AMC 10 · 2002 · #7

Grade 5 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen

fraction-arithmeticlcmdivisibility-rules bound-inequality-then-enumerate ↑ Prerequisites: fraction-arithmeticlcm

📏 Medium solution 💡 2 insights

Problem

A positive whole number n is chosen so that 1/2+1/3+1/7+1/n comes out to a whole number. Among five statements about n, find the one that is not true.

Pick an answer.

(A)

$\ 2\ \text{divides\ }n$

(B)

$\ 3\ \text{divides\ }n$

(C)

$\ 6\ \text{divides\ }n$

(D)

$\ 7\ \text{divides\ }n$

(E)

$\ n > 84$

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

You cannot judge the five statements until you know n, so the real work is finding n. First (Tool #7, Identify Subproblems) add the three known fractions to get one fixed number, 41/42. Then the key move is Tool #14 (Extreme Principle): because 1/n is small and positive, the running total is squeezed between two limits and can only land on the whole number 1 — that boundary argument forces n=42. With n known, Tool #3 (Eliminate Possibilities) tests each statement and keeps the one that fails.

1STEP 1

Add the three known fractions

Put the three n-free fractions over the common denominator 42: (21+14+6)/42=41/42.

1/2+1/3+1/7=(21+14+6)/42=41/42

2STEP 2

Trap the total between two limits

The total is 41/42+1/n. With 1/n positive and at most 1, it is trapped between 41/42 and 2, so the sum must be 1.

41/42 ＜ 41/42+1/n ＜ 41/42+1 ＜ 2 → sum=1

3STEP 3

Solve for n

So 1/n is the gap up to 1: 1/n=1-41/42=1/42, and equal unit fractions need equal denominators, giving n=42.

1/n=1-41/42=1/42 → n=42

4STEP 4

Test each statement against n = 42

Since 42=2×3×7, statements (A)–(D) all hold, but 42 is not greater than 84, so the false one is (E).

42=2×3×7 → 2 ∣ 42, 3 ∣ 42, 6 ∣ 42, 7 ∣ 42; 42 ≯ 84 → (E)

Answer

n ＞ 84

Check the found value directly: 1/2+1/3+1/7+1/42=(21+14+6+1)/42=42/42=1, a whole number, so n=42 is correct. Four of the statements (A, B, C, D) are then true and exactly one, (E), is false — matching the question's promise that a single statement fails. The trap answer would be to pick a divisibility statement, but 42 really is a multiple of 2, 3, 6, and 7; only the size comparison n ＞ 84 breaks.

💡Key takeaway

Add the fractions you can, and if what's left is a tiny positive piece, the total can only climb to the next whole number — here that pins n=42, so the false claim is n ＞ 84.

Add the three known fractions
Trap the total between two limits
Solve for n
Test each statement against n = 42

Add the three known fractions

Trap the total between two limits

Solve for n

Test each statement against n = 42

More like this