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2021 AMC 10A Problem 18

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2021 AMC 10A Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Functions · difficulty 3 of 5 · about 4 min · position P16-20

Let ff be a function defined on the set of positive rational numbers with the property that f(a⋅b)=f(a)+f(b)f(a\cdot b)=f(a)+f(b) for all positive rational numbers aa and bb . Furthermore, suppose that ff also has the property that f(p)=pf(p)=p for every prime number pp . For which of the following numbers xx is f(x)<0f(x)<0 ?

  1. A)

    1732\frac{17}{32}

  2. B)

    1116\frac{11}{16}

  3. C)

    79\frac{7}{9}

  4. D)

    76\frac{7}{6}

  5. E)

    2511\frac{25}{11}

Answer

E

Key insight

Multiplicativity makes f(n) the sum of n's prime factors with multiplicity and f(m/n) = f(m) - f(n); only 25/11 gives 10 - 11 < 0.

Solution

Setting a=b=1a = b = 1 gives f(1)=2f(1)f(1) = 2f(1), so f(1)=0f(1) = 0. Then f(n)+f(1/n)=f(1)=0f(n) + f(1/n) = f(1) = 0, so f(1/n)=−f(n)f(1/n) = -f(n) and in general f(m/n)=f(m)−f(n)f(m/n) = f(m) - f(n).

For a positive integer, repeated use of f(ab)=f(a)+f(b)f(ab) = f(a) + f(b) on its prime factorization shows f(n)f(n) is the sum of the prime factors of nn, counted with multiplicity. For example f(32)=f(25)=5⋅2=10f(32) = f(2^5) = 5 \cdot 2 = 10 and f(9)=3+3=6f(9) = 3 + 3 = 6.

Evaluate each choice as (sum for numerator) minus (sum for denominator):
- (A) 1732\frac{17}{32}: 17−10=717 - 10 = 7;
- (B) 1116\frac{11}{16}: 11−8=311 - 8 = 3;
- (C) 79\frac{7}{9}: 7−6=17 - 6 = 1;
- (D) 76\frac{7}{6}: 7−5=27 - 5 = 2;
- (E) 2511\frac{25}{11}: 10−11=−110 - 11 = -1.

Only (E) is negative. The answer is (E) 2511\boxed{\textbf{(E)}\ \frac{25}{11}}.

Why this works

A function turning products into sums is a logarithm in spirit: it is determined by its values on primes, and its value on a fraction is numerator minus denominator. Here the "log" of a prime is the prime itself, so ff measures the total weight of prime factors. Compute f(1)f(1) and f(1/n)f(1/n) from the equation before touching the choices.

The trap

Forgetting multiplicity (f(32) = 10, not 2) or thinking f(1/n) = 1/f(n) instead of -f(n).

Common mistakes

  • Forgetting multiplicity (f(32) = 10, not 2) or thinking f(1/n) = 1/f(n) instead of -f(n).
  • Assuming f(x)<0f(x) < 0 whenever x<1x < 1; fractions (A) through (C) are less than 11 yet have positive ff.

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Set up the equation/formula and compute; no special trick needed