2000 AMC 10 — Problems & Online Timed Exam

Timed

1:15:00

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 $I + M + O?$

$23$

$55$

$99$

$111$

$671$

Answer: E

Concepts:prime factorization optimization

Difficulty rating: 960

Small Hint:

Factor $2001$ into primes

Big Hint:

To make the sum large, keep one factor as small as possible so another can be as large as possible

Solution:

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

If one factor is $1,$ the possible pairs for the other two factors are $(3,667),$ $(23,87),$ and $(29,69).$ Their corresponding sums with $1$ are $671,$ $111,$ and $99.$ (The pair $(1,2001)$ would repeat the factor $1.$ )

If no factor is $1,$ all three prime factors must be split among the three integers, giving only $3,23,29$ and a much smaller sum. Therefore, the largest possible sum is $1 + 3 + 667 = 671.$

Thus, the correct answer is E.

2.

Which of the following is equal to $2000 \cdot 2000^{2000}?$

$2000^{2001}$

$4000^{2000}$

$2000^{4000}$

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

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

Answer: A

Concepts:exponent

Difficulty rating: 870

Small Hint:

Write the leading $2000$ as $2000^1$

Big Hint:

Multiplying powers with the same base adds the exponents

Solution:

Write the factor $2000$ as $2000^1.$ Then $\begin{aligned} 2000^1 \cdot 2000^{2000} &= 2000^{1 + 2000} \\ &= 2000^{2001}. \end{aligned}$

Each of the other options is larger than $2000^{2001}.$

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?

$40$

$50$

$55$

$60$

$75$

Answer: B

Concepts:percentage work backwards

Difficulty rating: 960

Small Hint:

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

Big Hint:

If $x$ is the original amount, then $(0.8)^2 x = 32$

Solution:

Since Jenny eats $20\%$ each day, $80\%$ remain at the end of each day.

If $x$ is the original number, then $(0.8)^2 x = 0.64x = 32,$ so $x = 50.$

Thus, the correct answer is B.

4.

Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $\$12.48,$ but in January her bill was $\$17.54$ because she used twice as much connect time as in December. What is the fixed monthly fee?

$\$2.53$

$\$5.06$

$\$6.24$

$\$7.42$

$\$8.77$

Answer: D

Concepts:system of equations linear equation

Difficulty rating: 1030

Small Hint:

The extra charge in January comes entirely from one additional December’s worth of connect time

Big Hint:

The difference $\$17.54 - \$12.48$ equals December’s connect-time cost

Solution:

January doubled only the connect time, so the increase $\$17.54 - \$12.48 = \$5.06$ equals December’s connect-time cost.

The fixed monthly fee is therefore $\$12.48 - \$5.06 = \$7.42.$

Thus, the correct answer is D.

5.

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB.$ As $P$ moves along a line that is parallel to side $AB,$ how many of the four quantities listed below change?

(a) the length of the segment $MN;$ (b) the perimeter of $\triangle PAB;$ (c) the area of $\triangle PAB;$ (d) the area of trapezoid $ABNM.$

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:triangle area midpoint parallel lines

Difficulty rating: 1240

Small Hint:

$MN$ is a midsegment of the triangle, so $MN = \tfrac12 AB$

Big Hint:

Since $P$ stays on a line parallel to $AB,$ the height from $P$ to $AB$ never changes

Solution:

Since $MN$ is a midsegment, $MN = \tfrac12 AB,$ which is fixed.

The base $AB$ and the height from $P$ to $AB$ are both constant as $P$ slides along the parallel line, so the area of $\triangle PAB$ does not change. The trapezoid $ABNM$ is the triangle minus $\triangle PMN,$ both of whose areas are constant, so its area does not change either.

Only the perimeter changes, since $PA$ and $PB$ vary as $P$ moves. So exactly one quantity changes.

Thus, the correct answer is B.

6.

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?

$0$

$4$

$6$

$7$

$9$

Answer: C

Concepts:Fibonacci units digit pattern recognition

Difficulty rating: 1240

Small Hint:

Track only the units digit of each term, adding the two previous units digits modulo $10$

Big Hint:

List the units digits and note which digit is slowest to first appear

Solution:

Recording only the units digits gives the sequence $\begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered}$

Scanning for the first appearance of each digit, the digit $6$ is the last of the ten digits to show up.

