2003 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

$0$

$1$

$2$

$2003$

$4006$

Answer: D

Concepts:summation pairing and grouping

Difficulty rating: 860

Small Hint:

Pair each even number with the odd number just below it

Big Hint:

Each of the $2003$ pairs contributes a difference of $1$

Solution:

The $k$ th even number $2k$ is exactly $1$ more than the $k$ th odd number $2k-1.$

Summing this difference over all $2003$ pairs gives $2003 \cdot 1 = 2003.$

Thus, the correct answer is D.

2.

Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $\$4$ per pair and each T-shirt costs $\$5$ more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $\$2366,$ how many members are in the League?

$77$

$91$

$143$

$182$

$286$

Answer: B

Concepts:money ratio and proportion

Difficulty rating: 1050

Small Hint:

A T-shirt costs $\$4 + \$5 = \$9$

Big Hint:

Each member needs two pairs of socks and two shirts

Solution:

Each T-shirt costs $\$4 + \$5 = \$9.$

Each member needs two pairs of socks and two shirts, costing $2 \cdot 4 + 2 \cdot 9 = \$26.$

The number of members is $2366 \div 26 = 91.$

Thus, the correct answer is B.

3.

A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?

$4.5$

$9$

$12$

$18$

$24$

Answer: D

Concepts:volume percentage cube geometry

Difficulty rating: 1050

Small Hint:

Each removed cube has volume $3^3$

Big Hint:

There are $8$ corners, and the original volume is $15 \cdot 10 \cdot 8$

Solution:

The eight removed cubes have total volume $8 \cdot 3^3 = 216$ cubic centimeters.

The original box has volume $15 \cdot 10 \cdot 8 = 1200$ cubic centimeters.

The percent removed is $\dfrac{216}{1200} \cdot 100\% = 18\%.$

Thus, the correct answer is D.

4.

It takes Mary $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?

$3$

$3.125$

$3.5$

$4$

$4.5$

Answer: A

Concepts:distance rate and time harmonic mean

Difficulty rating: 1130

Small Hint:

Average speed is total distance divided by total time, not the average of the two speeds

Big Hint:

She covers $2$ km in $30 + 10 = 40$ minutes

Solution:

Mary walks a total of $2$ km in $30 + 10 = 40$ minutes.

Since $40$ minutes is $\dfrac{2}{3}$ hour, her average speed is $2 \div \dfrac{2}{3} = 3$ km/hr.

Thus, the correct answer is A.

5.

Let $d$ and $e$ denote the solutions of $2x^2 + 3x - 5 = 0.$ What is the value of $(d - 1)(e - 1)?$

$-\dfrac{5}{2}$

$0$

$3$

$5$

$6$

Answer: B

Concepts:quadratic factoring

Difficulty rating: 1200

Small Hint:

Expand $(d-1)(e-1)$ in terms of $d+e$ and $de$

Big Hint:

Use Vieta’s formulas to find $d+e$ and $de$ without solving for the roots

Solution:

Factoring gives $2x^2 + 3x - 5 = (2x + 5)(x - 1),$ so the roots are $-\dfrac{5}{2}$ and $1.$

Since one root equals $1,$ one of the two factors $d-1$ and $e-1$ equals $0,$ making the product $0.$

Thus, the correct answer is B.

6.

Define $x \heartsuit y$ to be $|x - y|$ for all real numbers $x$ and $y.$ Which of the following statements is not true?

$x \heartsuit y = y \heartsuit x$ for all $x$ and $y$

$2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$

$x \heartsuit 0 = x$ for all $x$

$x \heartsuit x = 0$ for all $x$

$x \heartsuit y \gt 0$ if $x \ne y$

Answer: C

Concepts:custom operation absolute value counterexample

Difficulty rating: 1200

Small Hint:

Rewrite each statement using $x \heartsuit y = |x-y|$

Big Hint:

Pay special attention when one input is $0$ and the other is negative

Solution:

Statement (C) claims $x \heartsuit 0 = x,$ but $x \heartsuit 0 = |x - 0| = |x|,$ which fails for negative $x.$ For example, $-1 \heartsuit 0 = 1 \ne -1.$

The remaining statements all follow directly from the properties of absolute value.

Thus, the correct answer is C.

7.

