AMC 8 · 2011 · #15
Grade 6 arithmeticPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Computing 4⁵ · 5¹⁰ as a brute number is painful, but the bases 4 and 5 are hiding a 2 inside 4. Tool #16 (Change Focus) says: rewrite 4 as 2² so the bases become 2 and 5. Now the 2s and 5s pair up perfectly into 10s, turning the product into a clean power of 10. Tool #9 (Easier Related Problem) then takes over: counting digits in 10¹⁰ is far easier than in 4⁵ · 5¹⁰, and we already know the rule 10ⁿ has n+1 digits.
Since 4 = 2², the power-of-a-power rule gives 4⁵ = (2²)⁵ = 2¹⁰.
Changing the base from 4 to 2 lets the exponents talk to the 5¹⁰ next door.
6.EE.A.1Count The ComplementNow both factors share exponent 10, so 2¹⁰ · 5¹⁰ = (2·5)¹⁰ = 10¹⁰.
Every 2 finds a partner 5 and together they form a 10 — the whole product collapses into 10¹⁰.
6.EE.A.1Count The ComplementSwitch to an easier question — the digits of 10¹⁰; the rule is 10ⁿ has n+1 digits, a 1 followed by n zeros.
Multiplying by 10 tacks on a zero — a Grade 5 place-value pattern.
5.NBT.A.2Solve An Easier Related ProblemApply n = 10: 10¹⁰ has 10 + 1 = 11 digits.
One leading 1 plus ten trailing zeros equals eleven digits total.
5.NBT.A.2Solve An Easier Related ProblemWhenever you see 2s and 5s with matching exponents, pair them into 10s — the answer just falls out as a power of 10.