Thus, the correct answer is C.

7.

In rectangle $ABCD,$ $AD = 1,$ $P$ is on $\overline{AB},$ and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC.$ What is the perimeter of $\triangle BDP?$

$3 + \dfrac{\sqrt3}{3}$

$2 + \dfrac{4\sqrt3}{3}$

$2 + 2\sqrt2$

$\dfrac{3 + 3\sqrt5}{2}$

$2 + \dfrac{5\sqrt3}{3}$

Answer: B

Concepts:special right triangle rectangle perimeter

Difficulty rating: 1390

Small Hint:

Angle $ADC = 90^\circ$ is split into three $30^\circ$ angles

Big Hint:

Triangles $ADP$ and $ADB$ are both $30$ - $60$ - $90$ right triangles with leg $AD = 1$

Solution:

The right angle $\angle ADC = 90^\circ$ is trisected into three $30^\circ$ angles, so $\angle ADP = 30^\circ$ and $\angle ADB = 60^\circ.$

In right triangle $ADP,$ with $AD = 1,$ we get $DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}$ and $AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.$

In right triangle $ADB,$ with $AD = 1,$ we get $DB = \dfrac{1}{\cos 60^\circ} = 2$ and $AB = \tan 60^\circ = \sqrt3.$

Then $\begin{aligned} PB = AB - AP &= \sqrt3 - \dfrac{\sqrt3}{3} \\ &= \dfrac{2\sqrt3}{3}. \end{aligned}$

The perimeter of $\triangle BDP$ is $\begin{gathered} DP + PB + DB \\ = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 \\ = 2 + \dfrac{4\sqrt3}{3}. \end{gathered}$

Thus, the correct answer is B.

8.

At Olympic High School, $\tfrac25$ of the freshmen and $\tfrac45$ of the sophomores took the AMC $10.$ Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?

There are five times as many sophomores as freshmen.

There are twice as many sophomores as freshmen.

There are as many freshmen as sophomores.

There are twice as many freshmen as sophomores.

There are five times as many freshmen as sophomores.

Answer: D

Concepts:ratio and proportion linear equation

Difficulty rating: 1020

Small Hint:

Let $f$ and $s$ be the numbers of freshmen and sophomores, and set the two contestant counts equal

Big Hint:

Solve $\tfrac25 f = \tfrac45 s$ for $f$ in terms of $s$

Solution:

Let $f$ and $s$ be the numbers of freshmen and sophomores. The contestant counts are equal, so $\tfrac25 f = \tfrac45 s.$

Multiplying by $5$ gives $2f = 4s,$ so $f = 2s.$ There are twice as many freshmen as sophomores.

Thus, the correct answer is D.

9.

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

$-2$

$2$

$2 - 2p$

$2p - 2$

$|2p - 2|$

Answer: C

Concepts:absolute value algebraic manipulation

Difficulty rating: 1170

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 substitute into $x - p$

Solution:

Because $x \lt 2,$ 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.

10.

The sides of a triangle with positive area have lengths $4,$ $6,$ and $x.$ The sides of a second triangle with positive area have lengths $4,$ $6,$ and $y.$ What is the smallest positive number that is not a possible value of $|x - y|?$

$2$

$4$

$6$

$8$

$10$

Answer: D

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1370

Small Hint:

The triangle inequality forces each of $x$ and $y$ to lie strictly between $2$ and $10$

Big Hint:

Find the range of $|x - y|$ when $x$ and $y$ each range over that open interval

Solution:

By the triangle inequality, each of $x$ and $y$ can be any number strictly between $6 - 4 = 2$ and $6 + 4 = 10.$

Then $|x - y|$ can take any value with $0 \le |x - y| \lt 8.$

The smallest positive number not attainable is $10 - 2 = 8.$

Thus, the correct answer is D.

11.

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?

