AMC 10 · 2016 · #22

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-factorizationdivisor-countexponents casework ↑ Prerequisites: prime-factorization
📏 Long solution 💡 4 insights
Problem
For some positive integer n, the number 110n³ has exactly 110 positive divisors. Using that fact about n, find how many positive divisors 81n⁴ has.

Pick an answer.

(A)
110
(B)
191
(C)
261
(D)
325
(E)
425

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Write 110n³ = 2^a·5^b·11^c·…; then 'has 110 divisors' becomes (a+1)(b+1)(c+1)… = 110.

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

Split 110 into factors

Factor 110 = 2·5·11. Since 2, 5, 11 already divide 110n³, each factor is ≥ 2, and 110 splits into factors ≥ 2 only as 2·5·11.

110 = 2× 5× 11
3STEP 3

Read off the exponents

So 110n³ has only primes 2, 5, 11, and (a+1),(b+1),(c+1) = 2, 5, 11 in some order, so a, b, c = 1, 4, 10 in some order.

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

Recover the exponents of n

Remove 110 = 2·5·11, so n³ has exponents 0, 3, 9 (one less each); all are multiples of 3, so n has exponents 0, 1, 3.

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

Build and count 81n⁴

Raise n to the 4th: exponents 0, 4, 12 on 2, 5, 11; 81 = 3⁴ adds prime 3. Divisors: (4+1)(0+1)(4+1)(12+1) = 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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