Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #13
Grade 4 number-theoryPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only four parity pairings for (n, m): (even, even), (even, odd), (odd, even), (odd, odd). Tool #2 (Make a Systematic List) writes them all down so none are missed. Tool #3 (Eliminate Possibilities) then uses the given condition n²+m² even to throw out the parity pairs that violate it, leaving only the survivors. Checking each answer choice against the survivors tells us which choice is impossible. We deliberately avoid heavier tools (algebra, formal proof) — a small parity table is the cleanest path.
Table the parity cases
List all four even/odd combos for (n, m). Squaring never changes parity, so n² + m² matches the parity of n + m.
Building the parity table from scratch is the Grade 4 "even and odd, factors and multiples" idea applied systematically.
4.OA.B.4Make A Systematic ListKeep the even-sum cases
Apply n² + m² even: drop the two odd-result rows. Survivors are (even, even) and (odd, odd), so n and m must share the same parity.
Throwing out rows that contradict the given is the core eliminate-possibilities move, and the pattern "squares keep parity, so sum-of-squares parity matches n+m parity" is Grade 3 pattern reasoning.
3.OA.D.9Eliminate PossibilitiesTest choices A, B, C
Test the survivors: (A) both even and (B) both odd each occur, and (C) n + m even holds in both — so A, B, C are all possible.
Verifying each choice against the survivor list is a clean elimination check using even/odd rules.
4.OA.B.4Eliminate PossibilitiesTest choice D
Now (D): an odd n + m needs one even and one odd, but survivors share parity — so n + m can never be odd here.
Same-parity pairs always sum to an even number, so an odd sum is ruled out — the eliminate-possibilities tool delivers the answer directly.
Given that n²+m² is even, n+m can never be odd.
▸ Why?
n+m is odd only when n and m have opposite parity — one even and one odd — but the given condition forces them to share parity, so the mismatch never happens and the sum stays even.
▸ Why?
The condition forces n and m to share parity: a square has the same parity as its base, so n²+m² being even means n² and m² are both even or both odd, and therefore n and m are too.
▸ Why?
A square has the same parity as its base because k² is k multiplied by itself: even times anything is even and odd times odd stays odd, so n² matches n and m² matches m.
▸ Why?
Two numbers add to an even total only when they share parity: an even number splits into two equal groups with nothing left over, and a lone leftover on just one side would have no partner to pair with, so an even n²+m² forces n² and m² to match.
▸ Why?
When n and m share parity the sum splits cleanly into two equal groups: even + even leaves no leftover, and odd + odd pairs its two leftovers into one more group, so the total is even and never odd.
Rule out choice E
(D) is impossible while (A), (B), (C) are possible, so (E) is false — the unique answer is (D).
Exactly one choice survives the elimination — that is the impossible scenario the problem asks for.
4.OA.B.4Eliminate PossibilitiesThis AMC 8 problem only needs the Grade 4 even/odd rule — squaring keeps parity, so n²+m² and n+m are always the same parity!
- Table the parity cases
- Keep the even-sum cases
- Test choices A, B, C
- Test choice D
- Rule out choice E
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