2000 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

In the year $2001,$ the United States will host the International Mathematical Olympiad. Let $I,$ $M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001.$ What is the largest possible value of the sum $I + M + O?$

$23$

$55$

$99$

$111$

$671$

Concepts:prime factorization optimization

Difficulty rating: 1000

Small Hint:

Factor $2001$ into primes

Big Hint:

To maximize the sum, make one factor as large as possible and the others small

Solution:

Factoring gives $2001 = 3 \cdot 23 \cdot 29.$

To maximize the sum of three distinct positive integers with this product, take $I = 1,$ $M = 3,$ and $O = 23 \cdot 29 = 667.$

The largest sum is $1 + 3 + 667 = 671.$

Thus, the correct answer is E.

2.

What is $2000(2000^{2000})?$

$2000^{2001}$

$4000^{2000}$

$2000^{4000}$

$4{,}000{,}000^{2000}$

$2000^{4{,}000{,}000}$

Concepts:exponent

Difficulty rating: 950

Small Hint:

Write the leading factor as $2000^1$

Big Hint:

Multiplying powers of the same base adds the exponents

Solution:

Writing $2000 = 2000^1,$ we get $\begin{gathered} 2000^1 \cdot 2000^{2000} \\ = 2000^{1 + 2000} \\ = 2000^{2001}. \end{gathered}$

All of the other options are larger than this.

Thus, the correct answer is A.

3.

Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?

$40$

$50$

$55$

$60$

$75$

Concepts:percentage work backwards

Difficulty rating: 1080

Small Hint:

Eating $20\%$ leaves $80\%$ at the end of each day

Big Hint:

After two days, $(0.8)^2$ of the original amount remains

Solution:

Since $20\%$ is eaten each day, $80\%$ remains at the end of each day. If $x$ is the original number, then $(0.8)^2 x = 32.$

Solving gives $0.64x = 32,$ so $x = 50.$

Thus, the correct answer is B.

4.

The Fibonacci sequence $1,$ $1,$ $2,$ $3,$ $5,$ $8,$ $13,$ $21,$ $\ldots$ starts with two $1$ s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

$0$

$4$

$6$

$7$

$9$

Concepts:Fibonacci units digit systematic listing

Difficulty rating: 1240

Small Hint:

Track only the units digit of each term

Big Hint:

List the units digits until all ten digits have appeared, and note which shows up last

Solution:

The sequence of units digits begins $\begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered}$

Scanning this list, the digit $6$ is the last of the ten digits to appear.

Thus, the correct answer is C.

5.

If $|x - 2| = p,$ where $x \lt 2,$ then what is $x - p?$

$-2$

$2$

$2 - 2p$

$2p - 2$

$|2p - 2|$

Concepts:absolute value

Difficulty rating: 1150

Small Hint:

Since $x \lt 2,$ the quantity $x - 2$ is negative, so $|x - 2| = 2 - x$

Big Hint:

Solve $2 - x = p$ for $x,$ then compute $x - p$

Solution:

Since $x \lt 2,$ we have $|x - 2| = 2 - x = p,$ so $x = 2 - p.$

Then $x - p = (2 - p) - p = 2 - 2p.$

Thus, the correct answer is C.

6.

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$21$

$60$

$119$

$180$

$231$

Concepts:prime parity factoring

Difficulty rating: 1310

Small Hint:

The primes are $5, 7, 11, 13,$ and $17$ ; note that $xy - (x + y)$ $= (x - 1)(y - 1) - 1$

Big Hint:

Both primes are odd, so $(x-1)(y-1)-1 \equiv 3 \pmod 4;$ then test the remaining candidates

Solution:

The primes between $4$ and $18$ are $5, 7, 11, 13,$ and $17.$

For two such primes, $xy - (x + y)$ $= (x - 1)(y - 1) - 1$ is a product of two even numbers minus $1,$ hence it is $3 \pmod 4.$ This leaves $119$ and $231.$ The latter would require $(x-1)(y-1)=232,$ but no two distinct numbers in $\{4,6,10,12,16\}$ have product $232.$

Indeed, $\begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119. \end{aligned}$

Thus, the correct answer is C.

7.

