AMC 10 · 2014 · #22
Grade 8 number-theorycountingPick an answer.
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.
Two or three powers of two per block
A block holds only two or three.
Because 5 sits between 4 and 8, one step up in fives equals two or three steps up in twos.
Because five sits between four and eight, one step up in fives equals two or three steps up in twos.
▸ Why?
An exponent counts how many times its base is used, so the two ladders climb at fixed but different rates.
▸ Why?
Since five is strictly between two squared and two cubed, each block of fives can hold only two or three doublings.
Rewrite the target as a 3-power block
The target is exactly a three-block count.
A block roomy enough for three doublings gives exactly one pair, so count roomy blocks instead of pairs.
6.EE.B.5Change Focus Count The ComplementCount the blocks and the powers of two
Counting both totals gives two equations.
Line up the powers of 5 and 2 in order, then just read off how many of each fall in the span.
6.NS.C.7Identify SubproblemsSet up two equations and solve
Solving gives 279, choice (B).
Two totals — how many blocks and how many powers — pin down the two unknown counts exactly.
8.EE.C.8Introduce A VariableEach 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