2014 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is $10 \cdot \left(\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10}\right)^{-1}?$

$3$

$8$

$\dfrac{25}{2}$

$\dfrac{170}{3}$

$170$

Answer: C

Concepts:fraction exponent

Difficulty rating: 870

Small Hint:

Add the three fractions inside the parentheses first

Big Hint:

$\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{8}{10}$ , then take the reciprocal and multiply by $10$

Solution:

The sum inside the parentheses is $\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}.$

Its reciprocal is $\dfrac54,$ so the expression equals $10\cdot\dfrac54=\dfrac{25}{2}.$

Thus, the correct answer is C.

2.

At the theater children get in for half price. The price for $5$ adult tickets and $4$ child tickets is $\$24.50.$ How much would $8$ adult tickets and $6$ child tickets cost?

$\$35$

$\$38.50$

$\$40$

$\$42$

$\$42.50$

Answer: B

Concepts:ratio and proportion

Difficulty rating: 1020

Small Hint:

A child ticket costs half an adult ticket, so measure everything in adult tickets

Big Hint:

$5+\tfrac12\cdot4=7$ adult tickets cost $\$24.50$ ; find the cost of $8+\tfrac12\cdot6=11$ adult tickets

Solution:

Since a child ticket is half an adult ticket, $5$ adult and $4$ child tickets equal $5+\tfrac12\cdot4=7$ adult tickets, so one adult ticket costs $\dfrac{24.50}{7}=\$3.50.$

The second purchase equals $8+\tfrac12\cdot6=11$ adult tickets, costing $11\cdot 3.50=\$38.50.$

Thus, the correct answer is B.

3.

Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:arrangements with restrictions casework

Difficulty rating: 1200

Small Hint:

Split into cases by whether the orange house or the blue house comes first

Big Hint:

Because blue and yellow cannot be adjacent, at least one house must separate them

Solution:

If orange comes first, then blue and yellow cannot be adjacent, forcing the order orange, blue, red, yellow.

If blue comes first, yellow can be in the third or fourth position (never second, to avoid adjacency), giving blue, orange, yellow, red and blue, orange, red, yellow.

These are the only $3$ valid orderings.

Thus, the correct answer is B.

4.

Suppose that $a$ cows give $b$ gallons of milk in $c$ days. At this rate, how many gallons of milk will $d$ cows give in $e$ days?

$\dfrac{bde}{ac}$

$\dfrac{ac}{bde}$

$\dfrac{abde}{c}$

$\dfrac{bcde}{a}$

$\dfrac{abc}{de}$

Answer: A

Concepts:rate ratio and proportion

Difficulty rating: 1200

Small Hint:

Find how much milk one cow gives in one day

Big Hint:

One cow gives $\dfrac{b}{ac}$ gallons per day; scale by $d$ cows and $e$ days

Solution:

The rate is $\dfrac{b}{ac}$ gallons per cow per day.

So $d$ cows over $e$ days produce $\dfrac{b}{ac}\cdot d\cdot e=\dfrac{bde}{ac}$ gallons.

Thus, the correct answer is A.

5.

On an algebra quiz, $10\%$ of the students scored $70$ points, $35\%$ scored $80$ points, $30\%$ scored $90$ points, and the rest scored $100$ points. What is the difference between the mean and the median of the students’ scores on this quiz?

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:mean median (data)percentage

Difficulty rating: 1270

Small Hint:

The median is the score of the middle student after ordering everyone

Big Hint:

Compute the weighted mean $0.10(70)+0.35(80)$ $+0.30(90)+0.25(100)$ and compare it to the median

Solution:

The remaining $25\%$ scored $100.$ Since $45\%$ scored at most $80$ and $75\%$ scored at most $90,$ the median is $90.$

The mean is $\begin{aligned} &0.10(70)+0.35(80)\\ &\quad{}+0.30(90)+0.25(100)\\ &\quad{}=7+28+27+25=87. \end{aligned}$

The difference is $90-87=3.$

Thus, the correct answer is C.

