Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #17
Grade 6 number-theoryPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The expression 2w + 3x + 5y + 7z depends on four unknown exponents, so the problem really has two clean subproblems (Tool #7): (1) find the prime factorization of 588, and (2) read off w, x, y, z by matching exponents, then plug in. Tool #11 (Work Backwards) captures the reverse-engineering move: we are handed the product 588 and have to recover the exponents that built it. We deliberately avoid Tool #13 (Algebra) because no equation manipulation is needed — pulling out prime factors and matching is enough.
Factor 588 into primes
Pull out 2 twice from 588, then 3 from 147, leaving 49 = 7 · 7, so 588 = 2² · 3¹ · 7².
Finding factor pairs and identifying prime factors is exactly the Grade 4 standard 4.OA.B.4 — recognize that whole numbers are products of primes.
4.OA.B.4Identify SubproblemsMatch the exponents
Match same bases to read the exponents; base 5 is absent so y = 0, giving w = 2, x = 1, z = 2.
Reading off exponents from a factorization is the Grade 6 "whole-number exponents" standard 6.EE.A.1 in reverse — you know the value, you recover the exponent.
Comparing 2^w · 3^x · 5^y · 7^z against 588 forces each of the four exponents to exactly one whole number.
▸ Why?
For the primes 2, 3, and 7 that do appear in 588, the left side must repeat each of them the same number of times 588 does, since a whole number breaks into primes in only one way.
▸ Why?
Every whole number above 1 splits into primes in exactly one way apart from order, so however you pull the primes out of 588 you always land on the same primes the same number of times — the exponent on each base is decided by 588 alone and cannot be anything else.
▸ Why?
For the prime 5, which never appears when 588 is broken down, the left side must carry no factor of 5, so 5^y has to equal 1 and y must count down to zero.
▸ Why?
A factor of 1 multiplies by nothing and leaves the product unchanged, so the 5-part fits 588 only while it stays equal to 1 — slipping in even one 5 would change the number.
Plug in and compute
Substitute w = 2, x = 1, y = 0, z = 2 into 2w + 3x + 5y + 7z to get 21.
Evaluating a numerical expression with multiplication and addition follows Grade 5 order-of-operations standard 5.OA.A.1.
5.OA.A.1Identify SubproblemsThis AMC 8 problem only needs Grade 6 exponent reasoning — break 588 into primes, read off the exponents, plug in — that you already know!
- Factor 588 into primes
- Match the exponents
- Plug in and compute
A parent dashboard for the family lives at sensimlab.com.