How many positive integers $b$ have the property that $\log_b 729$ is a positive integer?

$0$

$1$

$2$

$3$

$4$

Concepts:logarithm factor counting perfect power

Difficulty rating: 1370

Small Hint:

If $\log_b 729 = n,$ then $b^n = 729 = 3^6$

Big Hint:

For $b$ to be an integer, $n$ must be a positive divisor of $6$

Solution:

If $\log_b 729 = n,$ then $b^n = 729 = 3^6,$ so $n$ must be a positive divisor of $6.$

The possibilities $n = 1, 2, 3, 6$ give $b = 3^6, 3^3, 3^2, 3^1,$ that is, $729, 27, 9,$ and $3.$

There are $4$ such values of $b.$

Thus, the correct answer is E.

8.

Figures $0,$ $1,$ $2,$ and $3$ consist of $1,$ $5,$ $13,$ and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure $100?$

$10401$

$19801$

$20201$

$39801$

$40801$

Concepts:pattern recognition perfect square

Difficulty rating: 1370

Small Hint:

Each figure is a diamond whose rows are the odd numbers $1, 3, 5, \ldots$ up and back down

Big Hint:

Figure $n$ has $n^2 + (n + 1)^2$ unit squares

Solution:

Figure $n$ is a diamond whose row lengths increase through the odd numbers and back down, giving a total of $n^2 + (n + 1)^2$ unit squares. This matches $1, 5, 13, 25$ for $n = 0, 1, 2, 3.$

Therefore figure $100$ has $\begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned}$ unit squares.

Thus, the correct answer is C.

9.

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were $71,$ $76,$ $80,$ $82,$ and $91.$ What was the last score Mrs. Walter entered?

$71$

$76$

$80$

$82$

$91$

Concepts:divisibility modular arithmetic mean

Difficulty rating: 1580

Small Hint:

The sum of the first $k$ scores entered must be divisible by $k$

Big Hint:

Use divisibility by $3$ then by $4$ ; the residues of the scores modulo $3$ are $2, 1, 2, 1, 1$

Solution:

The total is $71 + 76 + 80 + 82 + 91 = 400,$ which is divisible by $5.$ The sum of the first three scores must be divisible by $3.$

Modulo $3,$ the scores are $2, 1, 2, 1, 1.$ The only triple summing to a multiple of $3$ is $76 + 82 + 91 = 249,$ so these are the first three (with $91$ third, since the first two, $76$ and $82,$ must have equal parity).

Since $249 \equiv 1 \pmod 4,$ the fourth score must be $\equiv 3 \pmod 4,$ which is $71.$ That leaves $80$ as the last score entered.

Thus, the correct answer is C.

10.

The point $P = (1, 2, 3)$ is reflected in the $xy$ -plane, then its image $Q$ is rotated by $180^\circ$ about the $x$ -axis to produce $R,$ and finally, $R$ is translated by $5$ units in the positive $y$ direction to produce $S.$ What are the coordinates of $S?$

$(1, 7, -3)$

$(-1, 7, -3)$

$(-1, -2, 8)$

$(-1, 3, 3)$

$(1, 3, 3)$

Concepts:transformation 3D geometry coordinate geometry

Difficulty rating: 1390

Small Hint:

Reflecting in the $xy$ -plane negates the $z$ -coordinate

Big Hint:

A $180^\circ$ turn about the $x$ -axis negates both the $y$ - and $z$ -coordinates

Solution:

Reflecting $(1, 2, 3)$ in the $xy$ -plane gives $Q = (1, 2, -3).$

Rotating $180^\circ$ about the $x$ -axis negates $y$ and $z,$ giving $R = (1, -2, 3).$

Translating $5$ units in the positive $y$ direction gives $S = (1, 3, 3).$

Thus, the correct answer is E.

11.

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b.$ Find a possible value of $\frac{a}{b} + \frac{b}{a} - ab.$

$-2$

$-\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$2$

Concepts:algebraic manipulation substitution

Difficulty rating: 1530

Small Hint:

Combine over a common denominator: $\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}$

Big Hint:

Replace $ab$ with $a - b$ in the numerator and expand

Solution:

Combining over a common denominator, $\frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}.$

Replacing $ab$ with $a - b$ in the numerator, $\begin{gathered} a^2 + b^2 - (a - b)^2 \\ = a^2 + b^2 \\ {}- (a^2 - 2ab + b^2) \\ = 2ab. \end{gathered}$

Therefore the expression equals $\dfrac{2ab}{ab} = 2.$

Thus, the correct answer is E.