$21$

$60$

$119$

$180$

$231$

Answer: C

Concepts:prime parity bounding to limit cases

Difficulty rating: 1370

Small Hint:

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

Big Hint:

Note that $xy - (x + y)$ $= (x - 1)(y - 1) - 1,$ which is odd and increases with $x$ and $y$

Solution:

The primes between $4$ and $18$ are $5,$ $7,$ $11,$ $13,$ and $17.$ The product of two of them is odd and the sum is even, so $xy - (x + y)$ is odd.

Since $xy - (x + y)$ $= (x - 1)(y - 1) - 1$ increases as either prime increases, the result ranges from $5 \cdot 7 - 12 = 23$ up to $13 \cdot 17 - 30 = 191.$

The only odd option in $[23, 191]$ is $119 = 11 \cdot 13 - (11 + 13).$

Thus, the correct answer is C.

12.

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 $100?$

$10401$

$19801$

$20201$

$39801$

$40801$

Answer: C

Concepts:sum of first n odd numbers perfect square pattern recognition

Difficulty rating: 1240

Small Hint:

The differences $1, 5, 13, 25, \ldots$ grow by $4, 8, 12, \ldots$

Big Hint:

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

Solution:

Figure $n$ can be split into the sum of the first $n$ odd numbers and the first $n+1$ odd numbers, giving $n^2 + (n+1)^2$ unit squares.

For figure $100,$ this is $\begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201. \end{aligned}$

Thus, the correct answer is C.

13.

There are $5$ yellow pegs, $4$ red pegs, $3$ green pegs, $2$ blue pegs, and $1$ orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

$0$

$1$

$5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!$

$\frac{15!}{5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!}$

$15!$

Answer: B

Concepts:arrangements with restrictions logical deduction

Difficulty rating: 1370

Small Hint:

There are exactly five rows and five columns, and five yellow pegs

Big Hint:

Each color that has as many pegs as available lines must occupy exactly one peg in each such line, forcing its positions

Solution:

The board has five rows and five columns. To avoid two yellow pegs in a row or column, there must be exactly one yellow peg in each row, forcing the yellow pegs onto the long diagonal.

The four red pegs must then each go in rows $2$ through $5,$ and the only positions left force them into a single diagonal as well. Continuing with green, blue, and orange, every color is forced into a unique position.

Hence there is exactly one valid arrangement.

Thus, the correct answer is B.

14.

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?

$71$

$76$

$80$

$82$

$91$

Answer: C

Concepts:divisibility modular arithmetic mean

Difficulty rating: 1560

Small Hint:

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

Big Hint:

Use the residues of the scores modulo $3$ to pin down which three could be entered first

Solution:

The residues of $71, 76, 80, 82, 91$ modulo $3$ are $2, 1, 2, 1, 1.$ The sum of the first three scores must be divisible by $3,$ and the only such triple is $76 + 82 + 91 = 249,$ so the third score entered is $91$ and the first two are $76$ and $82.$

Since $249$ is one more than a multiple of $4,$ the fourth score must be three more than a multiple of $4,$ which only $71$ satisfies. That leaves $80$ as the fifth and last score.

Indeed $76, 158, 249, 320, 400$ are divisible by $1, 2, 3, 4, 5.$

Thus, the correct answer is C.

15.

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

$-2$

$-\dfrac12$

$\dfrac13$

$\dfrac12$

$2$

Answer: E

Concepts:algebraic manipulation substitution

Difficulty rating: 1420

Small Hint:

Combine $\dfrac{a}{b} + \dfrac{b}{a}$ over the common denominator $ab$

Big Hint:

Replace $ab$ with $a - b$ inside $(ab)^2$

Solution:

Over the common denominator $ab,$ $\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}.$

Substituting $ab = a - b$ gives $\dfrac{a^2 + b^2 - (a - b)^2}{ab} = \dfrac{2ab}{ab} = 2.$

Thus, the correct answer is E.

16.