How many non-congruent triangles with perimeter $7$ have integer side lengths?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:triangle inequality systematic listing

Difficulty rating: 1250

Small Hint:

The three sides are positive integers summing to $7$

Big Hint:

The longest side must be less than $3.5,$ so it is at most $3$

Solution:

The longest side cannot exceed $3,$ since otherwise the other two sides could not reach it.

The only possibilities are side lengths $1,$ $3,$ $3$ and $2,$ $2,$ $3,$ giving $2$ triangles.

Thus, the correct answer is B.

8.

What is the probability that a randomly drawn positive factor of $60$ is less than $7?$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: E

Concepts:factor basic probability

Difficulty rating: 1250

Small Hint:

List all positive divisors of $60$

Big Hint:

Count how many of the $12$ divisors are less than $7$

Solution:

The factors of $60$ are $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $10,$ $12,$ $15,$ $20,$ $30,$ and $60.$

Six of these twelve factors are less than $7,$ so the probability is $\dfrac{6}{12} = \dfrac{1}{2}.$

Thus, the correct answer is E.

9.

Simplify $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}.$

$\sqrt{x}$

$\sqrt[3]{x^2}$

$\sqrt[27]{x^2}$

$\sqrt[54]{x}$

$\sqrt[81]{x^{80}}$

Answer: A

Concepts:radical exponent

Difficulty rating: 1350

Small Hint:

Rewrite each radical using fractional exponents, working from the inside out

Big Hint:

The innermost $x\sqrt{x} = x^{\frac{3}{2}},$ and each cube root divides the exponent by $3$

Solution:

Working outward, $x\sqrt{x} = x^{\frac{3}{2}},$ and its cube root is $x^{\frac{1}{2}}.$

Then $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is again $x^{\frac{1}{2}}.$

Repeating once more, $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is $x^{\frac{1}{2}} = \sqrt{x}.$

Thus, the correct answer is A.

10.

The polygon enclosed by the solid lines in the figure consists of $4$ congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?

$2$

$3$

$4$

$5$

$6$

Answer: E

Concepts:net (3D geometry)cube geometry

Difficulty rating: 1410

Small Hint:

The four squares already wrap into four side faces of a cube

Big Hint:

The fifth square must fold up as one of the two remaining faces without landing on a face already covered

Solution:

Fold the four shaded squares first. They occupy four distinct faces of the cube, leaving two faces open.

Checking the numbered attachments, positions $1,$ $2,$ and $3$ fold the new square onto a face already occupied by one of the shaded squares. Positions $4,$ $5,$ $6,$ $7,$ $8,$ and $9$ fold it onto one of the two open faces. Therefore exactly $6$ of the $9$ polygons work.

Thus, the correct answer is E.

11.

The sum of the two $5$ -digit numbers $AMC10$ and $AMC12$ is $123422.$ What is $A + M + C?$

$10$

$11$

$12$

$13$

$14$

Answer: E

Concepts:cryptarithm place value

Difficulty rating: 1310

Small Hint:

Write each number as $100 \cdot \overline{AMC}$ plus its last two digits

Big Hint:

Their sum is $200 \cdot \overline{AMC} + 22 = 123422$

Solution:

The two numbers equal $100 \cdot \overline{AMC} + 10$ and $100 \cdot \overline{AMC} + 12,$ so their sum is $200 \cdot \overline{AMC} + 22 = 123422.$

Then $200 \cdot \overline{AMC} = 123400,$ so $\overline{AMC} = 617.$

Therefore $A + M + C = 6 + 1 + 7 = 14.$

Thus, the correct answer is E.

12.

A point $(x, y)$ is randomly picked from inside the rectangle with vertices $(0, 0),$ $(4, 0),$ $(4, 1),$ and $(0, 1).$ What is the probability that $x \lt y?$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

Answer: A

Concepts:geometric probability triangle area

Difficulty rating: 1410

Small Hint:

The favorable region lies above the line $y = x$

Big Hint:

Within the rectangle, that region is a triangle with legs of length $1$

Solution:

The condition $x \lt y$ holds in the triangle bounded by $y = x,$ $y = 1,$ and $x = 0,$ which has vertices $(0, 0),$ $(0, 1),$ and $(1, 1).$

This triangle has area $\dfrac{1}{2},$ while the rectangle has area $4.$

The probability is $\dfrac{\frac{1}{2}}{4} = \dfrac{1}{8}.$

Thus, the correct answer is A.

