AMC 10 · 2018 · #21

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-factorizationfactorsdivisibility-rules caseworkidentify-subproblems ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 3 insights
Problem
An even four-digit number n has all of its divisors written out in increasing order. One of those divisors is 323. Find the smallest possible value of the divisor that comes immediately after 323 in that list.

Pick an answer.

(A)
324
(B)
330
(C)
340
(D)
361
(E)
646

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Tool #9 (Solve an Easier Related Problem): the whole problem is about what divides n, so factoring 323 and tracking which prime factors must also divide n is the engine. Tool #4 (Introduce a Variable): naming the next divisor d lets us argue about it before we know its value. Tool #7 (Identify Subproblems): split the work into 'what must d look like' and then 'which choice is smallest and actually achievable'. Tool #3 (Eliminate Possibilities): once we know d must be a multiple of 17 or 19, two of the answer choices fail that test and drop out immediately.

1STEP 1

Name the next divisor and list the facts

Call the divisor right after 323 d, the smallest divisor of n above 323. Recall n is even, four-digit (1000 ≤ n ≤ 9999), and 323 divides it.

d = smallest divisor of n with d ＞ 323, 1000 ≤ n ≤ 9999
2STEP 2

Factor 323 to see what must divide n

Factor 323 = 17 × 19, both prime. Since 323 divides n, n is a multiple of 17, 19, and (being even) 2.

323 = 17 × 19 → 17 ∣ n and 19 ∣ n
3STEP 3

Show the next divisor must share a factor of 17 or 19

If d shares no factor with 323, then 323d divides n, so n ≥ 323 × 324 = 104652 — six digits. So d must be a multiple of 17 or 19.

gcd(d,323)=1 → 323 d ∣ n → n ≥ 323 × 324 = 104652 ＞ 9999
4STEP 4

Find the smallest multiple of 17 or 19 above 323

Next multiple of 17 is 17 × 20 = 340; of 19 is 19 × 18 = 342. Among the choices only 340, 361, 646 pass, and 340 is smallest.

17× 20 = 340, 19× 18 = 342 → min = 340
5STEP 5

Check that 340 is actually reachable

Take n = lcm(323, 340) = 6460, even and four-digit; its divisors go …323, 340… with nothing between. So 340 is reachable — answer (C).

lcm(323,340) = 2²· 5· 17· 19 = 6460 → (C) 340
Answer
340
Test the witness directly: 6460 = 323 × 20 is even and four-digit, 323 and 340 both divide it, and its divisors jump from 323 straight to 340. The two smaller choices fail for a clean reason — 324 and 330 are coprime to 323, so any n holding both them and 323 would need at least 323× 324, far more than four digits. That leaves 340 as the smallest achievable next divisor, matching (C).
💡Key takeaway

Since 323=17×19, the next divisor has to share a 17 or a 19, and the first one past 323 is 17×20=340.

  • Name the next divisor and list the facts
  • Factor 323 to see what must divide n
  • Show the next divisor must share a factor of 17 or 19
  • Find the smallest multiple of 17 or 19 above 323
  • Check that 340 is actually reachable

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