AMC 8 · 2016 · #15

Grade 6 number-theoryalgebra
prime-factorizationexponentsfactors identify-subproblemseasier-related-problem ↑ Prerequisites: prime-factorizationexponents
📏 Medium solution 💡 3 insights
Problem
Find the biggest power of 2 (that is, 2^k for the largest possible k) that divides the number 13⁴ - 11⁴ exactly, with no remainder.

Pick an answer.

(A)
$mbox{ }8$
(B)
$mbox{ }16$
(C)
$mbox{ }32$
(D)
$mbox{ }64$
(E)
$mbox{ }128$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Computing 13⁴ - 11⁴ = 13920 directly is doable but messy, and we then still have to factor 13920. Tool #7 (Identify Subproblems) gives a cleaner path: use the difference-of-squares identity a² - b² = (a-b)(a+b) twice to break 13⁴ - 11⁴ into three small factors — (13-11), (13+11), and (13² + 11²) — and count the 2s in each factor separately. Tool #9 (Easier Related Problem) backs this up: testing a² - b² = (a-b)(a+b) on a tiny case like 5² - 3² makes the identity obvious, so the factoring move is trustworthy. We avoid Tool #13 (Algebra) and brute-force division because the factored form does the work for us.

1STEP 1

Test the rule a² - b² = (a-b)(a+b) on a tiny case: 5² - 3² gives 16 both ways, so it is safe to use on 13 and 11.

5² - 3² = 16 and (5-3)(5+3) = 2 × 8 = 16
2STEP 2

See 13⁴ and 11⁴ as (13²)² and (11²)², then apply the identity once to split it into (13² - 11²)(13² + 11²).

13⁴ - 11⁴ = (13²)² - (11²)² = (13² - 11²)(13² + 11²)
3STEP 3

The first factor 13² - 11² is itself a difference of squares, so split it again: (13-11)(13+11) = 2 × 24.

13² - 11² = (13 - 11)(13 + 11) = 2 × 24, so 13⁴ - 11⁴ = 2 × 24 × (13² + 11²)
4STEP 4

The third factor is a plain sum: 13² + 11² = 169 + 121 = 290, so the full product is 2 × 24 × 290.

13² + 11² = 169 + 121 = 290, so 13⁴ - 11⁴ = 2 × 24 × 290
5STEP 5

Count the 2s per factor: 2 = 2¹, 24 = 2³ × 3, 290 = 2¹ × 145. Add exponents 1 + 3 + 1 = 5, so the largest power of 2 is 2⁵.

2¹ × (2³ × 3) × (2¹ × 145) = 2¹⁺³⁺¹ × 3 × 145 = 2⁵ × 435 = 32 × 435 → (C)
Answer
mbox{ }32
Direct check: 13⁴ = 28561, 11⁴ = 14641, so 13⁴ - 11⁴ = 13920. Halve repeatedly: 13920 → 6960 → 3480 → 1740 → 870 → 435. We divided by 2 exactly 5 times before reaching an odd number, so 13920 = 2⁵ × 435 = 32 × 435. That matches the factored answer 32 exactly, so choice (C) is correct. Choice (D) 64 would have needed 6 halvings, and choice (B) 16 would have stopped after only 4.
💡Key takeaway

This AMC 8 problem only needs the Grade 6 idea of prime factorization plus one factoring trick — a² - b² = (a-b)(a+b) — that you can verify with tiny numbers like 5² - 3²!