AMC 10 · 2005 · #21

Grade 7 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

triangular-numbersdivisibility-rulesfactors convert-to-algebrasystematic-enumeration ↑ Prerequisites: triangular-numbersdivisibility-rulesfactors

📏 Medium solution 💡 2 insights

Problem

Add up the whole numbers from 1 to n. For how many positive whole numbers n does that running total divide 6n with no remainder?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The word "total from 1 to n" is awkward to test one value at a time, so Tool #4 (Introduce a Variable) leads: replace the sum with the closed formula (n(n+1))/2 so the whole question becomes a single expression in n. Dividing 6n by that expression and simplifying collapses the messy ratio down to 12/(n+1) — now the problem is just "when is 12/(n+1) a whole number?" That turns it into a factor hunt, so Tool #2 (Make a Systematic List) is used to list every factor of 12, and Tool #3 (Eliminate Possibilities) throws out the one value that breaks the rule n ≥ 1. What is left is the count.

1STEP 1

Replace the sum with a formula

Adding term by term is slow, so use the shortcut 1+2+…+n = $\frac{n(n+1)}{2}$ — pair the first term with the last and every pair sums to n+1.

1+2+3+…b+n = (n(n+1))/2

2STEP 2

Divide 6n by the total

"Evenly divides" means the quotient is a whole number, so flip and multiply: $6n \div \frac{n(n+1)}{2} = 6n \cdot \frac{2}{n(n+1)} = \frac{12n}{n(n+1)}$ .

6n ÷ (n(n+1))/2 = 6n · 2/(n(n+1)) = 12n/(n(n+1))

3STEP 3

Cancel n to reveal the real condition

Since n ≥ 1, n is not zero, so cancel it: $\frac{12n}{n(n+1)} = \frac{12}{n+1}$ , whole exactly when n+1 is a factor of 12.

12n/(n(n+1)) = 12/(n+1) ⟹ (n+1) ∣ 12

4STEP 4

List every factor of 12

The positive factors of 12 are 1, 2, 3, 4, 6, 12; subtracting 1 from each gives n = 0, 1, 2, 3, 5, 11.

n+1 ∈ {1, 2, 3, 4, 6, 12} → n ∈ {0, 1, 2, 3, 5, 11}

5STEP 5

Drop n = 0 and count

n must be positive, so throw out n = 0; that leaves n = 1, 2, 3, 5, 11 — five values, choice (B).

n ∈ {1, 2, 3, 5, 11} → 5 values → (B)

Answer

Spot-check the endpoints of the list by hand. For n = 1: the total is 1 and 6n = 6, and 1 divides 6. For n = 11: the total is (11 · 12)/2 = 66 and 6n = 66, and 66 divides 66. For n = 5: the total is 15 and 6n = 30, and 15 divides 30. Now check that a skipped value fails, say n = 4: the total is 10 and 6n = 24, and 24 ÷ 10 = 2.4 is not whole — correctly excluded because n+1 = 5 is not a factor of 12. The five survivors all check out, so 5 is right.

💡Key takeaway

Swap the sum for (n(n+1))/2, and the whole problem shrinks to "n+1 must be a factor of 12" — count those and you are done.

Replace the sum with a formula
Divide 6n by the total
Cancel n to reveal the real condition
List every factor of 12
Drop n = 0 and count

Replace the sum with a formula

Divide 6n by the total

Cancel n to reveal the real condition

List every factor of 12

Drop n = 0 and count

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