6.

The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two-digit number and its reverse?

$44$

$55$

$77$

$99$

$110$

Answer: D

Concepts:place value digits

Difficulty rating: 1270

Small Hint:

Write the number as $10a+b$ and its reverse as $10b+a$

Big Hint:

The difference is $9(a-b)$ , and setting it equal to $5(a+b)$ gives $2a=7b$

Solution:

Let the larger number be $10a+b.$ Then $\begin{aligned} &(10a+b)-(10b+a)=9(a-b)\\ &\quad{}=5(a+b), \end{aligned}$ which simplifies to $2a=7b.$

The only nonzero digits satisfying this are $a=7$ and $b=2,$ so the number is $72$ and its reverse is $27.$

Their sum is $72+27=99.$

Thus, the correct answer is D.

7.

The first three terms of a geometric progression are $\sqrt{3},$ $\sqrt[3]{3},$ and $\sqrt[6]{3}.$ What is the fourth term?

$1$

$\sqrt[7]{3}$

$\sqrt[8]{3}$

$\sqrt[9]{3}$

$\sqrt[10]{3}$

Answer: A

Concepts:geometric sequence exponent

Difficulty rating: 1340

Small Hint:

Rewrite each term as a power of $3$ : $3^{\frac{1}{2}},$ $3^{\frac{1}{3}},$ $3^{\frac{1}{6}}$

Big Hint:

The common ratio is $3^{\frac{1}{3}-\frac{1}{2}}=3^{-\frac{1}{6}}$ ; multiply the third term by it

Solution:

Writing the terms as powers of $3,$ they are $3^{\frac{1}{2}},$ $3^{\frac{1}{3}},$ $3^{\frac{1}{6}}.$ The common ratio is $\dfrac{3^{\frac{1}{3}}}{3^{\frac{1}{2}}}=3^{-\frac{1}{6}}.$

The fourth term is $3^{\frac{1}{6}}\cdot3^{-\frac{1}{6}}=3^{0}=1.$

Thus, the correct answer is A.

8.

A customer who intends to purchase an appliance has three coupons, only one of which may be used:

Coupon $1:$ $10\%$ off the listed price if the listed price is at least $\$50$

Coupon $2:$ $\$20$ off the listed price if the listed price is at least $\$100$

Coupon $3:$ $18\%$ off the amount by which the listed price exceeds $\$100$

For which of the following listed prices will coupon $1$ offer a greater price reduction than either coupon $2$ or coupon $3?$

$\$179.95$

$\$199.95$

$\$219.95$

$\$239.95$

$\$259.95$

Answer: C

Concepts:inequality percentage

Difficulty rating: 1440

Small Hint:

Let $P$ be the listed price and write each coupon’s reduction as an expression in $P$

Big Hint:

Coupon $1$ beats coupon $2$ when $P\gt200$ and beats coupon $3$ when $P\lt225$

Solution:

For a price $P\gt100,$ the reductions are $\dfrac{P}{10},$ $20,$ and $\dfrac{18}{100}(P-100).$

Coupon $1$ beats coupon $2$ when $\dfrac{P}{10}\gt20,$ that is $P\gt200.$ Coupon $1$ beats coupon $3$ when $\dfrac{P}{10}\gt\dfrac{18}{100}(P-100),$ that is $P\lt225.$

The only listed price in $(200,225)$ is $\$219.95.$

Thus, the correct answer is C.

9.

Five positive consecutive integers starting with $a$ have average $b.$ What is the average of $5$ consecutive integers that start with $b?$

$a+3$

$a+4$

$a+5$

$a+6$

$a+7$

Answer: B

Concepts:mean arithmetic sequence

Difficulty rating: 1270

Small Hint:

The average of consecutive integers is the middle one

Big Hint:

From the first set $b=a+2$ , and the new average is $b+2$

Solution:

The integers $a,a+1,a+2,a+3,a+4$ have average $a+2,$ so $b=a+2.$

The integers starting at $b$ have average $b+2=(a+2)+2=a+4.$

Thus, the correct answer is B.

