AMC 8 · 2014 · #22
Grade 3 algebraPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The answer choices give only 5 possible units digits (1, 3, 5, 7, 9), so we can test each one directly. For each candidate units digit b, pick a simple tens digit (say a = 2) and check whether 10a + b equals a × b + a + b. Tool #6 (Guess and Check) does the testing; Tool #3 (Eliminate) discards every candidate that fails. After two or three checks a clear pattern jumps out (Tool #5), confirming the winning digit without needing algebra.
Test b = 1 with a = 2 (number 21): (2 × 1) + (2 + 1) = 5 ≠ 21, so eliminate (A).
Multiplying and adding single-digit numbers is a Grade 3 fluency skill.
3.OA.C.7Guess And CheckTest b = 3 with a = 2 (number 23): (2 × 3) + (2 + 3) = 11 ≠ 23, so eliminate (B).
Each failed check rules out one answer choice — the classic Tool #3 (Eliminate) move on a multiple-choice problem.
3.OA.C.7Eliminate PossibilitiesLook for a pattern: the number minus (product + sum) equals a × (9 - b), which is zero only when b = 9.
Spotting the common factor a in the gap is a Grade 3 distributive-property observation, not full algebra.
3.OA.B.5Look For A PatternQuick-check the rest: (2 × 5) + (2 + 5) = 17 ≠ 25 and (2 × 7) + (2 + 7) = 23 ≠ 27 — eliminate (C), (D).
The gap formula predicts 2 × (9-5) = 8 and 2 × (9-7) = 4, matching 25 - 17 and 27 - 23 — the pattern holds.
3.OA.C.7Eliminate PossibilitiesTest b = 9: 29 = (2 × 9) + (2 + 9) and 59 = (5 × 9) + (5 + 9) both hold, so the units digit is 9 → (E).
Verifying with two different tens digits confirms b = 9 works for every valid a — Grade 3 multi-step word-problem checking.
3.OA.D.8Guess And CheckThis AMC 8 problem only needs Grade 3 multiplication-and-addition checking — just test each answer choice and the right units digit shows itself!