AMC 10 · 2013 · #24

Grade 6 number-theory
divisor-sumdivisor-countprime-factorization caseworksystematic-enumeration ↑ Prerequisites: prime-factorizationdivisor-count
📏 Long solution 💡 3 insights
Problem
Call a positive integer n "nice" if some positive integer m has exactly four positive divisors (including 1 and m) whose sum equals n. Count how many of the ten integers 2010, 2011, 2012, …, 2019 are nice.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The word "nice" hides a divisor-sum condition, so the first move is to name the primes. Tool #4 (Introduce a Variable): write m = pq and turn its divisor sum 1 + p + q + pq into the factored form (1+p)(1+q) — this converts the murky question "is n nice?" into the sharp test "can n be split into two factors that are each one-more-than-a-prime?" Tool #7 (Identify Subproblems) first splits four-divisor numbers into the two shapes p³ and pq. Tool #3 (Eliminate Possibilities) uses parity to knock out all five odd numbers at once. Tool #2 (Make a Systematic List) then tests each surviving even number through its factor pairs.

1STEP 1

Two shapes of a four-divisor number

Exactly four divisors happens only two ways: m = p³ with divisors 1, p, p², p³, or m = pq for distinct primes with 1, p, q, pq.

m = p³ → {1, p, p², p³} or m = pq → {1, p, q, pq}
2STEP 2

Factor the divisor sum

For m = pq the four divisors add to 1 + p + q + pq, which factors as (1+p)(1+q) — n is nice when both factors are one more than a prime.

1 + p + q + pq = (1+p)(1+q)
3STEP 3

The cube shape overshoots

The cube sums 1 + p + p² + p³ jump from 1464 at p = 11 straight to 2380 at p = 13, skipping 2010–2019 entirely — only pq survives.

1 + 11 + 11² + 11³ = 1464, 1 + 13 + 13² + 13³ = 2380
4STEP 4

Odd numbers cannot be nice

At least one prime is odd, so a factor 1+p is even and n = (1+p)(1+q) is even — only 2010, 2012, 2014, 2016, 2018 remain.

n = (1+p)(1+q) is even → n ∈ {2010, 2012, 2014, 2016, 2018}
5STEP 5

Test the five even candidates

Need a split a · b with a-1, b-1 both prime: 2010, 2012, 2014 fail (334, 66, 502, 52 are composite) and 2018 splits only as 2 · 1009.

2010, 2012, 2014, 2018: no split a · b with a-1, b-1 both prime
6STEP 6

2016 is the one nice number

2016 = 4 · 504 = (1+3)(1+503) with 3 and 503 prime, so m = 1509 works — exactly one of the ten is nice, choice (A).

2016 = 4 · 504 = (1+3)(1+503), 1 + 3 + 503 + 1509 = 2016
Answer
1
The count 1 matches choice (A). Double-check 2016: m = 1509 = 3 × 503 has divisors 1, 3, 503, 1509 summing to 2016, and 503 is prime (no prime up to 22 divides it). The parity filter is airtight — every nice number must be even — and among the five evens only 2016 survived, because it is the one that factors as (small even) × (one-more-than-a-large-prime) with both parts one-more-than-a-prime.
💡Key takeaway

A number with exactly four divisors is pq, and its divisor sum factors as (1+p)(1+q) — so "nice" just means you can split n into two parts that are each one-more-than-a-prime; only 2016 = 4 · 504 = (1+3)(1+503) works, so the answer is 1, choice (A).

  • Two shapes of a four-divisor number
  • Factor the divisor sum
  • The cube shape overshoots
  • Odd numbers cannot be nice
  • Test the five even candidates
  • 2016 is the one nice number