12.

Let $A,$ $M,$ and $C$ be nonnegative integers such that $A + M + C = 12.$ What is the maximum value of $\begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}$

$62$

$72$

$92$

$102$

$112$

Concepts:factoring AM-GM Inequality optimization

Difficulty rating: 1650

Small Hint:

$AMC + AM + MC + CA$ $= (A + 1)(M + 1)(C + 1)$ $- (A + M + C) - 1$

Big Hint:

The three factors $A+1, M+1, C+1$ sum to $15,$ so their product is largest when they are equal

Solution:

Observe that $\begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1. \end{aligned}$

Since $A + M + C = 12,$ this equals $(A + 1)(M + 1)(C + 1) - 13.$ The three factors sum to $15,$ so their product is maximized when each equals $5,$ giving $5^3 = 125.$

The maximum value is $125 - 13 = 112.$

Thus, the correct answer is E.

13.

One morning each member of Angela’s family drank an $8$ -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

$3$

$4$

$5$

$6$

$7$

Concepts:fraction linear equation bounding to limit cases

Difficulty rating: 1710

Small Hint:

Suppose Angela’s own cup contains $c$ units of coffee and $m$ units of milk with $c + m = 1$

Big Hint:

Then the family total is $6c$ coffee and $4m$ milk, so the number of people is $6c + 4m$

Solution:

Measure amounts in $8$ -ounce cups, so Angela’s cup holds $c$ coffee and $m$ milk with $c + m = 1.$

Since Angela drank a sixth of the coffee, the total coffee is $6c$ ; since she drank a quarter of the milk, the total milk is $4m.$ The number of people equals the total number of cups, $\begin{aligned} 6c + 4m &= 6c + 4(1 - c) \\ &= 4 + 2c. \end{aligned}$

This is an integer only when $2c$ is an integer, and since $0 \lt c \lt 1$ this forces $c = \tfrac12,$ giving $4 + 1 = 5$ people.

Thus, the correct answer is C.

14.

When the mean, median, and mode of the list $10, 2, 5, 2, 4, 2, x$

are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x?$

$3$

$6$

$9$

$17$

$20$

Concepts:mean median (data)mode casework

Difficulty rating: 1840

Small Hint:

The mode is $2,$ and the mean is $\dfrac{25 + x}{7}$

Big Hint:

Split into cases by where $x$ falls relative to the other numbers, and require a non-constant progression

Solution:

The six fixed numbers sum to $25,$ so the mean is $\dfrac{25 + x}{7},$ and the mode is $2.$ If $x \le 2,$ then $2$ is both median and mode, forcing a constant progression, so $x \gt 2.$

Case $2 \lt x \lt 4$ : the median is $x.$ Requiring $2, x, \dfrac{25 + x}{7}$ to form an arithmetic progression yields $x = 3$ as the only value in this range.

Case $x \ge 4$ : the median is $4,$ and the progression $2, 4, 6$ forces the mean to be $6,$ so $\dfrac{25 + x}{7} = 6,$ giving $x = 17.$

The sum of all possible values is $3 + 17 = 20.$

Thus, the correct answer is E.

15.

Let $f$ be a function for which $f\!\left(\dfrac{x}{3}\right) = x^2 + x + 1.$ Find the sum of all values of $z$ for which $f(3z) = 7.$

$-\dfrac{1}{3}$

$-\dfrac{1}{9}$

$0$

$\dfrac{5}{9}$

$\dfrac{5}{3}$

Concepts:function substitution Vieta’s Formulas

Difficulty rating: 1650

Small Hint:

To evaluate $f(3z),$ set $\dfrac{x}{3} = 3z,$ so $x = 9z$

Big Hint:

This gives a quadratic in $z$ ; use the sum-of-roots formula $-\dfrac{b}{a}$

Solution:

Setting $\dfrac{x}{3} = 3z$ gives $x = 9z,$ so $\begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned}$

This rearranges to $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac{1}{9}.$

Thus, the correct answer is B.