10.

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1.$ The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

$\dfrac{\sqrt3}{4}$

$\dfrac{\sqrt3}{3}$

$\dfrac{2}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{\sqrt3}{2}$

Answer: B

Concepts:equilateral triangle triangle area Pythagorean Theorem

Difficulty rating: 1560

Small Hint:

Each isosceles triangle has base $1$ ; let its height be $h$

Big Hint:

Three triangles of area $\tfrac12 h$ sum to $\tfrac{\sqrt3}{4}$ ; solve for $h$ , then use the Pythagorean theorem

Solution:

The equilateral triangle has area $\dfrac{\sqrt3}{4}.$ Each isosceles triangle has base $1$ and height $h,$ so $3\cdot\dfrac12 h=\dfrac{\sqrt3}{4},$ giving $h=\dfrac{\sqrt3}{6}.$

A congruent side is the hypotenuse from the apex to a base endpoint: $\begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered}$

Thus, the correct answer is B.

11.

David drives from his home to the airport to catch a flight. He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed. He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early. How many miles is the airport from his home?

$140$

$175$

$210$

$245$

$280$

Answer: C

Concepts:distance rate and time system of equations

Difficulty rating: 1440

Small Hint:

Let $t$ be the time still needed after the first hour, and write the remaining distance two ways

Big Hint:

The leftover distance $d$ satisfies $d=35(t+1)$ and $d=50\left(t-\tfrac12\right)$

Solution:

Let $d$ be the remaining distance after one hour and $t$ the remaining time until the flight. At $35$ mph he would be an hour late, so $d=35(t+1).$ At $50$ mph he is half an hour early, so $d=50\left(t-\tfrac12\right).$

Setting these equal gives $35t+35=50t-25,$ so $t=4$ and $d=175.$

The total distance is $175+35=210$ miles.

Thus, the correct answer is C.

12.

Two circles intersect at points $A$ and $B.$ The minor arcs $AB$ measure $30^\circ$ on one circle and $60^\circ$ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?

$2$

$1+\sqrt3$

$3$

$2+\sqrt3$

$4$

Answer: D

Concepts:chord trigonometry area ratio

Difficulty rating: 1630

Small Hint:

The common chord $AB$ equals $2R\sin15^\circ$ in one circle and $2r\sin30^\circ$ in the other

Big Hint:

The area ratio is $\left(\dfrac{R}{r}\right)^2=\dfrac{1}{4\sin^2 15^\circ}$ , and $4\sin^2 15^\circ=2-\sqrt3$

Solution:

Let the circles have radii $R$ (with the $30^\circ$ arc) and $r$ (with the $60^\circ$ arc). The common chord has length $2R\sin15^\circ=2r\sin30^\circ,$ so $\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.$

The smaller central angle gives the larger radius, so $R\gt r.$ The area ratio is $\begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered}$

Thus, the correct answer is D.

13.

A fancy bed and breakfast inn has $5$ rooms, each with a distinctive color-coded decor. One day $5$ friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than $2$ friends per room. In how many ways can the innkeeper assign the guests to the rooms?

$2100$

$2220$

$3000$

$3120$

$3125$

Answer: B

Concepts:casework combinations multiplication principle

Difficulty rating: 1660

Small Hint:

Split into cases by the room occupancies: all singles, one pair, or two pairs

Big Hint:

Count each case as (ways to form the groups) times (ways to place the groups into distinct rooms)

Solution:

All singles: assign $5$ friends to $5$ rooms in $5!=120$ ways.

One pair: choose the pair in $\binom52=10$ ways, then place the $4$ groups into rooms in $5\cdot4\cdot3\cdot2=120$ ways, giving $10\cdot120=1200.$

Two pairs: choose the solo friend in $5$ ways and split the rest into two pairs in $3$ ways ( $15$ groupings), then place the $3$ groups into rooms in $5\cdot4\cdot3=60$ ways, giving $15\cdot60=900.$

The total is $120+1200+900=2220.$

Thus, the correct answer is B.

