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2021 AMC 10A Problem 18

2021 AMC 10A Problem 18AMC 10 Step by Step

Name: Date: October 28, 2026

Functions · difficulty 3 of 5 · about 4 min · position P16-20

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

A) $\frac{17}{32}$
B) $\frac{11}{16}$
C) $\frac{7}{9}$
D) $\frac{7}{6}$
E) $\frac{25}{11}$

Answer

Key insight

The rule turns products into sums, so f of a whole number is the sum of its prime factors, repeats included, and f of a fraction is top minus bottom.

Solution

1. Start with whole numbers and use the rule on small examples:
- $f(4) = f(2 \times 2) = f(2) + f(2) = 2 + 2 = 4$ ;
- $f(8) = f(4 \times 2) = 4 + 2 = 6$ ;
- $f(6) = f(2 \times 3) = 2 + 3 = 5$ .

So $f$ of a whole number is the sum of the primes in its factorization, with repeats counted. For example $f(32) = f(2^5) = 5 \times 2 = 10$ and $f(25) = 5 + 5 = 10$ .
2. Next find $f(1)$ . Since $1 = 1 \times 1$ , the rule gives $f(1) = f(1) + f(1)$ . The only number equal to twice itself is $0$ , so $f(1) = 0$ .
3. Now fractions. Since $n \times \frac1n = 1$ , we get $f(n) + f\!\left(\frac1n\right) = f(1) = 0$ , so $f\!\left(\frac1n\right) = -f(n)$ . Then
$f\!\left(\frac{m}{n}\right) = f(m) + f\!\left(\frac1n\right) = f(m) - f(n).$
In words: $f$ of the top minus $f$ of the bottom.
4. Check each choice:

choice	top	$f$ (top)	bottom	$f$ (bottom)	$f(x)$
(A)	$17$	$17$	$32 = 2^5$	$10$	$7$
(B)	$11$	$11$	$16 = 2^4$	$8$	$3$
(C)	$7$	$7$	$9 = 3^2$	$6$	$1$
(D)	$7$	$7$	$6 = 2 \times 3$	$5$	$2$
(E)	$25 = 5^2$	$10$	$11$	$11$	$-1$

Only (E) is negative.

The answer is $\boxed{\textbf{(E)}\ \frac{25}{11}}$ .

Why this works

A rule that turns multiplying into adding means everything is decided by the values on primes, because every whole number is built from primes by multiplying. Always work out $f(1)$ and $f\!\left(\frac1n\right)$ from the rule before looking at the choices; after that, each choice is a short sum.

The trap

Counting each prime only once, like f(32) = 2 instead of 2 + 2 + 2 + 2 + 2 = 10, or thinking f(1/n) is 1/f(n) when it is really -f(n).

Concise solution

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

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

Evaluate each choice as (sum for numerator) minus (sum for denominator):
- (A) $\frac{17}{32}$ : $17 - 10 = 7$ ;
- (B) $\frac{11}{16}$ : $11 - 8 = 3$ ;
- (C) $\frac{7}{9}$ : $7 - 6 = 1$ ;
- (D) $\frac{7}{6}$ : $7 - 5 = 2$ ;
- (E) $\frac{25}{11}$ : $10 - 11 = -1$ .

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

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)

Techniques

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