AMC 10 · 2003 · #18

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesmodular-arithmeticmultiples caseworkguess-and-check ↑ Prerequisites: divisibility-rules
📏 Long solution 💡 3 insights
Problem
The five factors n+1, n+3, n+5, n+7, n+9 multiply to one number. As n runs through the positive even integers 2,4,6,…, this product keeps changing. Find the largest whole number that divides the product no matter which positive even n you pick.

Pick an answer.

(A)
3
(B)
5
(C)
11
(D)
15
(E)
165

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

The choices are a fixed short list, so the goal is to keep the largest one that always divides and rule out the rest (Tool #3). First reorganize the expression: since n is even, the five factors are just five consecutive odd integers (Tool #15), which turns the whole question into a fact about odd numbers in a row. Then look for the repeating structure of multiples (Tool #5): among five consecutive odds there is always exactly one multiple of 5 and always at least one multiple of 3, so 3 and 5 — and hence 15 — divide every product. The only bigger choices (11 and 165) both need a factor of 11, so test a single well-chosen even value (Tool #6): if one product has no multiple of 11, both are eliminated. What survives and is largest is the answer.

1STEP 1

See five consecutive odd numbers

n is even, so all five factors are odd and rise by 2: five consecutive odd numbers. n=2 gives 3,5,7,9,11.

n=2: 3,5,7,9,11 n=8: 9,11,13,15,17
2STEP 2

One factor is always a multiple of 5

Adding 2 walks the remainders mod 5 through 0,1,2,3,4, so exactly one factor is a multiple of 5: in 3,5,7,9,11 it is 5.

remainders mod 5: {0,1,2,3,4} → exactly one multiple of 5
3STEP 3

One factor is always a multiple of 3

Any three odds in a row cover all remainders mod 3, so a multiple of 3 is there too; 3 and 5 are coprime, so 15 always divides.

3 ∣ product and 5 ∣ product → 15 ∣ product
4STEP 4

Test 11 with one example

The bigger choices 11 and 165 = 3·5·11 both need a factor of 11, but n=12 gives 13,15,17,19,21 — no multiple of 11. One example kills both.

n=12: 13,15,17,19,21 — none divisible by 11
5STEP 5

Keep the largest divisor that always works

3 and 5 are smaller, 11 and 165 fail at n=12, so the largest divisor that always works is 15, choice (D).

A,B ＜ 15; C,E need 11 (fails) → (D) 15
Answer
15
15 = 3·5, and we showed a multiple of 3 and a multiple of 5 always sit among the five odd factors, so 15 genuinely divides every product — not just a lucky fit. It is also the largest choice that can: 11 and 165 die at n=12 (factors 13,15,17,19,21, no multiple of 11), while 3 and 5 are smaller than 15. Nothing above 15 is forced either — n=10 gives 11·13·15·17·19, whose only factor of 3 and only factor of 5 both come from the single 15, so no second 3 or 5 is guaranteed. That pins the answer at exactly 15.
💡Key takeaway

Five odd numbers in a row always hide a multiple of 3 and a multiple of 5, so 15 always divides — but 11 can go missing, so nothing bigger is guaranteed.

  • See five consecutive odd numbers
  • One factor is always a multiple of 5
  • One factor is always a multiple of 3
  • Test 11 with one example
  • Keep the largest divisor that always works

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