The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E.$ Find the length of segment $AE.$

$\dfrac{4\sqrt5}{3}$

$\dfrac{5\sqrt5}{3}$

$\dfrac{12\sqrt5}{7}$

$2\sqrt5$

$\dfrac{5\sqrt{65}}{9}$

Answer: B

Concepts:coordinate geometry distance formula system of equations

Difficulty rating: 1690

Small Hint:

Assign coordinates $A = (0, 3),$ $B = (6, 0),$ $C = (4, 2),$ $D = (2, 0)$

Big Hint:

Find the equations of lines $AB$ and $CD$ and solve for their intersection $E$

Solution:

Place the points at $A = (0, 3),$ $B = (6, 0),$ $C = (4, 2),$ $D = (2, 0).$

Line $AB$ is $x + 2y = 6$ and line $CD$ is $x - y = 2.$ Solving simultaneously gives $E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right).$

Then $\begin{aligned} AE &= \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} \\ &= \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} \\ &= \dfrac{5\sqrt5}{3}. \end{aligned}$

Thus, the correct answer is B.

17.

Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?

$\$3.63$

$\$5.13$

$\$6.30$

$\$7.45$

$\$9.07$

Answer: D

Concepts:invariant modular arithmetic money

Difficulty rating: 1750

Small Hint:

A quarter for five nickels and a nickel for five pennies each leave the total value unchanged

Big Hint:

Only the penny-for-five-quarters trade changes the total, always adding $\$1.24$

Solution:

Trading a quarter for five nickels or a nickel for five pennies does not change the total value. Only trading a penny for five quarters changes it, adding $5 \cdot 25 - 1 = 124$ cents.

Starting from $1$ cent, Boris always has $1 + 124n$ cents for some nonnegative integer $n.$

Only $\$7.45$ has this form, since $745 = 1 + 124 \cdot 6.$

Thus, the correct answer is D.

18.

Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?

$24$

$27$

$39$

$40$

$42$

Answer: C

Concepts:area decomposition circle area estimation

Difficulty rating: 1820

Small Hint:

Split the visible region into the part inside the square and the part outside it

Big Hint:

Outside, the region is four $5 \times 1$ rectangles plus four quarter circles of radius $1$

Solution:

Inside the square, Charlyn sees everything except a central square of side $5 - 2 = 3,$ an area of $25 - 9 = 16$ square kilometers.

Outside the square, the region is four rectangles each $5 \times 1$ plus four quarter circles of radius $1,$ an area of $4 \cdot 5 + \pi = 20 + \pi$ square kilometers.

The total area is $36 + \pi \approx 39$ square kilometers.

Thus, the correct answer is C.

19.

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. The ratio of the area of the other small right triangle to the area of the square is

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

$m$

$1 - m$

$\dfrac{1}{4m}$

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

Answer: D

Difficulty rating: 1750

Small Hint:

Let the square have side $1,$ and let the two small triangles have legs matching the square’s sides

Big Hint:

The two small triangles are similar, so their leg lengths are reciprocals

Solution:

Let the square have side $1.$ One small triangle has legs $1$ and $r,$ with area $\tfrac12 r = m,$ so $r = 2m.$

The two small triangles are similar, so the other has legs $1$ and $\tfrac1r,$ with area $\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.$

Since the square has area $1,$ the desired ratio is $\dfrac{1}{4m}.$

Thus, the correct answer is D.

20.

Let $A,$ $M,$ and $C$ be nonnegative integers such that $A + M + C = 10.$ What is the maximum value of $\begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}$

$49$

$59$

$69$

$79$

$89$

Answer: C

Concepts:factoring optimization

Difficulty rating: 1820

Small Hint:

Add $1$ to each variable and expand $(A+1)(M+1)(C+1)$

Big Hint:

The expression equals $(A+1)(M+1)(C+1) - 11,$ so maximize a product of three positive integers summing to $13$

Solution:

Notice that $\begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11. \end{gathered}$

We maximize a product of three positive integers summing to $13.$ The most balanced split is $4, 4, 5,$ giving $4 \cdot 4 \cdot 5 = 80.$

The maximum is $80 - 11 = 69.$

Thus, the correct answer is C.

21.

If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?

I. All alligators are creepy crawlers.

II. Some ferocious creatures are creepy crawlers.

III. Some alligators are not creepy crawlers.

$\mathrm{I}$ only

$\mathrm{II}$ only

$\mathrm{III}$ only

$\mathrm{II}$ and $\mathrm{III}$ only

None must be true

Answer: B

Concepts:logical deduction Venn Diagram

Difficulty rating: 1510

Small Hint:

Some creepy crawlers are alligators, and every alligator is ferocious

Big Hint:

Test statements I and III by drawing a diagram where alligators sit entirely inside the creepy crawlers

Solution:

Some creepy crawlers are alligators, and all alligators are ferocious, so those creatures are both creepy crawlers and ferocious. Hence some ferocious creatures are creepy crawlers, making II true.