14.

Let $a\lt b\lt c$ be three integers such that $a,$ $b,$ $c$ is an arithmetic progression and $a,$ $c,$ $b$ is a geometric progression. What is the smallest possible value for $c?$

$-2$

$1$

$2$

$4$

$6$

Answer: C

Concepts:arithmetic sequence geometric sequence Diophantine Equation

Difficulty rating: 1630

Small Hint:

Let the common difference be $d$ : $b=a+d$ and $c=a+2d$

Big Hint:

Since $a,c,b$ is geometric, $c^2=ab$ ; this simplifies to $3a+4d=0$

Solution:

Let $d=b-a\gt0,$ so $b=a+d$ and $c=a+2d.$ Since $a,c,b$ is geometric, $\begin{gathered} \dfrac{c}{a}=\dfrac{b}{c}\\ \implies(a+2d)^2=a(a+d), \end{gathered}$ which simplifies to $3ad+4d^2=0,$ so $3a+4d=0.$

Then $a=-4k$ and $d=3k$ for a positive integer $k,$ giving $c=a+2d=2k.$ The smallest value is $c=2$ (with $a=-4,$ $b=-1,$ $c=2$ ).

Thus, the correct answer is C.

15.

A five-digit palindrome is a positive integer with respective digits $abcba,$ where $a$ is not zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S?$

$9$

$18$

$27$

$36$

$45$

Answer: B

Concepts:palindrome place value summation

Difficulty rating: 1660

Small Hint:

A palindrome $\overline{abcba}$ equals $10001a+1010b+100c$

Big Hint:

Each $a$ appears in $10\cdot10$ palindromes, each $b$ in $9\cdot10$ , each $c$ in $9\cdot10$ ; combine with the digit sum $45$

Solution:

Write $\overline{abcba}=10001a+1010b+100c.$ Summing over all palindromes, each value of $a\in\{1,\dots,9\}$ occurs with $10\cdot10=100$ choices of $b,c,$ and each value of $b$ or $c$ occurs with $9\cdot10=90$ choices of the other two digits.

Using $\sum a=\sum b=\sum c=45,$ $\begin{gathered} \scriptsize S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)\\ =45\cdot1{,}100{,}000\\ =49{,}500{,}000. \end{gathered}$

The sum of the digits of $S$ is $4+9+5=18.$

Thus, the correct answer is B.

16.

The product $(8)(888\ldots8),$ where the second factor has $k$ digits, is an integer whose digits have a sum of $1000.$ What is $k?$

$901$

$911$

$919$

$991$

$999$

Answer: D

Concepts:digits pattern recognition small cases

Difficulty rating: 1660

Small Hint:

Multiply small cases: $8\cdot8=64,$ $8\cdot88=704,$ $8\cdot888=7104$

Big Hint:

The product is $7\underbrace{1\cdots1}_{k-2}04,$ whose digit sum is $11+(k-2)$

Solution:

By carrying out the multiplication, $8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04,$ which has $k-2$ ones.

The digit sum is $7+(k-2)+0+4=k+9.$ Setting $k+9=1000$ gives $k=991.$

Thus, the correct answer is D.

17.

A $4\times4\times h$ rectangular box contains a sphere of radius $2$ and eight smaller spheres of radius $1.$ The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is $h?$

$2+2\sqrt7$

$3+2\sqrt5$

$4+2\sqrt7$

$4\sqrt5$

$4\sqrt7$

Answer: A

Concepts:3D geometry sphere Pythagorean Theorem

Difficulty rating: 1800

Small Hint:

The four top small-sphere centers form a square of side $2$ , directly below the big sphere’s center

Big Hint:

With center distance $3$ (sum of radii) and horizontal offset $\sqrt2$ , the vertical gap is $\sqrt{3^2-\sqrt2^2}=\sqrt7$ ; then add $1$ at the top and bottom

Solution:

Place the box with a corner at the origin. Each small sphere sits in a corner with center $1$ unit from three faces. The four top small-sphere centers form a square of side $2,$ whose center lies on the box axis; a corner of that square is $\sqrt2$ from the center.

