AMC 10 · 2023 · #22

Grade 9 number-theory
functional-equationprime-factorizationdivisor-sumfactors easier-related-problemsystematic-enumerationidentify-subproblems ↑ Prerequisites: prime-factorizationfactors
📏 Long solution 💡 3 insights
Problem
There is exactly one function on the positive integers with this property: for every positive integer, running through its divisors, multiplying each by the function's value at the complementary quotient, and adding gives one. Find the value this function takes at 2023.

Pick an answer.

(A)
-1536
(B)
96
(C)
108
(D)
116
(E)
144
How to solve
Strategy Solve an Easier Related Problem

There is no formula for f to plug 2023 into, so the only handle is the family of equations itself. Two observations turn that family into something finite. First, rewriting the sum so the d=1 term sits alone (Organize Information in More Ways) shows that the equation for n contains f(n) exactly once, with everything else evaluated at strictly smaller divisors of n. That single rearrangement is why f exists and is unique, and it turns the condition into a recipe that computes f(n) from values already known. Second, the equation holds for every n, so nothing stops you from using tiny values of n first (Solve an Easier Related Problem) and climbing upward. Since only divisors of 2023 ever appear along the way, and 2023 = 7 · 17² has just six divisors, the climb is six short steps (Identify Subproblems). Along the way a pattern shows up for prime powers (Look for a Pattern), which is worth noticing but is not needed — the six-step climb reaches the answer on its own, and it avoids the multiplicativity shortcut that quietly produces one of the wrong choices (Eliminate Possibilities).

1STEP 1

Isolate the term

Isolate the term you want.

Σ_d ∣ n d · f (n/d) = 1 · f(n) + Σ_d ∣ n, d > 1 d · f (n/d) = 1 ⟹ f(n) = 1 - Σ_d ∣ n, d > 1 d · f (n/d)
2STEP 2

List every divisor

There are only six divisors.

2023 = 7 · 289 = 7 · 17², divisors of 2023: 1, 7, 17, 119, 289, 2023
3STEP 3

Start at the bottom

Start from the smallest.

n = 1: 1 · f (1/1) = 1 ⟹ f(1) = 1
4STEP 4

The values at the primes

Compute the values at both primes.

f(7) + 7 f(1) = 1 ⟹ f(7) = 1 - 7 = -6, f(17) + 17 f(1) = 1 ⟹ f(17) = 1 - 17 = -16
5STEP 5

At the square of a prime

Compute it at the prime's square.

f(289) + 17 f(17) + 289 f(1) = 1 ⟹ f(289) - 272 + 289 = 1 ⟹ f(289) = 1 - 17 = -16
6STEP 6

At the product of two primes

Compute it at their product.

f(119) + 7 f(17) + 17 f(7) + 119 f(1) = 1 ⟹ f(119) - 112 - 102 + 119 = 1 ⟹ f(119) - 95 = 1 ⟹ f(119) = 96
7STEP 7

Finish at the target

The final value is 96.

f(2023) + 7 f(289) + 17 f(119) + 119 f(17) + 289 f(7) + 2023 f(1) = 1 ; f(2023) - 112 + 1632 - 1904 - 1734 + 2023 = 1 ; f(2023) - 95 = 1 ⟹ f(2023) = 96
Answer
96
The strongest check is to feed the six values back into the defining condition and confirm every equation really does total 1. With f(1)=1, f(7)=-6, f(17)=-16, f(119)=96, f(289)=-16, f(2023)=96: for n=7, -6 + 7 = 1; for n=17, -16 + 17 = 1; for n=289, -16 - 272 + 289 = 1; for n=119, 96 - 112 - 102 + 119 = 1; and for n=2023, 96 - 112 + 1632 - 1904 - 1734 + 2023 = 1. All six hold, so the values are consistent and the answer is 96. The wrong choices are worth reading too. Choice (A)~-1536 is (-6)·(-16)·(-16) — what you get by treating 2023 = 7 · 17 · 17 as three separate prime factors and multiplying f over all of them, that is, by assuming f(17²) = f(17)². Step 5 computed f(289) = -16 directly, not 256, so that assumption is false and -1536 can be crossed off. The sign also checks out: 96 is positive because the two negative factors in the structure pair up.
💡Key takeaway

When a rule holds for every number, try the smallest numbers first — each equation hands you one new value, and you climb from there.

  • Isolate the term holding f(n)
  • Only six numbers ever matter
  • Start at the bottom: n=1
  • Climb to the primes 7 and 17
  • Square of a prime: f(289)
  • Product of two primes: f(119)
  • Finish at n = 2023