13.

The sum of three numbers is $20.$ The first is $4$ times the sum of the other two. The second is seven times the third. What is the product of all three?

$28$

$40$

$100$

$400$

$800$

Answer: A

Concepts:system of equations substitution

Difficulty rating: 1310

Small Hint:

Let the third number be $c;$ then the second is $7c$

Big Hint:

The first number equals $4(b + c),$ so $4(b + c) + (b + c) = 20$

Solution:

Let the numbers be $a,$ $b,$ $c.$ Since $a = 4(b + c),$ we get $4(b + c) + (b + c) = 20,$ so $b + c = 4$ and $a = 16.$

With $b = 7c,$ we have $7c + c = 4,$ so $c = \dfrac{1}{2}$ and $b = \dfrac{7}{2}.$

The product is $16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28.$

Thus, the correct answer is A.

14.

Let $n$ be the largest integer that is the product of exactly $3$ distinct prime numbers, $d,$ $e,$ and $10d + e,$ where $d$ and $e$ are single digits. What is the sum of the digits of $n?$

$12$

$15$

$18$

$21$

$24$

Answer: A

Concepts:prime digits systematic listing

Difficulty rating: 1500

Small Hint:

The single-digit primes are $2,$ $3,$ $5,$ $7,$ and $10d + e$ must also be prime

Big Hint:

Test candidate pairs in descending order of $d,$ discarding any pair for which $10d+e$ is composite

Solution:

Both $d$ and $e$ are distinct members of $\{2,3,5,7\},$ and $10d+e$ must also be prime.

Start with the largest possible tens digit. For $d=7,$ the choices $e=5$ and $e=2$ give the composite numbers $75$ and $72,$ while $e=3$ gives the prime $73.$ This produces $n=7\cdot3\cdot73=1533.$

For $d=5,$ the choices $e=7$ and $e=2$ give the composite numbers $57$ and $52,$ while $e=3$ gives only $5\cdot3\cdot53=795.$ Every case with $d\le3$ is at most $3\cdot7\cdot37=777.$ Hence $1533$ is the largest valid value.

The sum of its digits is $1 + 5 + 3 + 3 = 12.$

Thus, the correct answer is A.

15.

What is the probability that an integer in the set $\{1, 2, 3, \ldots, 100\}$ is divisible by $2$ and not divisible by $3?$

$\dfrac{1}{6}$

$\dfrac{33}{100}$

$\dfrac{17}{50}$

$\dfrac{1}{2}$

$\dfrac{18}{25}$

Answer: C

Concepts:basic probability multiple counting integers in a range

Difficulty rating: 1310

Small Hint:

Count the multiples of $2,$ then remove those also divisible by $3$

Big Hint:

There are $50$ even numbers and $16$ multiples of $6$

Solution:

Of the $100$ integers, $50$ are divisible by $2.$

Among those, the ones also divisible by $3$ are the multiples of $6,$ of which there are $16.$

So $50 - 16 = 34$ qualify, giving probability $\dfrac{34}{100} = \dfrac{17}{50}.$

Thus, the correct answer is C.

16.

What is the units digit of $13^{2003}?$

$1$

$3$

$7$

$8$

$9$

Answer: C

Concepts:units digit modular arithmetic pattern recognition

Difficulty rating: 1350

Small Hint:

The units digit of $13^n$ equals the units digit of $3^n$

Big Hint:

Units digits of powers of $3$ cycle $3,$ $9,$ $7,$ $1$ with period $4$

Solution:

The units digit of $13^{2003}$ matches that of $3^{2003}.$

Powers of $3$ have units digits cycling $3,$ $9,$ $7,$ $1$ with period $4.$

Since $2003 = 4 \cdot 500 + 3,$ the units digit is the third in the cycle, which is $7.$

Thus, the correct answer is C.

17.

