AMC 10 · 2014 · #25

Grade 8 number-theorycounting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

exponentspattern-recognitioninterval-arithmetic convert-to-algebracomplementary-countingidentify-subproblems ↑ Prerequisites: exponents

📏 Medium solution 💡 3 insights

Problem

The number

5^{867}

lies between

2^{2013}

and

2^{2014}

. How many pairs of integers

(m,n)

are there with

1 \le m \le 2012

and

5^n \lt 2^m \lt 2^{m+2} \lt 5^{n+1}

Pick an answer.

(A)

278

(B)

279

(C)

280

(D)

281

(E)

282

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Chasing individual pairs (m,n) across a range of thousands is hopeless, so the plan is to convert the whole question into a small system of equations. Tool #16 (Change Focus) reframes 'count pairs' as 'count the blocks (5ⁿ,5ⁿ⁺¹) that hold three powers of two,' because each such block yields exactly one pair. Tool #7 (Identify Subproblems) supplies the two facts that make the count finite: every block holds only 2 or 3 powers of two (since 4 ＜ 5 ＜ 8), and the given anchor 2²⁰¹³ ＜ 5⁸⁶⁷ ＜ 2²⁰¹⁴ fixes how many blocks and how many powers of two are in play. Tool #4 (Introduce a Variable) then names the two block-types a and b, writes 'total blocks' and 'total powers of two' as two linear equations, and solves them. The whole difficulty collapses into subtracting one equation from another.

1STEP 1

Two or three powers of two per block

Call one block the stretch from $5^n$ to $5^{n+1}$ . Since $2^2 \lt 5 \lt 2^3$ , it holds 2 or 3 powers of two — never fewer, never more.

2²=4 ＜ 5 ＜ 8=2³

2STEP 2

Rewrite the target as a 3-power block

The chain $5^n \lt 2^m \lt 2^{m+2} \lt 5^{n+1}$ just says that block holds 3 powers of two, and then m is forced.

5ⁿ ＜ 2^m ＜ 2^m+1 ＜ 2^m+2 ＜ 5ⁿ⁺¹⇔ block (5ⁿ,5ⁿ⁺¹) holds 3 powers of two

3STEP 3

Count the blocks and the powers of two

Blocks $(5^0,5^1)$ through $(5^{866},5^{<span class="hl-ask">867</span>})$ number 867; the powers of two inside, $2^1$ up to $2^{<span class="hl-ask">2013</span>}$ , number 2013.

867 blocks; 2¹,2²,…,2²⁰¹³→ 2013 powers of two

4STEP 4

Set up two equations and solve

With a blocks of 3 and b of 2, $a+b=867$ and $3a+2b=2013$ ; subtracting twice the first leaves 279 — choice (B).

a+b=867, 3a+2b=2013 → a=2013-2 · 867=279 → (B)

Answer

279

The solution a=279 forces b=867-279=588, and both totals check out: 279+588=867 blocks and 3 · 279+2 · 588=837+1176=2013 powers of two. The average block holds 2013/867≈ 2.32 powers of two, comfortably between 2 and 3 and nearer 2, matching the fact that 2-blocks (588) outnumber 3-blocks (279). Neighbouring choices like 278 or 280 would break one of the two exact totals, so 279 is the only consistent value. Every counted m is at most 2011, so the bound m ≤ 2012 is respected.

💡Key takeaway

Each gap between powers of 5 fits either 2 or 3 powers of 2; two totals — 867 gaps and 2013 powers — pin down exactly 279 of them as the roomy 3-power gaps.

Two or three powers of two per block
Rewrite the target as a 3-power block
Count the blocks and the powers of two
Set up two equations and solve

Two or three powers of two per block

Rewrite the target as a 3-power block

Count the blocks and the powers of two

Set up two equations and solve

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