AMC 10 · 2004 · #3
Grade 6 number-theoryPick an answer.
One equation carries two unknown exponents, which looks tangled until you notice the two primes never mix. That splits the problem into two independent counts — how many 2's hide in 1296, and how many 3's — which is Tool #7 (Identify Subproblems). Each count is done with Tool #11 (Work Backwards): the finished product 1296 is given, so divide it back down to recover the exponents. Tool #3 (Eliminate Possibilities) then closes the gap the counting leaves open, showing no other pair (x,y) can work, which is what makes x+y a single number.
Split into two separate counts
The two primes cannot substitute for each other, so the equation splits in two.
Different primes never stand in for each other, so one messy equation becomes two clean counts.
Different primes can never substitute for one another, so one messy equation splits into two clean counts.
▸ Why?
A number has only one prime recipe, so the count of 2s and the count of 3s in it are each fixed once and for all.
▸ Why?
Two matching powers of the same base force their exponents to agree, so each side hands over one exact number.
Peel off the factors of 2
Halving until odd peels off exactly four factors of two.
Halve until you land on an odd number; the number of halvings is the exponent of 2.
4.OA.B.4Work BackwardsPeel off the factors of 3
The leftover 81 divides down to give four factors of three.
Keep dividing until you reach 1; the count of divisions is the exponent.
6.EE.A.1Work BackwardsShow no other pair fits
Parity rules out every other pair, so x plus y is 8, choice (A).
Powers of 3 are always odd, so every factor of 2 has to come from the 2^x side — and that pins x down to one value.
6.EE.B.5Eliminate PossibilitiesBreak a number into its primes, then match one prime at a time across the equation — the exponents can only line up one way.
- Split into two separate counts
- Peel off the factors of 2
- Peel off the factors of 3
- Show no other pair fits