16.

A checkerboard of $13$ rows and $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,$ $2,$ $\ldots,$ $17,$ the second row $18,$ $19,$ $\ldots,$ $34,$ and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,$ $2,$ $\ldots,$ $13,$ the second column $14,$ $15,$ $\ldots,$ $26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).

$222$

$333$

$444$

$555$

$666$

Concepts:Diophantine Equation systematic listing

Difficulty rating: 1770

Small Hint:

The square in row $m,$ column $n$ is numbered $17(m - 1) + n$ originally and $13(n - 1) + m$ after

Big Hint:

Setting the two expressions equal gives $4m - 3n = 1$ ; find all valid $(m, n)$

Solution:

The square $(m, n)$ is numbered $17(m - 1) + n$ originally and $13(n - 1) + m$ after renumbering. Setting these equal gives $4m - 3n = 1.$

The solutions with $1 \le m \le 13$ and $1 \le n \le 17$ are $(1, 1),$ $(4, 5),$ $(7, 9),$ $(10, 13),$ and $(13, 17).$

These squares hold the numbers $1, 56, 111, 166,$ and $221,$ whose sum is $555.$

Thus, the correct answer is D.

17.

A circle centered at $O$ has radius $1$ and contains the point $A.$ Segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta.$ If point $C$ lies on $\overline{OA}$ and $BC$ bisects $\angle ABO,$ then what is $OC?$

$\sec^2\theta - \tan\theta$

$\dfrac{1}{2}$

$\dfrac{\cos^2\theta}{1 + \sin\theta}$

$\dfrac{1}{1 + \sin\theta}$

$\dfrac{\sin\theta}{\cos^2\theta}$

Concepts:angle bisector theorem tangent line trigonometry

Difficulty rating: 1870

Small Hint:

Since $OA = 1$ and $\angle OAB = 90^\circ,$ we have $BA = \tan\theta$ and $OB = \sec\theta$

Big Hint:

The bisector gives $\dfrac{OC}{CA} = \dfrac{OB}{BA},$ and $OC + CA = OA = 1$

Solution:

Because $OA = 1$ and $AB$ is tangent at $A,$ angle $OAB$ is right, so $BA = \tan\theta, \qquad OB = \sec\theta.$

Since $BC$ bisects $\angle ABO,$ the angle bisector theorem gives $\dfrac{OC}{CA} = \dfrac{OB}{BA}.$ Using $OC + CA = OA = 1,$ $\begin{aligned} OC &= \frac{OB}{OB + BA} \\ &= \frac{\sec\theta}{\sec\theta + \tan\theta}. \end{aligned}$

Multiplying numerator and denominator by $\cos\theta$ gives $OC = \dfrac{1}{1 + \sin\theta}.$

Thus, the correct answer is D.

18.

In year $N,$ the $300$ th day of the year is a Tuesday. In year $N + 1,$ the $200$ th day is also a Tuesday. On what day of the week did the $100$ th day of year $N - 1$ occur?

Thursday

Friday

Saturday

Sunday

Monday

Concepts:date and time modular arithmetic

Difficulty rating: 1870

Small Hint:

Two dates fall on the same weekday exactly when the number of days between them is a multiple of $7$

Big Hint:

Use the two given Tuesdays to decide which of $N - 1, N, N + 1$ is a leap year

Solution:

From day $300$ of year $N$ to day $200$ of year $N + 1,$ the number of days is $365 - 300 + 200 = 265$ if $N$ is not a leap year. But $265 = 7 \cdot 37 + 6,$ which would land on a Monday, not a Tuesday.

So year $N$ is a leap year, and the gap is $266 = 7 \cdot 38$ days, giving a Tuesday as stated. It follows that year $N - 1$ is not a leap year.

The $100$ th day of year $N - 1$ precedes the Tuesday on day $300$ of year $N$ by $365 - 100 + 300 = 565$ days. Since $565 = 7 \cdot 80 + 5,$ that day is $5$ weekdays before Tuesday, namely Thursday.

