AMC 10 · 2011 · #5

Grade 6 number-theory
lcmmultiplesdigit-sum systematic-enumeration ↑ Prerequisites: lcm
📏 Medium solution 💡 2 insights
Problem
A number is divisible by every one of the first six counting numbers. Add the digits of the second smallest.

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
9
How to solve
Strategy Extreme Principle

The word "second" is the whole difficulty: finding one qualifying number is easy, but ranking them needs the full list. Tool #14 (Extreme Principle) supplies it. Take the smallest qualifying number L and use its minimality as a lever: any qualifying number leaves a remainder on division by L, and that remainder would itself qualify while being smaller than L unless it is 0. So the qualifying numbers are exactly L, 2L, 3L, …, and the second smallest is 2L. Tool #9 (Solve an Easier Related Problem) trims six divisibility conditions down to three. Tool #2 (Make a Systematic List) finds L by a short finite scan instead of a formula. Tool #7 (Identify Subproblems) splits the job into: find L, prove the list, then add digits.

1STEP 1

Trim six conditions to three

Six conditions trim down to three.

4 ∣ n, 5 ∣ n, 6 ∣ n ⇔ 1,2,3,4,5,6 all divide n
2STEP 2

Scan for the smallest one

The smallest such number is 60.

60 = 2² · 3 · 5, L = 60
3STEP 3

Prove the list is 60, 120, 180, …

Every other one is a multiple of it.

n = 60q + r, 0 ≤ r < 60 → r = 0 → N = 120
4STEP 4

Add the digits of N

The digits of the second add to 3, choice (A).

1 + 2 + 0 = 3 → (A)
Answer
3
Test N = 120 against the original wording: 120 ÷ 1 = 120, ÷ 2 = 60, ÷ 3 = 40, ÷ 4 = 30, ÷ 5 = 24, ÷ 6 = 20, all whole, so 120 qualifies. Is it really second? The only smaller qualifying number is 60, and Step 3 showed every qualifying number is a multiple of 60, so nothing sits strictly between 60 and 120. Digit sum 1+2+0=3 matches (A). The size is sensible too: a number divisible by 4, 5, and 6 cannot be small, and 120 is the second such number. Note that the digit sum 6 of the smallest number 60 appears as choice (D), confirming the trap is deliberate rather than a sign of an error.
💡Key takeaway

Find the smallest number that fits every rule, show the leftovers force every other one to be a multiple of it, then just count down the list: 60, then 120.

  • Trim six conditions to three
  • Scan for the smallest one
  • Prove the list is 60, 120, 180, …
  • Add the digits of N