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

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

$\dfrac{3\sqrt{3}}{\pi}$

$\sqrt{3}$

$\dfrac{6}{\pi}$

$\sqrt{3}\pi$

Answer: B

Concepts:equilateral triangle circumcircle, circumcenter, and circumradius special right triangle

Difficulty rating: 1600

Small Hint:

For an equilateral triangle of side $s,$ the circumradius is $R = \dfrac{s}{\sqrt{3}}$

Big Hint:

Set the perimeter $3s$ equal to the area $\pi R^2,$ then substitute $s = R\sqrt{3}$

Solution:

Let the side length be $s$ and the circumradius be $R.$ From a $30$ - $60$ - $90$ triangle formed by the center and a side, $R = \dfrac{s}{\sqrt{3}},$ so $s = R\sqrt{3}.$

The perimeter is $3s = 3R\sqrt{3}$ and the circle’s area is $\pi R^2.$

Setting them equal, $3R\sqrt{3} = \pi R^2,$ so $R = \dfrac{3\sqrt{3}}{\pi}.$

Thus, the correct answer is B.

18.

What is the sum of the reciprocals of the roots of the equation $\dfrac{2003}{2004}x + 1 + \dfrac{1}{x} = 0?$

$-\dfrac{2004}{2003}$

$-1$

$\dfrac{2003}{2004}$

$1$

$\dfrac{2004}{2003}$

Answer: B

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1440

Small Hint:

Multiply through by $x$ to get a quadratic $ax^2 + x + 1 = 0$ with $a = \dfrac{2003}{2004}$

Big Hint:

If the roots are $r$ and $s,$ their reciprocal sum is $\dfrac{r + s}{rs},$ and by Vieta $r + s = -\dfrac{1}{a},$ $rs = \dfrac{1}{a}$

Solution:

Let $a = \dfrac{2003}{2004}.$ Multiplying the equation by $x$ gives $ax^2 + x + 1 = 0.$

If the roots are $r$ and $s,$ then by Vieta’s formulas $r + s = -\dfrac{1}{a}$ and $rs = \dfrac{1}{a}.$

The sum of reciprocals is $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-\frac{1}{a}}{\frac{1}{a}} = -1.$

Thus, the correct answer is B.

19.

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2,$ as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

$\dfrac{1}{6}\pi - \dfrac{\sqrt{3}}{4}$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{12}\pi$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{12}\pi$

Answer: C

Concepts:sector area decomposition equilateral triangle

Difficulty rating: 1660

Small Hint:

The lune equals a triangle plus the small semicircle, minus a sector of the large circle

Big Hint:

The chord of length $1$ subtends a $60^\circ$ arc of the large semicircle

Solution:

The small semicircle’s diameter is a chord of length $1$ in the large circle. Joining its endpoints to the large circle’s center gives an equilateral triangle of side $1$ and area $\dfrac{\sqrt{3}}{4}.$

The region between the chord and the small arc, taken together with that triangle, has area $\dfrac{\sqrt{3}}{4} + \dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^2 = \dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8}.$

Subtracting the $60^\circ$ sector of the large circle, of area $\dfrac{1}{6}\pi(1)^2 = \dfrac{\pi}{6},$ leaves the lune: $\dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8} - \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{24}.$

Thus, the correct answer is C.

20.

A base- $10$ three-digit number $n$ is selected at random. Which of the following is closest to the probability that the base- $9$ representation and the base- $11$ representation of $n$ are both three-digit numerals?

$0.3$

$0.4$

$0.5$

$0.6$

$0.7$

Answer: E

Concepts:number base basic probability counting integers in a range

Difficulty rating: 1660

Small Hint:

A number is three-digit in base $9$ when $81 \le n \le 728,$ and three-digit in base $11$ when $121 \le n \le 1330$

Big Hint:

Among the $900$ three-digit base- $10$ numbers, count those with $121 \le n \le 728$

Solution:

The largest three-digit base- $9$ number is $9^3 - 1 = 728,$ and the smallest three-digit base- $11$ number is $11^2 = 121.$

So both conditions hold exactly when $121 \le n \le 728,$ giving $608$ integers.

Out of $900$ three-digit numbers, the probability is $\dfrac{608}{900} \approx 0.7.$

Thus, the correct answer is E.

21.

Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?

$22$

$25$

$27$

$28$

$729$

Answer: D

Concepts:stars and bars

Difficulty rating: 1600

Small Hint:

Count the nonnegative integer solutions to $a + b + c = 6$

Big Hint:

By stars and bars this is $\dbinom{6 + 2}{2}$

Solution:

An assortment is determined by how many of each type are chosen, so we count nonnegative integer solutions to $a + b + c = 6.$

By stars and bars, placing $2$ dividers among $8$ slots gives $\dbinom{8}{2} = 28$ assortments.