The big sphere’s center is on the axis, at distance $2+1=3$ from each top small center. The vertical gap between them is $\sqrt{3^2-\sqrt2^2}=\sqrt7.$

The big center is at height $\dfrac h2$ and the top small centers at height $h-1,$ so $(h-1)-\dfrac h2=\sqrt7,$ giving $\dfrac h2=1+\sqrt7$ and $h=2+2\sqrt7.$

Thus, the correct answer is A.

18.

The domain of the function $\tiny f(x)=\log_{\frac{1}{2}}\!\left(\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)\right)$ is an interval of length $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$19$

$31$

$271$

$319$

$511$

Answer: C

Concepts:logarithm inequality

Difficulty rating: 1910

Small Hint:

Peel the logarithms from the outside in, requiring each argument to be positive

Big Hint:

Remember that bases below $1$ reverse inequalities; translate each successive positivity condition before moving to the next inner logarithm

Solution:

Working from the outside, $f$ is defined exactly when $\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)$ $\gt0,$ which is equivalent to $\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\gt1.$

Since the base $\tfrac14\lt1,$ this means $0\lt\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\lt\tfrac14,$ hence $1\lt\log_{\frac{1}{16}}x\lt16^{\frac{1}{4}}=2.$

As $\tfrac{1}{16}\lt1,$ this reverses to $\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1,$ i.e. $\tfrac{1}{256}\lt x\lt\tfrac{1}{16}.$ The length is $\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256},$ so $m+n=15+256=271.$

Thus, the correct answer is C.

19.

There are exactly $N$ distinct rational numbers $k$ such that $|k|\lt200$ and $5x^2+kx+12=0$ has at least one integer solution for $x.$ What is $N?$

$6$

$12$

$24$

$48$

$78$

Answer: E

Concepts:quadratic bounding to limit cases

Difficulty rating: 1990

Small Hint:

If $x$ is an integer root, solve for $k=-\dfrac{5x^2+12}{x}$ $=-\left(5x+\dfrac{12}{x}\right)$

Big Hint:

Require $|k|=5|x|+\dfrac{12}{|x|}\lt200$ , and check the resulting values of $k$ are all distinct

Solution:

If an integer $x$ is a root, then $k=-\left(5x+\dfrac{12}{x}\right),$ so $x\ne0.$ For $x\ge2,$ $|k|=5|x|+\dfrac{12}{|x|}$ increases, and $|x|=39$ gives $|k|\approx195.3\lt200,$ while $|x|=40$ gives $|k|\gt200.$

Thus $x$ ranges over $\pm1,\pm2,\dots,\pm39,$ which is $78$ values. If two integers $a\ne b$ gave the same $k,$ then $5a+\tfrac{12}{a}=5b+\tfrac{12}{b}$ forces $5ab=12,$ which has no integer solutions, so all $78$ values of $k$ are distinct.

Thus, the correct answer is E.

20.

In $\triangle BAC,$ $\angle BAC=40^\circ,$ $AB=10,$ and $AC=6.$ Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC},$ respectively. What is the minimum possible value of $BE+DE+CD?$

$6\sqrt3+3$

$\dfrac{27}{2}$

$8\sqrt3$

$14$

$3\sqrt3+9$

Answer: D

Concepts:reflection (geometry)law of cosines optimization

Difficulty rating: 2110

Small Hint:

Reflect $B$ over line $AC$ to $B'$ and $C$ over line $AB$ to $C'$

Big Hint:

The path becomes $B'E+ED+DC'$ , minimized as the straight segment $B'C'$ ; use the Law of Cosines with $\angle B'AC'=120^\circ$

Solution:

Reflect $B$ across line $AC$ to get $B',$ and reflect $C$ across line $AB$ to get $C'.$ Then $BE=B'E$ and $CD=C'D,$ so $BE+DE+CD$ $=B'E+ED+DC',$ a broken path from $B'$ to $C'.$

This is minimized when the path is the straight segment $B'C'.$ We have $AB'=AB=10,$ $AC'=AC=6,$ and $\angle B'AC'=3\cdot40^\circ=120^\circ.$

By the Law of Cosines, $\begin{gathered} B'C'^2=10^2+6^2\\ {}-2\cdot10\cdot6\cos120^\circ\\ =136+60\\ =196, \end{gathered}$ so $B'C'=14.$

Thus, the correct answer is D.

