AMC 10 · 2021 · #18
Grade 7 number-theoryProblem
Let f be a function defined on the set of positive rational numbers with the property that f(a⋅b)=f(a)+f(b) for all positive rational numbers a and b. 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?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $f(ab) = f(a) + f(b)$ for all positive rationals $a, b$; $f(p) = p$ whenever $p$ is prime; Choices: (A) $\tfrac{17}{32}$, (B) $\tfrac{11}{16}$, (C) $\tfrac{7}{9}$, (D) $\tfrac{7}{6}$, (E) $\tfrac{25}{11}$
Unknowns: Which choice $x$ has $f(x) < 0$
Understand
Restated: 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.
Givens: $f(ab) = f(a) + f(b)$ for all positive rationals $a, b$; $f(p) = p$ whenever $p$ is prime; Choices: (A) $\tfrac{17}{32}$, (B) $\tfrac{11}{16}$, (C) $\tfrac{7}{9}$, (D) $\tfrac{7}{6}$, (E) $\tfrac{25}{11}$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
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 $f\bigl(\tfrac{\prod p_i^{a_i}}{\prod q_j^{b_j}}\bigr) = \sum a_i p_i - \sum b_j q_j$, 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.
Execute — Answer: E
6.EE.A.4 Step 1 - First derive $f(1)$.
- Set $a = b = 1$: $f(1) = f(1) + f(1) = 2f(1)$, so $f(1) = 0$.
- Next, for any positive rational $x$, set $b = \tfrac{1}{x}$ and $a = x$: $f(1) = f(x) + f(\tfrac{1}{x})$, so $f(\tfrac{1}{x}) = -f(x)$.
💡 A function that turns multiplication into addition acts like a logarithm — $f(1) = 0$ and reciprocals flip sign.
6.NS.B.4 Step 2 - Repeated application of the rule on a prime power: $f(p^k) = f(p) + f(p) + \cdots + f(p) = k \cdot p$.
- Combining with the reciprocal rule, for any rational $\tfrac{N}{D}$ with prime factorizations $N = \prod p_i^{a_i}$ and $D = \prod q_j^{b_j}$, we get a clean formula.
💡 Factor into primes; add primes upstairs, subtract primes downstairs.
7.NS.A.3 Step 3 - Apply the formula to choice (A) $\tfrac{17}{32}$.
- Here $17$ is prime ($a_1 = 1, p_1 = 17$) and $32 = 2^5$ ($b_1 = 5, q_1 = 2$).
- So $f\!\left(\tfrac{17}{32}\right) = 17 - 5 \cdot 2 = 17 - 10 = 7$.
- Positive — eliminate.
💡 One $17$ minus five $2$'s — still positive.
7.NS.A.3 Step 4 - Choice (B) $\tfrac{11}{16}$: $11$ is prime, $16 = 2^4$.
- So $f = 11 - 4 \cdot 2 = 11 - 8 = 3$.
- Positive — eliminate.
💡 Eleven beats four $2$'s.
7.NS.A.3 Step 5 - Choice (C) $\tfrac{7}{9}$: $7$ is prime, $9 = 3^2$.
- So $f = 7 - 2 \cdot 3 = 7 - 6 = 1$.
- Positive — eliminate.
💡 Just barely positive.
7.NS.A.3 Step 6 - Choice (D) $\tfrac{7}{6}$: $7$ is prime, $6 = 2 \cdot 3$.
- So $f = 7 - (2 + 3) = 7 - 5 = 2$.
- Positive — eliminate.
💡 Seven still beats $2 + 3 = 5$.
7.NS.A.3 Step 7 - Choice (E) $\tfrac{25}{11}$: $25 = 5^2$, $11$ is prime.
- So $f = 2 \cdot 5 - 11 = 10 - 11 = -1$.
- Negative — this is the answer.
💡 Only this fraction has "smaller primes upstairs, bigger prime downstairs".
6.EE.A.4 First derive $f(1)$. Set $a = b = 1$: $f(1) = f(1) + f(1) = 2f(1)$, so $f(1) = 0 6.NS.B.4 Repeated application of the rule on a prime power: $f(p^k) = f(p) + f(p) + \cdot 7.NS.A.3 Apply the formula to choice (A) $\tfrac{17}{32}$. Here $17$ is prime ($a_1 = 1, 7.NS.A.3 Choice (B) $\tfrac{11}{16}$: $11$ is prime, $16 = 2^4$. So $f = 11 - 4 \cdot 2 = 7.NS.A.3 Choice (C) $\tfrac{7}{9}$: $7$ is prime, $9 = 3^2$. So $f = 7 - 2 \cdot 3 = 7 - 7.NS.A.3 Choice (D) $\tfrac{7}{6}$: $7$ is prime, $6 = 2 \cdot 3$. So $f = 7 - (2 + 3) = 7.NS.A.3 Choice (E) $\tfrac{25}{11}$: $25 = 5^2$, $11$ is prime. So $f = 2 \cdot 5 - 11 = Review
Reasonableness: The big idea — $f$ acts like a weighted prime counter — predicts when $f(\tfrac{N}{D}) < 0$: the prime sum of $D$ must exceed the prime sum of $N$. Only (E) flips that comparison: $N = 25 = 5 \cdot 5$ 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.
Alternative: Tool #16 (Change Focus / Complement) — frame the question as "which choice has a bigger denominator-prime-sum than numerator-prime-sum?" rather than computing each $f$ value. Skim the five choices and look for the one where the denominator is a large prime and the numerator is a small prime to a small power: only (E) fits the shape, so it must be the answer without arithmetic.
CCSS standards used (min grade 7)
6.EE.A.4Identify when two expressions are equivalent (Deriving $f(1) = 0$ and $f(1/x) = -f(x)$ from the functional rule by treating it as an algebraic identity.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Factoring each numerator and denominator into primes ($32 = 2^5$, $16 = 2^4$, $9 = 3^2$, $6 = 2 \cdot 3$, $25 = 5^2$) to apply the formula.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Computing each $f$-value as a signed integer sum like $17 - 10 = 7$ or $10 - 11 = -1$ and comparing to zero.)
⭐ $f$ 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 $\tfrac{25}{11}$, giving $f = -1$ — so the answer is $\textbf{(E)}$.
⭐ $f$ 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 $\tfrac{25}{11}$, giving $f = -1$ — so the answer is $\textbf{(E)}$.
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