Thus, the correct answer is D.

22.

In rectangle $ABCD,$ we have $AB = 8,$ $BC = 9,$ $H$ is on $\overline{BC}$ with $BH = 6,$ $E$ is on $AD$ with $DE = 4,$ line $EC$ intersects line $AH$ at $G,$ and $F$ is on line $AD$ with $\overline{GF} \perp \overline{AF}.$ Find the length $\overline{GF}.$

$16$

$20$

$24$

$28$

$30$

Answer: B

Concepts:coordinate geometry linear equation

Difficulty rating: 1800

Small Hint:

Place $D$ at the origin with $A$ on the positive $x$ -axis

Big Hint:

Find where line $AH$ and line $EC$ meet, then $\overline{GF}$ is the height of $G$ above line $AD$

Solution:

Place $D = (0, 0),$ $A = (9, 0),$ $B = (9, 8),$ $C = (0, 8),$ $H = (3, 8),$ and $E = (4, 0).$

Line $AH$ has equation $y = -\dfrac{4}{3}x + 12,$ and line $EC$ has equation $y = -2x + 8.$

Setting them equal gives $x = -6$ and $y = 20,$ so $G = (-6, 20).$ Since $\overline{GF}$ is perpendicular to line $AD$ (the $x$ -axis), its length is the height $20.$

Thus, the correct answer is B.

23.

A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?

$1{,}004{,}004$

$1{,}005{,}006$

$1{,}507{,}509$

$3{,}015{,}018$

$6{,}021{,}018$

Answer: C

Concepts:summation pattern recognition

Difficulty rating: 1730

Small Hint:

A base row of $2003$ small triangles means $2n - 1 = 2003,$ so there are $n$ rows

Big Hint:

Row $k$ needs $3k$ toothpicks, so the total is $3(1 + 2 + \cdots + n)$

Solution:

A triangle with $n$ rows has $2n - 1$ small triangles in its base row, so $2n - 1 = 2003$ gives $n = 1002.$

Each row $k$ requires $3k$ toothpicks, so the total is $3(1 + 2 + \cdots + 1002).$

This equals $3 \cdot \dfrac{1002 \cdot 1003}{2} = 1{,}507{,}509.$

Thus, the correct answer is C.

24.

Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6.$ She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

$8$

$9$

$10$

$11$

$12$

Answer: E

Concepts:divisibility logical deduction

Difficulty rating: 1840

Small Hint:

Consider which blue cards each red card can neighbor, since the red number must divide the blue number

Big Hint:

Red $4$ and red $5$ each divide only one blue card, forcing the ends of the stack

Solution:

Among blue cards $3,$ $4,$ $5,$ $6,$ red $5$ divides only $5$ and red $4$ divides only $4,$ so those pairs must sit at the ends.

Red $2$ divides only $4$ and $6,$ and red $3$ divides only $3$ and $6.$ Chaining these forces the stack $R4,$ $B4,$ $R2,$ $B6,$ $R3,$ $B3,$ $R1,$ $B5,$ $R5.$

The middle three cards are $B6,$ $R3,$ $B3,$ summing to $6 + 3 + 3 = 12.$

Thus, the correct answer is E.

25.

Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100.$ For how many values of $n$ is $q + r$ divisible by $11?$

$8180$

$8181$

$8182$

$9000$

$9090$

Answer: B

Concepts:divisibility modular arithmetic counting integers in a range

Difficulty rating: 2070

Small Hint:

Write $n = 100q + r,$ so $n = (q + r) + 99q$

Big Hint:

Since $99q$ is divisible by $11,$ $q + r$ is divisible by $11$ exactly when $n$ is

Solution:

Write $n = 100q + r = (q + r) + 99q.$

Since $99q$ is a multiple of $11,$ $q + r$ is divisible by $11$ if and only if $n$ is.

The $5$ -digit multiples of $11$ satisfy $10000 \le n \le 99999,$ and there are

$\begin{aligned} &\left\lfloor\dfrac{99999}{11}\right\rfloor - \left\lfloor\dfrac{9999}{11}\right\rfloor \\ &= 9090 - 909 \\ &= 8181. \end{aligned}$

Thus, the correct answer is B.