21.

For every real number $x,$ let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x,$ and let $f(x)=\lfloor x\rfloor\left(2014^{\,x-\lfloor x\rfloor}-1\right).$ The set of all numbers $x$ such that $1\le x\lt2014$ and $f(x)\le1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?

$1$

$\dfrac{\log2015}{\log2014}$

$\dfrac{\log2014}{\log2013}$

$\dfrac{2014}{2013}$

$2014^{\frac{1}{2014}}$

Answer: A

Concepts:floor and ceiling functions logarithm telescoping

Difficulty rating: 2170

Small Hint:

Write $x=n+r$ with $n=\lfloor x\rfloor$ and $0\le r\lt1,$ so $f(x)=n\left(2014^{\,r}-1\right)$

Big Hint:

The condition $f(x)\le1$ gives $0\le r\le\log_{2014}\dfrac{n+1}{n}$ ; sum these lengths and telescope

Solution:

Write $x=n+r$ with integer $n$ ( $1\le n\le2013$ ) and $0\le r\lt1.$ Then $f(x)=n\left(2014^{\,r}-1\right),$ and $f(x)\le1$ becomes $2014^{\,r}\le1+\dfrac1n,$ i.e. $0\le r\le\log_{2014}\dfrac{n+1}{n}.$

Each $n$ contributes an interval of length $\log_{2014}\dfrac{n+1}{n},$ so the total is $\begin{gathered} \sum_{n=1}^{2013}\log_{2014}\dfrac{n+1}{n}\\ =\log_{2014}\!\left(\dfrac21\cdot\dfrac32\cdots\dfrac{2014}{2013}\right)\\ =\log_{2014}2014=1. \end{gathered}$

Thus, the correct answer is A.

22.

The number $5^{867}$ is between $2^{2013}$ and $2^{2014}.$ How many pairs of integers $(m,n)$ are there such that $1\le m\le2012$ and $5^n\lt2^m\lt2^{m+2}\lt5^{n+1}?$

$278$

$279$

$280$

$281$

$282$

Answer: B

Concepts:exponent system of equations

Difficulty rating: 2270

Small Hint:

Between consecutive powers $5^n$ and $5^{n+1}$ there are either two or three powers of $2$

Big Hint:

The inequality asks for three such powers; if $d,t$ count the two-power and three-power gaps, then $d+t=867$ and $2d+3t=2013$

Solution:

Because $2^2\lt5\lt2^3,$ each interval $(5^n,5^{n+1})$ contains either two or three powers of $2.$ The chain $5^n\lt2^m\lt2^{m+2}\lt5^{n+1}$ holds exactly when the interval contains three consecutive powers of $2,$ and then there is a unique such $m.$

Let $d$ and $t$ be the numbers of intervals $(5^n,5^{n+1})$ for $0\le n\le866$ containing two and three powers of $2,$ respectively. Since $2^{2013}\lt5^{867}\lt2^{2014}$ there are $2013$ powers of $2$ in total, giving $d+t=867$ and $2d+3t=2013.$

Solving, $t=2013-2\cdot867=279.$

Thus, the correct answer is B.

23.