Statement I fails because not every alligator need be a creepy crawler, and III fails because it is possible that all alligators are creepy crawlers. Only II must hold.

Thus, the correct answer is B.

22.

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?

$3$

$4$

$5$

$6$

$7$

Answer: C

Concepts:weighted mean bounding to limit cases mixture ratio and proportion

Difficulty rating: 1900

Small Hint:

Angela’s cup is a fraction $\tfrac1n$ of everyone’s total drink

Big Hint:

Her cup is a mix of $\tfrac14$ of the milk and $\tfrac16$ of the coffee, so $\tfrac1n$ lies strictly between $\tfrac16$ and $\tfrac14$

Solution:

Let there be $n$ people, drinking $8n$ ounces total, split into milk $M$ and coffee $C.$ Angela drank one cup, so $\tfrac14 M + \tfrac16 C = \tfrac1n (M + C).$

The left side is a weighted average of $\tfrac14$ and $\tfrac16,$ so $\tfrac1n$ lies strictly between $\tfrac16$ and $\tfrac14.$ That forces $4 \lt n \lt 6,$ so $n = 5.$

Thus, the correct answer is C.

23.

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 $x?$

$3$

$6$

$9$

$17$

$20$

Answer: E

Concepts:mean median (data)arithmetic sequence casework

Difficulty rating: 1950

Small Hint:

The mode is $2$ and the mean is $\dfrac{25 + x}{7};$ the median depends on where $x$ falls

Big Hint:

Split into cases by the size of $x,$ and require the three ordered values to be equally spaced

Solution:

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

If $x \leq 2,$ the median is also $2.$ Two equal values cannot belong to a non-constant three-term arithmetic progression, so this case gives no solutions.

If $2 < x < 4,$ the values occur in the order $2 < x < \mu.$ They form an arithmetic progression exactly when $x-2 = \mu-x.$ Substituting for $\mu$ gives $x-2 = \dfrac{25+x}{7}-x,$ whose solution is $x=3.$

If $x \geq 4,$ the median is $4$ and $\mu>4.$ Thus the condition is $4-2 = \mu-4,$ so $\mu=6$ and $x=17.$

The two possible values are therefore $3$ and $17,$ whose sum is $20.$

Thus, the correct answer is E.

24.

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 $f(3z) = 7.$

$-\dfrac13$

$-\dfrac19$

$0$

$\dfrac59$

$\dfrac53$

Answer: B

Concepts:function substitution Vieta’s Formulas

Difficulty rating: 1690

Small Hint:

To make $\dfrac{x}{3} = 3z,$ set $x = 9z$

Big Hint:

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

Solution:

To evaluate $f(3z),$ set $\dfrac{x}{3} = 3z,$ so $x = 9z.$ Then $\begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1. \end{aligned}$

Setting this equal to $7$ gives $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac19.$

Thus, the correct answer is B.

25.

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

Answer: A

Concepts:date and time modular arithmetic

Difficulty rating: 1860

Small Hint:

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

Big Hint:

Count the days from day $300$ of year $N$ to day $200$ of year $N+1$ to decide whether $N$ is a leap year

Solution:

From day $300$ of year $N$ to day $200$ of year $N + 1$ is $(L - 300) + 200$ days, where $L$ is the length of year $N.$ If $N$ were not a leap year, this is $265 \equiv 6 \pmod 7,$ giving a Monday, not a Tuesday. So year $N$ is a leap year, and the count is $266 = 7 \cdot 38,$ consistent with Tuesday.

Then years $N - 1$ and $N + 1$ are not leap years.

The $100$ th day of year $N - 1$ precedes the Tuesday (day $300$ of year $N$ ) by $(365 - 100) + 300 = 565$ days. Since $565 = 7 \cdot 80 + 5,$ that day is $5$ days earlier in the week than Tuesday, which is a Thursday.

Thus, the correct answer is A.