AMC 10 · 2009 · #18
Grade 8 number-theoryPick an answer.
The question asks for a maximum, so Tool #14 (Extreme Principle) sets the target: find where the count of 2s is pushed as high as possible, which turns out to be a single boundary case. To get there, Tool #4 (Introduce a Variable) rewrites the digit-picture I_k as the clean formula 10^k+2+64 and then as 2^k+25^k+2+2⁶, exposing the two separate sources of 2s. Tool #7 (Identify Subproblems) splits the work by comparing the two powers k+2 and 6: when they differ, the smaller one wins and no extra 2s appear; only when they are equal can the two pieces combine to release more 2s. Tool #6 (Guess and Check) then tests that balance point directly (k=4) to read off the exact count.
Write the number as a formula
A formula replaces the digit description.
Each zero you add just shifts the leading 1 one more place, which multiplies its place value by another 10.
5.NBT.A.2Introduce A VariableSplit each piece into 2s and 5s
Each piece splits into powers of two and five.
Powers of 10 are just paired-up 2s and 5s, so pulling them apart shows exactly where the 2s live.
8.EE.A.1Introduce A VariableFactor out the smaller power of 2
Factoring out the smaller power splits it into cases.
A sum is divisible by 2^m only as far as both parts are, so the weaker term caps the shared 2s.
A sum is divisible by a power of two only as far as both parts are, so the weaker term caps the shared twos.
▸ Why?
Every number has one prime recipe, so the twos in each term are fixed before they are added.
▸ Why?
Once the smaller stack of twos is pulled out, one part is odd and one is even, so the sum is odd and gains nothing more.
Case k ＜ 4: the smaller side wins
One side dominates below the boundary, capping the count at 5.
One odd plus one even is always odd, so the leftover part contributes zero extra 2s.
4.OA.B.4Identify SubproblemsCase k ＞ 4: capped at 64's six 2s
Above it the other side caps the count at 6.
Once 64 is the smaller stack of 2s, adding an even number to it leaves an odd remainder, so it caps the count at 6.
4.OA.B.4Identify SubproblemsBoundary case k = 4: the two sides balance
At the boundary the two balance and an extra factor appears.
Only when the two equal stacks of 2s meet does adding them release a fresh 2.
6.NS.B.3Guess And CheckTake the maximum
The maximum is therefore 7, choice (B).
The peak sits exactly at the boundary where the two powers of two match, not at the extremes of k.
6.EE.A.2Extreme PrincipleSplit the number into its 2s from 10^k+2 and its 2s from 64; you gain an extra factor of 2 only when those two stacks are exactly equal, which happens at k=4 and gives the maximum N=7, choice (B).
- Write the number as a formula
- Split each piece into 2s and 5s
- Factor out the smaller power of 2
- Case k ＜ 4: the smaller side wins
- Case k ＞ 4: capped at 64's six 2s
- Boundary case k = 4: the two sides balance
- Take the maximum