The fraction $\dfrac{1}{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},$ where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}?$

$874$

$883$

$887$

$891$

$892$

Answer: B

Concepts:repeating decimal digits pattern recognition

Difficulty rating: 2380

Small Hint:

Since $\dfrac{1}{99^2}=\dfrac{1}{9801}$ , the repeating block satisfies $10^n-1=9801\cdot\overline{b_{n-1}\ldots b_0}$

Big Hint:

Group the repeating decimal into base- $100$ digits and use $(100-1)^{-2};$ follow the carries until the first base- $100$ digit repeats

Solution:

Reading the block in pairs of digits (base $100$ ), $\dfrac{1}{9801}=\dfrac{1}{99^2}$ expands as $00,01,02,\ldots,$ since $\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.$ Let $a_j$ be the $j$ th base- $100$ digit of the repeating block. Multiplying the block by $99^2$ shows first that $a_0=99.$ The resulting carry gives $a_1=97,$ after which there is no carry and successively $a_j=98-j$ for $1\le j\le98.$ The next digit is again $99,$ so the period is $00,01,02,\ldots,96,97,99,$ with $98$ omitted.

If the blocks $00$ through $99$ all appeared, the digit sum would be $(0+1+\cdots+9)\cdot20=900.$ Removing the missing $98$ subtracts $9+8,$ giving $900-9-8=883.$

Thus, the correct answer is B.

24.

Let $f_0(x)=x+|x-100|$ $-|x+100|,$ and for $n\ge1,$ let $f_n(x)=|f_{n-1}(x)|-1.$ For how many values of $x$ is $f_{100}(x)=0?$

$299$

$300$

$301$

$302$

$303$

Answer: C

Concepts:absolute value recursion counting intersections

Difficulty rating: 2520

Small Hint:

If $f_0(x)=\pm k$ for a nonnegative integer $k,$ then $f_k(x)=0$ and afterward the values cycle $0,-1,0,-1,\ldots$

Big Hint:

So $f_{100}(x)=0$ needs $f_0(x)=2k$ with $-50\le k\le50$ ; graph $f_0$ and count intersections with each line $y=2k$

Solution:

If $f_{n-1}(x)=\pm k,$ then $f_n(x)=k-1.$ So if $f_0(x)=\pm k$ for a nonnegative integer $k,$ then $f_k(x)=0,$ after which the sequence alternates $0,-1,0,\ldots$ Thus $f_{100}(x)=0$ exactly when $f_0(x)=2k$ for some integer $-50\le k\le50.$

Now $f_0(x)=x+|x-100|$ $-|x+100|$ equals $x+200$ for $x\lt-100,$ $-x$ for $-100\le x\lt100,$ and $x-200$ for $x\ge100.$ Its graph is piecewise linear with turning points $(-100,100)$ and $(100,-100).$

A line $y=2k$ meets this graph three times for $-49\le k\le49$ and twice for $k=\pm50.$ The total is $99\cdot3+2\cdot2=301.$

Thus, the correct answer is C.

25.

The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3).$ For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\le1000?$

$38$

$40$

$42$

$44$

$46$

Answer: B

Concepts:parabola lattice point Diophantine Equation

Difficulty rating: 2650

Small Hint:

$(0,0)$ is the midpoint of $(4,3)$ and $(-4,-3),$ so this chord is the latus rectum and the directrix is parallel to it

Big Hint:

The directrix is $4y-3x+25=0$ ; parametrize the lattice points and reduce $|4x+3y|$ to $|50u+25|\le1000$

Solution:

Since $(0,0)$ is the midpoint of $A=(4,3)$ and $B=(-4,-3),$ the segment $AB$ is the latus rectum, so the directrix is parallel to $AB$ at distance $5$ on the far side, namely $4y-3x+25=0.$

Equating distances to focus and directrix gives $(4x+3y)^2$ $=25\big(25+2(4y-3x)\big).$ Writing $4x+3y=5s$ forces $s$ to be a multiple of $5,$ and $s=5t$ forces $t$ odd; with $t=2u+1$ the integer points are $\begin{aligned} x&=-6u^2+2u+4,\\ y&=8u^2+14u+3. \end{aligned}$

Then $|4x+3y|=|50u+25|\le1000$ iff $|2u+1|\le39,$ i.e. $-20\le u\le19.$ That gives $40$ lattice points.

Thus, the correct answer is B.