AMC 10 · 2022 · #3

Grade 4 arithmetic
paritydigit-countingsystematic-enumerationcombinations-basic caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: paritymulti-digit-arithmetic
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Problem
Among all three-digit numbers (from 100 to 999), how many have an odd number of even digits? Each digit is independently classified as even ({0, 2, 4, 6, 8}) or odd ({1, 3, 5, 7, 9}). "Odd number of even digits" means exactly 1 or exactly 3 even digits.

Pick an answer.

(A)
150
(B)
250
(C)
350
(D)
450
(E)
550

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

"How many three-digit numbers with __" is a counting problem. Tool #2 (Systematic List) gives us the order: list every digit-pattern by where the even digits sit. The valid patterns split into Tool #7 (Subproblems) — Case A: exactly one even digit (patterns EOO, OEO, OOE); Case B: all three even (EEE). For each pattern we use the multiplication principle, remembering the hundreds digit can't be 0. Tool #3 (Eliminate) gives a sanity range — the answer must be under 900 and one of the five listed values.

1STEP 1

Count choices per slot by parity: hundreds H has 4 even / 5 odd (no leading 0); tens and ones each have 5 even and 5 odd.

H-even = 4, H-odd = 5, T-even = U-even = 5, T-odd = U-odd = 5
2STEP 2

Case A, exactly one even digit: EOO gives 4·5·5 = 100, OEO gives 5·5·5 = 125, OOE gives 5·5·5 = 125.

EOO = 4 · 5 · 5 = 100, OEO = 5 · 5 · 5 = 125, OOE = 5 · 5 · 5 = 125
3STEP 3

Add the Case A patterns: 100 + 125 + 125 = 350 numbers with exactly one even digit.

100 + 125 + 125 = 350
4STEP 4

Case B, all three digits even (EEE): 4·5·5 = 100.

EEE = 4 · 5 · 5 = 100
5STEP 5

Add the two cases: 350 + 100 = 450, choice (D).

350 + 100 = 450 → (D)
Answer
450
Symmetry check. The total number of three-digit integers is 900. For each three-digit number the parity count of even digits is 0, 1, 2, or 3. Pair up complementary cases: numbers with 0 even digits (OOO = 5 · 5 · 5 = 125) plus numbers with 2 even digits (EEO + EOE + OEE = 100 + 100 + 125 = 325) gives 450 for the "even count of even digits" side. The remaining 900 - 450 = 450 must be the "odd count of even digits" side — exactly our answer (D). The two sides split 900 down the middle.
💡Key takeaway

This AMC 10 problem only needs Grade 4 "organize cases, multiply slot choices, then add" — exactly 1 even digit gives 350 numbers, exactly 3 even gives 100, and the total is 450.