AMC 10 · 2021 · #18

Grade 7 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

prime-factorizationfunction-evaluationfraction-arithmeticprime-numbers convert-to-algebraguess-and-check ↑ Prerequisites: prime-factorization

📏 Long solution 💡 2 insights

Problem

A function f is defined on positive rationals and turns products into sums: f(ab) = f(a) + f(b). It also satisfies f(p) = p for every prime p. Among the five fractions listed, find the one whose f-value is negative.

Pick an answer.

(A)

$\frac{17}{32}$

(B)

$\frac{11}{16}$

(C)

$\frac79$

(D)

$\frac76$

(E)

$\frac{25}{11}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

Tool #13 (Algebra) — the functional rule f(ab) = f(a) + f(b) together with f(p) = p fully determines f on positive rationals: factor each number into primes, sum up primes for the numerator, subtract the prime-sum for the denominator. Once we have this clean formula f((Π p_i^a_i)/(Π q_j^b_j)) = Σ a_i p_i - Σ b_j q_j, Tool #6 (Guess & Check) plugs each of the five choices in and Tool #3 (Eliminate) discards every positive value. This problem is multiple choice with only five candidates — checking them all is the right move.

1STEP 1

Set a = b = 1 to get f(1) = 2f(1), so f(1) = 0; then a = x, b = 1/x gives f(1/x) = -f(x).

f(1) = 0, f (1/x) = -f(x)

2STEP 2

Adding f(p) k times gives f(p^k) = k·p; with the reciprocal rule, f(N/D) = Σ a_i p_i - Σ b_j q_j.

f (N/D) = Σ a_i p_i - Σ b_j q_j

3STEP 3

For (A) $\frac{17}{32}$ : 17 is prime, 32 = 2⁵, so f = 17 - 5·2 = 7 — positive, eliminate.

f (

\frac{17}{32}

) = 17 - 10 = 7 ＞ 0

4STEP 4

For (B) $\frac{11}{16}$ : 11 is prime, 16 = 2⁴, so f = 11 - 4·2 = 3 — positive, eliminate.

f (

\frac{11}{16}

) = 11 - 8 = 3 ＞ 0

5STEP 5

For (C) $\frac{7}{9}$ : 7 is prime, 9 = 3², so f = 7 - 2·3 = 1 — positive, eliminate.

f (

\frac{7}{9}

) = 7 - 6 = 1 ＞ 0

6STEP 6

For (D) $\frac{7}{6}$ : 7 is prime, 6 = 2·3, so f = 7 - (2 + 3) = 2 — positive, eliminate.

f (

\frac{7}{6}

) = 7 - (2 + 3) = 2 ＞ 0

7STEP 7

For (E) $\frac{25}{11}$ : 25 = 5², 11 is prime, so f = 2·5 - 11 = -1 — negative, this is the answer.

f (

\frac{25}{11}

) = 10 - 11 = -1 ＜ 0 → (E)

Answer

\frac{25}{11}

The big idea — f acts like a weighted prime counter — predicts when f(N/D) ＜ 0: the prime sum of D must exceed the prime sum of N. Only (E) flips that comparison: N = 25 = 5 · 5 contributes 5 + 5 = 10, while D = 11 contributes 11, and 11 ＞ 10. The other four choices have a single large prime upstairs (17, 11, 7, 7) and small repeated primes downstairs, so the upstairs total wins.

💡Key takeaway

f acts like a recipe: chop the number into its prime ingredients, add up the primes upstairs, subtract the primes downstairs. The only choice where the downstairs prime 11 beats the upstairs primes 5+5=10 is $\frac{25}{11}$ , giving f = -1 — so the answer is (E).

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