Thus, the correct answer is A.

19.

In triangle $ABC,$ $AB = 13,$ $BC = 14,$ and $AC = 15.$ Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC.$ Which of the following is closest to the area of triangle $ADE?$

$2$

$2.5$

$3$

$3.5$

$4$

Concepts:Heron’s Formula angle bisector theorem triangle area

Difficulty rating: 1810

Small Hint:

By Heron’s formula the area of $\triangle ABC$ is $84,$ so the altitude from $A$ to $BC$ is $12$

Big Hint:

$D$ is at distance $7$ from $B$ ; the bisector foot $E$ satisfies $BE : EC = 13 : 15$

Solution:

By Heron’s formula, the area of $\triangle ABC$ is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the altitude from $A$ to $BC$ is $\dfrac{2 \cdot 84}{14} = 12.$

The midpoint $D$ is $7$ from $B.$ The bisector from $A$ meets $BC$ at $E$ with $BE : EC = AB : AC = 13 : 15,$ so $BE = 14 \cdot \dfrac{13}{28} = 6.5.$

Both $D$ and $E$ lie on $BC,$ so $\triangle ADE$ has base $DE = 7 - 6.5 = 0.5$ and altitude $12,$ giving area $\tfrac12 \cdot 0.5 \cdot 12 = 3.$

Thus, the correct answer is C.

20.

If $x,$ $y,$ and $z$ are positive numbers satisfying $x + \frac{1}{y} = 4,$ $y + \frac{1}{z} = 1,$ and $z + \frac{1}{x} = \frac{7}{3},$ then what is $xyz?$

$\dfrac{2}{3}$

$1$

$\dfrac{4}{3}$

$2$

$\dfrac{7}{3}$

Concepts:system of equations algebraic manipulation symmetry (algebra)

Difficulty rating: 1970

Small Hint:

Add the three equations, and separately multiply all three together

Big Hint:

Expanding the product produces $xyz + \dfrac{1}{xyz}$ plus the sum you already computed

Solution:

Adding the three equations gives $\begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered}$

Multiplying them gives $4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.$

Expanding the product, $\begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned}$ The middle group is the sum $\tfrac{22}{3},$ so $xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.$

Hence $(xyz - 1)^2 = 0,$ so $xyz = 1.$

Thus, the correct answer is B.

21.

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. What is the ratio of the area of the other small right triangle to the area of the square?

$\dfrac{1}{2m + 1}$

$m$

$1 - m$

$\dfrac{1}{4m}$

$\dfrac{1}{8m^2}$

Concepts:similarity area ratio

Difficulty rating: 1970

Small Hint:

Let the square have side $1$ ; one small triangle has a leg $r$ along the square, so its area is $\dfrac{r}{2} = m$

Big Hint:

The two small triangles are similar, so the other triangle’s matching leg is $\dfrac{1}{r}$

Solution:

Let the square have side $1.$ The small triangle sharing one side of the square has a perpendicular leg $r,$ so its area is $\tfrac12 \cdot 1 \cdot r = m,$ giving $r = 2m.$

The two small triangles are similar, so the other triangle’s leg along the square is $\dfrac1r,$ and its area is $\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}.$

Thus, the correct answer is D.

22.

The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest?

$P(-1)$

The product of the zeros of $P$

The product of the non-real zeros of $P$

The sum of the coefficients of $P$

The sum of the real zeros of $P$

Concepts:polynomial Vieta’s Formulas complex number

Difficulty rating: 2030

Small Hint:

The graph shows exactly two real zeros, both positive, so $P$ has two non-real (conjugate) zeros

Big Hint:

The product of all zeros equals $d,$ the $y$ -intercept; divide it by the product of the real zeros

Solution:

The graph crosses the $x$ -axis exactly twice, both times at positive values, so $P$ has two real zeros and two non-real (complex conjugate) zeros.

Reading off the graph: the sum of the coefficients is $P(1) \gt 3;$ $P(-1) \gt 4;$ the sum of the real zeros is greater than $4.5;$ and the product of all zeros is $d,$ the $y$ -intercept, which is less than $6.$

Let $R$ be the product of the real zeros, so $R \gt 4.5.$ The product of the non-real zeros is $\dfrac dR,$ which is less than $\dfrac{6}{4.5} \lt 2.$

This is smaller than every other listed quantity.

