AMC 8 · 2013 · #15

Grade 6 algebraarithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

exponentslinear-equations-one-varperfect-squares identify-subproblemsguess-and-check ↑ Prerequisites: exponentsmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

Three separate equations each hide one exponent: 3^p + 3⁴ = 90, 2^r + 44 = 76, and 5³ + 6^s = 1421. Find p, r, and s, then compute the product p · r · s.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The three equations have no shared variables, so Tool #7 (Identify Subproblems) is the obvious move — solve each equation by itself, then multiply at the end. Inside each subproblem, isolate the power of the base on one side and ask "which exponent of this base gives that number?" Listing b¹, b², b³, … until you hit the target is Tool #6 (Guess and Check) on a tiny ladder of values.

1STEP 1

Subtract 3⁴ = 81 from 90, leaving 3^p = 9 = 3², so p = 2.

3^p + 81 = 90 → 3^p = 9 = 3² → p = 2

2STEP 2

Subtract 44 from 76 to get 2^r = 32 = 2⁵, so r = 5.

2^r + 44 = 76 → 2^r = 32 = 2⁵ → r = 5

3STEP 3

Subtract 5³ = 125 from 1421 to get 6^s = 1296; since 6⁴ = 1296, s = 4.

125 + 6^s = 1421 → 6^s = 1296. Check: 6² = 36, 6³ = 216, 6⁴ = 1296. So s = 4.

4STEP 4

Multiply the three exponents: 2 × 5 × 4 = 40, choice (B).

p · r · s = 2 × 5 × 4 = 40 → (B)

Answer

Substitute back: 3² + 3⁴ = 9 + 81 = 90 ✓, 2⁵ + 44 = 32 + 44 = 76 ✓, 5³ + 6⁴ = 125 + 1296 = 1421 ✓. The product 2 · 5 · 4 = 40 matches choice (B), and it's the only answer choice that factors into three small positive integers consistent with these equations.

💡Key takeaway

Three equations, one trick: peel off the constant, then ask "which power of this base gives that number?" — all Grade 6 exponent work.

More like this