AMC 10 · 2018 · #7

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-factorizationexponentsdivisibility-rules convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: prime-factorizationexponents
📏 Medium solution 💡 2 insights
Problem
The expression 4000·(2/5)ⁿ uses an integer exponent n that may be positive, zero, or negative. Count how many integer values of n make the whole expression a whole number (an integer).

Pick an answer.

(A)
3
(B)
4
(C)
6
(D)
8
(E)
9

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

Tool #13 (Convert to Algebra): rewrite everything with prime powers so the vague phrase 'is an integer' becomes an exact algebraic condition on exponents. Tool #7 (Identify Subproblems): the only primes around are 2 and 5, so the single condition splits into one rule for the power of 2 and one rule for the power of 5. Tool #2 (Make a Systematic List): once those two rules pin n to a range, list the integers in that range and count them.

1STEP 1

Factor 4000 into primes

Break 4000 into primes: 4000 = 2⁵· 5³, so the whole expression is built from only the primes 2 and 5.

4000 = 2⁵· 5³
2STEP 2

Rewrite as one prime product

Apply exponent rules: (2/5)ⁿ = 2ⁿ· 5⁻ⁿ, so multiplying by 2⁵· 5³ and adding exponents gives 4000·(2/5)ⁿ = 2⁵⁺ⁿ· 5³⁻ⁿ.

4000·(2/5)ⁿ = 2⁵· 5³· 2ⁿ· 5⁻ⁿ = 2⁵⁺ⁿ· 5³⁻ⁿ
3STEP 3

Turn 'integer' into exponent rules

It is an integer only when neither prime falls into the denominator, that is 5+n ≥ 0 and 3-n ≥ 0.

2⁵⁺ⁿ· 5³⁻ⁿinZ⇔ 5+n ≥ 0 and 3-n ≥ 0
4STEP 4

Solve the two inequalities

Solving each gives n ≥ -5 and n ≤ 3, and both must hold at once, so -5 ≤ n ≤ 3.

n ≥ -5 and n ≤ 3 ⟹ -5 ≤ n ≤ 3
5STEP 5

Count the integers in range

Count the integers from -5 to 3 as 3-(-5)+1, giving 9 integer values of n — choice (E).

-5,-4,-3,-2,-1,0,1,2,3 → 3-(-5)+1 = 9 → (E)
Answer
9
Sanity-check the two ends. At n=3: 2⁸· 5⁰=256, an integer. At n=-5: 2⁰· 5⁸=390625, an integer. Step just past each end fails: n=4 gives 5⁻¹ (a fifth in the denominator) and n=-6 gives 2⁻¹ (a half in the denominator). So -5 and 3 really are the last values that work, and every integer between them works too. The count 9 matches choice (E), and it is the largest option, which fits a 'both signs allowed' problem that deliberately rewards remembering the negative side.
💡Key takeaway

Write 4000·(2/5)ⁿ as 2⁵⁺ⁿ· 5³⁻ⁿ; it stays a whole number only while both exponents are ≥ 0, so -5 ≤ n ≤ 3, which is 9 values — choice (E).

  • Factor 4000 into primes
  • Rewrite as one prime product
  • Turn 'integer' into exponent rules
  • Solve the two inequalities
  • Count the integers in range

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