Thus, the correct answer is C.

23.

Professor Gamble buys a lottery ticket, which requires that he pick six different integers from $1$ through $46,$ inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property -- the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$1$

Concepts:logarithm prime factorization basic probability

Difficulty rating: 2330

Small Hint:

The sum of logarithms is an integer exactly when the product of the six numbers is a power of $10$

Big Hint:

Only numbers of the form $2^a 5^b$ can appear; balance the total factors of $2$ and $5$

Solution:

The sum of the logarithms is an integer $k$ exactly when the product of the six numbers is $10^k.$ Since $10 = 2 \cdot 5,$ each chosen number must be of the form $2^a 5^b,$ so it comes from $1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40.$

For each, record the excess of factors of $2$ over factors of $5$ : $0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2.$ The product is a power of $10$ only if the six chosen values have equal totals of $2$ s and $5$ s, i.e. their excesses sum to $0.$

The only negative excesses are $-1$ for $5$ and $-2$ for $25.$ A six-number ticket with total excess $0$ must contain both: omitting either leaves too few zero and small positive excesses to reach six numbers. The other four numbers must therefore have total excess $3.$ There are two numbers of excess $0$ ( $1,10$ ), two of excess $1$ ( $2,20$ ), and two of excess $2$ ( $4,40$ ); any number of excess at least $3$ would leave too few zeros to complete a four-number selection. Thus we must take both excess- $0$ numbers, one excess- $1$ number, and one excess- $2$ number. This gives exactly four valid tickets: $\{1, 5, 10, 20, 25, 40\},$ $\{1, 2, 5, 10, 25, 40\},$ $\{1, 2, 4, 5, 10, 25\},$ and $\{1, 4, 5, 10, 20, 25\}.$

Professor Gamble holds one of these four, and only one matches the winning ticket, so the probability is $\dfrac14.$

Thus, the correct answer is B.

24.

If circular arcs $AC$ and $BC$ have centers at $B$ and $A,$ respectively, then there exists a circle tangent to both arc $AC$ and arc $BC,$ and to $\overline{AB}.$ If the length of arc $BC$ is $12,$ then what is the circumference of the circle?

$24$

$25$

$26$

$27$

$28$

Concepts:power of a point tangent circles arc

Difficulty rating: 2390

Small Hint:

Both arcs have radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral

Big Hint:

Use Power of a Point at $A$ with $AD = \tfrac12 AB$ to find the small circle’s radius $r = \tfrac38 AB$

Solution:

Each arc has radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral. Thus arc $BC$ subtends $60^\circ$ of a circle of radius $AB,$ whose full circumference is $6 \cdot 12 = 72.$

Let the small circle have radius $r$ and touch $\overline{AB}$ at its midpoint $D,$ where $AD = \tfrac12 AB.$ By Power of a Point, $AD^2 = AB(AB - 2r),$ so $\frac{AB^2}{4} = AB^2 - 2r\,AB,$ giving $2r = \tfrac34 AB,$ hence $r = \tfrac38 AB.$

The circumferences are in the ratio of the radii, so the small circle’s circumference is $\tfrac38 \cdot 72 = 27.$

Thus, the correct answer is D.

25.

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)

$210$

$560$

$840$

$1260$

$1680$

Concepts:Burnside’s Lemma permutations symmetry

Difficulty rating: 2440

Small Hint:

There are $8!$ ways to place the eight distinct colors before accounting for rotations

Big Hint:

The rotation group of the octahedron has $24$ elements, and distinct colors give each arrangement a trivial stabilizer

Solution:

There are $8!$ ways to assign the eight distinct colors to the eight faces. Two assignments give the same octahedron exactly when one is a rotation of the other.

The rotation group of a regular octahedron has $24$ elements. Because all eight colors are different, no nontrivial rotation fixes a coloring, so each distinguishable octahedron corresponds to exactly $24$ assignments.

Therefore the number of distinguishable octahedrons is $\frac{8!}{24} = \frac{40320}{24} = 1680.$

Thus, the correct answer is E.