(k + 1),$ giving $k = 2.$

Thus, the correct answer is C.

22.

A sequence of complex numbers $z_0,$ $z_1,$ $z_2,$ $\ldots$ is defined by the rule $z_{n+1} = \dfrac{i z_n}{\overline{z_n}},$ where $\overline{z_n}$ is the complex conjugate of $z_n$ and $i^2 = -1.$ Suppose that $|z_0| = 1$ and $z_{2005} = 1.$ How many possible values are there for $z_0?$

$1$

$2$

$4$

$2005$

$2^{2005}$

Answer: E

Concepts:complex number roots of unity recursion

Difficulty rating: 2170

Small Hint:

Since $|z_n| = 1,$ $\overline{z_n} = \dfrac{1}{z_n},$ so $z_{n+1} = i z_n^2$

Big Hint:

Iterating gives $z_n$ as a fixed constant times $z_0^{2^n}$ ; count roots of the resulting equation

Solution:

Because $|z_0| = 1,$ every $|z_n| = 1,$ so $\overline{z_n} = \dfrac{1}{z_n}$ and $z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2.$

Iterating, $z_1 = i z_0^2$ and $z_2 = i(i z_0^2)^2 = -i z_0^4.$ Moreover, if $z_n=-i z_0^{2^n},$ then $z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}.$ Thus $z_n=-i z_0^{2^n}$ for every $n\ge2.$

The condition $z_{2005} = 1$ is therefore $z_0^{2^{2005}} = i.$ Every nonzero complex equation $z_0^{N} = i$ has exactly $N$ distinct solutions, all on the unit circle.

Here $N = 2^{2005},$ so there are $2^{2005}$ possible values for $z_0.$

Thus, the correct answer is E.

23.

Let $S$ be the set of ordered triples $(x, y, z)$ of real numbers for which $\log_{10}(x + y) = z$ and $\log_{10}(x^2 + y^2) = z + 1.$ There are real numbers $a$ and $b$ such that for all ordered triples $(x, y, z)$ in $S$ we have $x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a + b?$

$\dfrac{15}{2}$

$\dfrac{29}{2}$

$15$

$\dfrac{39}{2}$

$24$

Answer: B

Concepts:sum and difference of cubes logarithm symmetry (algebra)

Difficulty rating: 2110

Small Hint:

Rewrite the conditions as $x + y = 10^z$ and $x^2 + y^2 = 10 \cdot 10^z$

Big Hint:

Use $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$ with $xy = \tfrac12\left[(x+y)^2 - (x^2+y^2)\right]$

Solution:

The conditions give $x + y = 10^z$ and $x^2 + y^2 = 10 \cdot 10^z.$ Then $\begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned}$ so $xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).$

Using $x^3 + y^3$ $= (x+y)^3 - 3xy(x+y),$ $\begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}. \end{aligned}$

So $a = -\dfrac12$ and $b = 15,$ giving $a + b = \dfrac{29}{2}.$ These coefficients are determined: setting $z=0$ gives real solutions $x,y=\dfrac{1\pm\sqrt{19}}{2},$ and their equation forces this same value of $a+b.$

Thus, the correct answer is B.

24.

All three vertices of an equilateral triangle are on the parabola $y = x^2,$ and one of its sides has a slope of $2.$ The $x$ -coordinates of the three vertices have a sum of $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n?$

$14$

$15$

$16$

$17$

$18$

Answer: A

Concepts:parabola trigonometric identity equilateral triangle

Difficulty rating: 2300

Small Hint:

The chord joining $(a, a^2)$ and $(b, b^2)$ has slope $a + b$

Big Hint:

The three side slopes are $\tan\theta$ and $\tan(\theta \pm 60^\circ)$ ; their sum is twice the vertex-sum

Solution:

For vertices $(a, a^2), (b, b^2), (c, c^2),$ the slope of a side is $\dfrac{b^2 - a^2}{b - a} = a + b.$ Adding the three side slopes, $\begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n}. \end{aligned}$

One side has slope $2 = \tan\theta.$ Because the triangle is equilateral, its sides make angles $\theta$ and $\theta \pm 60^\circ,$ so the other two slopes are $\begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11}. \end{aligned}$

The sum of the three slopes is $2 - \dfrac{8 + 5\sqrt3}{11}$ $- \dfrac{8 - 5\sqrt3}{11} =$ $\dfrac{22 - 16}{11} = \dfrac{6}{11}.$

Thus $a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11},$ so $m + n = 3 + 11 = 14.$

Thus, the correct answer is A.

25.

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?

$\dfrac{5}{256}$

$\dfrac{21}{1024}$

$\dfrac{11}{512}$

$\dfrac{23}{1024}$

$\dfrac{3}{128}$

Answer: A

Concepts:basic probability arrangements with restrictions casework

Difficulty rating: 2520

Small Hint:

There are $4^6$ move combinations; each vertex is non-adjacent only to its opposite

Big Hint:

A valid final assignment is a permutation with no vertex sent to itself or its opposite; split by whether opposite vertices’ images are opposite or adjacent

Solution:

There are $4^6$ equally likely combinations of moves. Label the vertices $A, B, C, A', B', C',$ where primed vertices are opposite the corresponding unprimed ones. An ant cannot move to its own vertex or the opposite one, so a valid outcome is a permutation $f$ with $f(A) \notin \{A, A'\},$ and similarly for each pair.

There are $4 \cdot 3 = 12$ ordered choices for $(f(A), f(A')).$ Of these, $f(A)$ and $f(A')$ are opposite in $4$ cases and adjacent in $8.$

If $f(A), f(A')$ are opposite, say $B, B',$ then $\{f(C), f(C')\} = \{A, A'\}$ and $\{f(B), f(B')\} = \{C, C'\},$ giving $4 \cdot 2 \cdot 2 = 16$ valid combinations.

If $f(A), f(A')$ are adjacent, say $B, C,$ then one of $f(B), f(B')$ must be $C'$ and there are $4$ ordered choices for $(f(B), f(B')),$ each leaving $2$ for $(f(C), f(C')):$ that is $8 \cdot 4 \cdot 2 = 64$ valid combinations.

Hence the probability is $\dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}.$

Thus, the correct answer is A.