AMC 10-style practice problems

Hint 1: Rewrite each term as a difference of two fractions.

Hint 2: Use 1/[k(k + 1)] = 1/k - 1/(k + 1).

Answer: C. 9/10

The sum is (1 - 1/2) + (1/2 - 1/3) + ... + (1/9 - 1/10). All intermediate fractions cancel, leaving 1 - 1/10 = 9/10.

Hint 1: Square x + y.

Hint 2: The expansion contains x^2 + y^2 plus 2xy.

Answer: C. 29

The identity (x + y)^2 = x^2 + y^2 + 2xy gives x^2 + y^2 = 49 - 20 = 29. The given conditions are consistent, for example x = 2 and y = 5.

Hint 1: The perpendicular from the center bisects the chord.

Hint 2: Use a right triangle with legs 5 and 6 and hypotenuse equal to the radius.

Answer: D. 61 pi

The half-chord has length 6. By the Pythagorean theorem, the radius satisfies r^2 = 5^2 + 6^2 = 61. Thus the circle has area pi r^2 = 61 pi.

Hint 1: Each remainder is 3 less than its divisor.

Hint 2: Add 3 to the unknown number; the result is divisible by 4, 5, and 6.

Answer: D. 57

If n is the required number, n + 3 is a positive common multiple of 4, 5, and 6. Their least common multiple is 60, so the smallest positive candidate is n = 60 - 3 = 57. It gives the three required remainders.

Hint 1: Count results whose entries are at most 5.

Hint 2: Subtract those whose entries are all at most 4.

Answer: C. 61/216

There are 6^3 = 216 equally likely ordered results. Exactly 5^3 - 4^3 = 125 - 64 = 61 have all entries at most 5 and at least one entry equal to 5. Therefore the probability is 61/216.

Hint 1: The sum and product of the roots are 5 and 3.

Hint 2: Expand (r + s)^3 and group the mixed terms as 3rs(r + s).

Answer: D. 80

From the coefficients, r + s = 5 and rs = 3. Thus r^3 + s^3 = (r + s)^3 - 3rs(r + s) = 125 - 45 = 80.

Hint 1: Use coordinates based on the two sides from A; area ratios do not depend on the triangle's shape.

Hint 2: Set A = (0, 0), B = (1, 0), C = (0, 1). Then D = (1/2, 1/2) and E = (0, 1/3).

Answer: B. 21

Use oblique coordinates with vectors AB and AC as basis vectors. Then D = (1/2, 1/2), E = (0, 1/3), and points on AD have equal coordinates. A point on BE has coordinates (1 - t, t/3). Equating them gives 1 - t = t/3, so t = 3/4 and P = (1/4, 1/4).

Triangles ABP and ABC share base AB. The altitude of P above AB is 1/4 the altitude of C, because the AC-coordinate is 1/4. Their area ratio is 1/4, so the required area is 84/4 = 21.

Hint 1: The possible occurrence-count patterns are (6,0,0), (4,2,0), and (2,2,2), in any order.

Hint 2: Count the choices of letters as well as their positions.

Answer: D. 183

For pattern (6,0,0), choose the repeated letter in 3 ways. For pattern (4,2,0), choose the letter used four times in 3 ways and the letter used twice in 2 ways; choose its two positions in 15 ways. This gives 3 x 2 x 15 = 90.

For pattern (2,2,2), choose the two A positions in 15 ways and the two B positions among the remaining four in 6 ways; C fills the rest. This gives 90. The disjoint cases total 3 + 90 + 90 = 183.

Hint 1: Group the digits by their remainder on division by 3.

Hint 2: Each remainder class has only two digits, so a valid triple must use one from each class.

Answer: C. 48

The remainder classes are {3,6}, {1,4}, and {2,5}. A sum of three remainders is 0 modulo 3 only if all three remainders are equal or all are different. Equal remainders would require three digits from a class of size two, which is impossible.

Choose one digit from each class in 2 x 2 x 2 = 8 ways. Each triple has 6 orders, all producing three-digit numbers, so the answer is 8 x 6 = 48.

Hint 1: First find xy + yz + zx by squaring x + y + z.

Hint 2: Square xy + yz + zx; the extra mixed term contains xyz(x + y + z).

Answer: D. 98

Squaring x + y + z = 0 gives xy + yz + zx = -7. Squaring this relation gives x^2 y^2 + y^2 z^2 + z^2 x^2 + 2xyz(x + y + z) = 49. The last term is zero, so the sum of pairwise squared products is 49.

Now 14^2 = (x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2 x 49. Hence x^4 + y^4 + z^4 = 196 - 98 = 98.

Hint 1: Calculate the next three terms.

Hint 2: Check that the terms 2, -1, and 1/2 repeat, and none makes a denominator zero.

Answer: E. 2

Starting at 2 gives a_2 = -1, a_3 = 1/2, and a_4 = 2. Since the recurrence is deterministic and all three denominators are nonzero, these three values repeat forever. Because 2026 = 3 x 675 + 1, a_2026 = a_1 = 2.

Hint 1: There are 70 equally likely color orders. Count the orders that first let blue move ahead.

Hint 2: For a bad order, interchange red and blue up to and including the first prefix with one more blue than red.

Answer: C. 1/5

There are 8 choose 4 = 70 color orders. For each bad order, find the first prefix with one more blue than red, and swap both colors throughout that prefix. The resulting order has 5 red and 3 blue counters.

This is a bijection: every order with 5 red and 3 blue first reaches a prefix with one more red than blue; swapping that prefix recovers a unique bad order with 4 of each color. There are 8 choose 3 = 56 such orders, so 70 - 56 = 14 are good. The probability is 14/70 = 1/5.