AMC 10 · 2005 · #13

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
exponentsinterval-arithmetic easier-related-problemidentify-subproblems ↑ Prerequisites: exponentsinterval-arithmetic
📏 Medium solution 💡 2 insights
Problem
A positive integer n must make (130n)⁵⁰ ＞ n¹⁰⁰ and n¹⁰⁰ ＞ 2²⁰⁰ true at the same time. How many such positive integers n are there? The exponents are enormous, so the real work is turning them into a plain range of allowed n.

Pick an answer.

(A)
0
(B)
7
(C)
12
(D)
65
(E)
125

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Numbers like 2²⁰⁰ are impossible to compare directly, so Tool #9 (Solve an Easier Related Problem) is the whole game: rewrite each power so both sides share one exponent, then compare the bases instead of the giant numbers. Tool #7 (Identify Subproblems) splits the double inequality into two independent inequalities — one giving an upper bound on n, one a lower bound. Tool #3 (Eliminate Possibilities) finishes by counting the integers in the surviving range and matching to the answer list.

1STEP 1

Match powers in the left inequality

The right side is a 50th power in disguise: n¹⁰⁰=(n²)⁵⁰, so (130n)⁵⁰ ＞ n¹⁰⁰ wears the exponent 50 on both sides.

(130n)⁵⁰ ＞ n¹⁰⁰=(n²)⁵⁰
2STEP 2

Compare the bases, not the giants

Same exponent, so the bigger base wins: 130n ＞ n². Dividing by the positive n gives 130 ＞ n, so n is at most 129.

130n ＞ n² → 130 ＞ n → n ≤ 129
3STEP 3

Match powers in the right inequality

The other half: fold the giant into a 100th power, 2²⁰⁰=(2²)¹⁰⁰=4¹⁰⁰, so n¹⁰⁰ ＞ 2²⁰⁰ becomes n¹⁰⁰ ＞ 4¹⁰⁰.

n¹⁰⁰ ＞ 2²⁰⁰=(2²)¹⁰⁰=4¹⁰⁰
4STEP 4

Get the lower bound on n

Compare bases again: n¹⁰⁰ ＞ 4¹⁰⁰ holds exactly when n ＞ 4, so the smallest integer allowed is 5.

n¹⁰⁰ ＞ 4¹⁰⁰ → n ＞ 4 → n ≥ 5
5STEP 5

Count the surviving integers

Both bounds at once give 5 ≤ n ≤ 129, and that run holds 129-5+1=125 integers — choice (E).

5 ≤ n ≤ 129 → 129-5+1=125 → (E)
Answer
125
The two bounds come from independent, honest steps: n² ＜ 130n caps n at 129, and n ＞ 4 floors it at 5. Spot-check the ends. At n=5: n¹⁰⁰=5¹⁰⁰ ＞ 4¹⁰⁰=2²⁰⁰ (true) and (130 · 5)⁵⁰=650⁵⁰ ＞ 25⁵⁰=5¹⁰⁰ (true), so 5 works. At n=129: 129 ＜ 130 keeps the left inequality, and 129 ＞ 4 keeps the right, so 129 works. Just outside, n=4 fails (4¹⁰⁰=2²⁰⁰ is not strictly greater than itself) and n=130 fails (130 · 130=130² is not strictly greater than 130²). So the range is exactly 5 through 129, giving 125 integers, matching choice (E).
💡Key takeaway

When powers are too big to compare, rewrite them to share one exponent, then just compare the bases.

  • Match powers in the left inequality
  • Compare the bases, not the giants
  • Match powers in the right inequality
  • Get the lower bound on n
  • Count the surviving integers

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