AMC 10 · 2020 · #13

Grade 11 algebra

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

exponentsfraction-arithmeticlinear-diophantinefactorsbound-inequality-then-enumerate convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: exponentsfraction-arithmetic

📏 Long solution 💡 3 insights

Problem

Three integers, each at least 2, are such that a triple-nested radical of a number equals the thirty-sixth root of that number to the twenty-fifth, for every number other than one. Find the middle integer.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

Tool #13 (Convert to Algebra): a radical is just a fractional exponent, so the whole tower collapses into one power of N and the picture becomes a single equation in a, b, c. Tool #7 (Subproblems): peel the tower one radical at a time — inner first — instead of trying to flatten all three layers at once. Tool #14 (Extreme Principle): the smallest legal value a = 2 turns the equation into an inequality that caps b, cutting five choices down to two. Tool #3 (Eliminate) and Tool #6 (Check): test the two survivors against the integer constraint and confirm the winner by substituting back.

1STEP 1

Turn radicals into exponents

Rewrite the radicals as exponents.

^k√(X) = X¹/k ⟹ ³⁶√(N²⁵) = N²5/36

2STEP 2

Peel the inner layer

Peel the innermost layer first.

^b√(N^c√(N)) = (N^(c+1)/c)¹/b = N^(c+1)/bc

3STEP 3

Peel the outer layer

The outer layer works the same way.

^a√(N^b√(Nsqrt[c]{N})) = N^(bc + c + 1)/abc

4STEP 4

Equate the exponents

Set the two exponents equal.

(bc + c + 1)/abc = 25/36

5STEP 5

Clear the fractions

Clear denominators for an integer equation.

36(bc + c + 1) = 25 abc, a, b, c ∈ Z, a, b, c ≥ 2

6STEP 6

Cap the middle integer

Being at least two gives a ceiling.

b ≤ (18(c+1))/7c ≤ 27/7 ＜ 4 ⟹ b ∈ {2, 3}

7STEP 7

Rule out the small candidate

The small candidate gives no integer solution.

b = 2 → c(25a - 54) = 18 → no integer c ≥ 2

8STEP 8

Solve the remaining case

What remains is 3.

b = 3 → c(25a - 48) = 12 → a = 2, c = 6 → b = 3 = (B)

Answer

Substitute (a, b, c) = (2, 3, 6) back into the collapsed exponent: the numerator is bc + c + 1 = 18 + 6 + 1 = 25 and the denominator is abc = 2 · 3 · 6 = 36, giving exactly N²5/36 = ³⁶√(N²⁵). The 25 and the 36 appear on the nose with no cancelling, which is a strong sign the triple is the intended one. A size check also fits: 25/36 ≈ 0.69 is a bit over 2/3, and the outer square root alone already halves everything, so the inner layers must give back nearly 1.39 — only small indices like 3 and 6 can do that.

💡Key takeaway

This AMC 12 problem needs only the Grade 11 rule ^k√(X) = X¹/k. Peel the tower from the inside out and it collapses into one power, N^(bc+c+1)/abc, so you just need (bc+c+1)/abc = 25/36. Since every index shrinks the exponent, a ≥ 2 forces b ＜ 4, and only b = 3 leaves an integer solution: a = 2, b = 3, c = 6, where bc+c+1 = 25 and abc = 36 land exactly on target. Answer (B).

Turn radicals into exponents
Peel the innermost layer
Peel the outer layer
Equate the two exponents
Clear the fractions
Use a ≥ 2 to cap b
Rule out b = 2
Solve the case b = 3