AMC 10 · 2016 · #18

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-factorizationdivisor-countexponents casework ↑ Prerequisites: prime-factorization
📏 Long solution 💡 4 insights
Problem
A multiple of one number has a known divisor count, which pins the unknown's prime shape. Count the divisors of another expression.

Pick an answer.

(A)
110
(B)
191
(C)
261
(D)
325
(E)
425
How to solve
Strategy Introduce a Variable

Divisor problems live in the exponents of a prime factorization, so Tool #4 (Introduce a Variable) names the exponents of 2, 5, and 11 inside 110n³ and turns 'has 110 divisors' into an equation about (exponent+1) products. Tool #3 (Eliminate Possibilities) then pins those exponents down, because 110=2 · 5 · 11 can be split into divisor-count factors in only one way once each factor must be at least 2. Tool #7 (Identify Subproblems) keeps the work orderly: first find the exponents of n, then rebuild 81n⁴ and count its divisors.

1STEP 1

Name the exponents

The divisor count is a product of exponents plus one.

d(p₁^e₁… p_k^e_k)=(e₁+1)…(e_k+1)
2STEP 2

Split 110 into factors

The given count splits into three factors.

110 = 2 × 5 × 11
3STEP 3

Read off the exponents

That names each exponent exactly.

{a+1, b+1, c+1}={2,5,11} → {a,b,c}={1,4,10}
4STEP 4

Recover the exponents of n

Dividing by three recovers the unknown.

n³=2⁰5³11⁹ (in some order) → n=2⁰5¹11³ (in some order)
5STEP 5

Build and count 81n⁴

Rebuilding gives 325, choice (D).

81n⁴=3⁴ 2⁰5⁴11¹² (primes in some order), (4+1)(0+1)(4+1)(12+1)=5·1·5·13=325
Answer
325
Pick one concrete n that fits, say n = 5 · 11³ = 6655 (exponents 0,1,3 on 2,5,11). Then 110n³ = 2¹ 5⁴ 11¹⁰, whose divisor count is 2 · 5 · 11 = 110, matching the hypothesis. For that same n, 81n⁴ = 3⁴ 5⁴ 11¹² has (4+1)(4+1)(12+1) = 5 · 5 · 13 = 325 divisors, confirming (D). The value also does not depend on which order the exponents 0,1,3 sit on, since multiplication is the same regardless of order.
💡Key takeaway

Count divisors by reading the exponents in a prime factorization: each prime gives one-more-than-its-exponent choices, and you just multiply them.

  • Name the exponents
  • Split 110 into factors
  • Read off the exponents
  • Recover the exponents of n
  • Build and count 81n⁴

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