Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #15
Grade 6 number-theoryalgebraPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Test the identity on small numbers
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.
Verifying an identity on small numbers first is the Tool #9 "easier problem" move — it builds confidence before applying the rule to harder numbers.
6.EE.A.1Solve An Easier Related ProblemSplit the fourth-power difference
See 13⁴ and 11⁴ as (13²)² and (11²)², then apply the identity once to split it into (13² - 11²)(13² + 11²).
Splitting one hard expression into two easier ones is the core Tool #7 "subproblems" move.
6.EE.A.1Identify SubproblemsFactor the difference again
The first factor 13² - 11² is itself a difference of squares, so split it again: (13-11)(13+11) = 2 × 24.
Re-applying the same easier-problem split to a leftover piece is exactly how subproblems compound — each call peels off a layer.
The expression 13⁴ - 11⁴ equals the product of three small whole numbers, (13 - 11)(13 + 11)(13² + 11²).
▸ Why?
A difference of two squares always splits as a² - b² = (a - b)(a + b); using this on 13⁴ - 11⁴ = (13²)² - (11²)², and again on the leftover 13² - 11², produces the three factors.
▸ Why?
Multiplying (a - b)(a + b) out, each piece of the first bracket multiplies each piece of the second, giving a² + ab - ba - b².
▸ Why?
The middle terms +ab and -ba are the same amount added and then taken back, a net change of zero, so only a² - b² is left.
▸ Why?
ab and ba count the same array of dots, so the two middle terms are equal in size.
▸ Why?
Taking away exactly what was just added returns you to the starting amount, erasing the middle terms.
▸ Why?
Reading 13⁴ as (13²)² and 11⁴ as (11²)² only regroups the four equal factors, and regrouping a product changes nothing, so the squares-of-squares form is the same number.
Add the two squares
The third factor is a plain sum: 13² + 11² = 169 + 121 = 290, so the full product is 2 × 24 × 290.
Each factor is now a number we can hold in our head, so the hard expression has shrunk to three friendly pieces.
5.NBT.B.5Identify SubproblemsCount the factors of 2
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.
Prime factorization tells you each factor's "2-content"; the exponent rule 2^a × 2^b = 2^a+b lets you total them in one step.
6.NS.B.4Identify SubproblemsThis 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!
- Test the identity on small numbers
- Split the fourth-power difference
- Factor the difference again
- Add the two squares
- Count the factors of 2
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