AMC 10 · 2005 · #21

Grade 8 number-theory

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Ratio Shift — Two State

divisor-countprime-factorizationexponents convert-to-algebraidentify-subproblems ↑ Prerequisites: divisor-countprime-factorization

📏 Long solution 💡 3 insights

Problem

A positive integer has exactly 60 divisors, and seven times it has exactly 80. Find the largest power of seven dividing the integer.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Nothing about n is known except two divisor counts, so the only workable move is to name the piece we care about. Tool #4 splits n into a power of 7 times a 7-free part, which makes the exponent k an actual variable. Tool #7 then counts divisors as two independent choices instead of one hard count. Tool #15 lines up the two resulting equations so that subtracting them erases the unwanted unknown. Tool #6 finishes by building one concrete n, so we know the forced value of k is genuinely achievable and not just algebraically consistent.

1STEP 1

Peel the sevens off n

Splitting off the sevens leaves a part with no sevens in it.

n = 7^k · m, 7 ∤ m

2STEP 2

Count divisors as two choices

Divisors then come from two independent choices.

d(n) = (k+1) · d(m) = 60

3STEP 3

Multiplying by 7 adds one slot

Multiplying by seven adds exactly one slot.

7n = 7^k+1 · m, d(7n) = (k+2) · d(m) = 80

4STEP 4

Subtract to isolate d(m)

Subtracting the two equations isolates the other count.

(k+2)d(m) - (k+1)d(m) = 80 - 60 ⟹ d(m) = 20

5STEP 5

Back-substitute for k

Back-substituting gives the exponent 2.

(k+1) · 20 = 60 ⟹ k+1 = 3 ⟹ k = 2

6STEP 6

Check such an n exists

An explicit number shows this really happens, so the answer is 2, choice (C).

n = 7² · 2¹⁹: d(n) = 3 · 20 = 60, d(7n) = 4 · 20 = 80

Answer

The two counts 60 and 80 are in the ratio 3 to 4, so the number of allowed powers of 7 inside a divisor goes from 3 to 4 — that is k+1 = 3, giving k = 2, the same value the elimination produced. Testing the other choices confirms it: k = 0 forces d(m) = 60 and d(7n) = 120; k = 1 forces d(m) = 30 and d(7n) = 90; k = 3 forces d(m) = 15 and d(7n) = 75; k = 4 forces d(m) = 12 and d(7n) = 72. Only k = 2 lands on 80.

💡Key takeaway

Counting divisors is counting choices: pick the power of 7, then pick a divisor of what is left, and multiply the two counts.

Peel the sevens off n
Count divisors as two choices
Multiplying by 7 adds one slot
Subtract to isolate d(m)
Back-substitute for k
Check such an n exists