AMC 8 · 2022 · #6
Grade 3 arithmeticalgebraPick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a multiple-choice problem where each candidate for the smallest number is a small whole number (4, 5, 6, 7, 8). For each candidate x, we can compute the largest number (4x) and then check whether 15 sits exactly in the middle of x and 4x (equal gaps). Tool #6 (Guess and Check) tests each candidate directly, and Tool #3 (Eliminate Possibilities) lets us cross off the ones that fail. This avoids reaching for algebra (Tool #13) when the bounded answer set makes plug-and-check the fastest, most age-appropriate path.
"Equally spaced" means 15 sits exactly halfway between the smallest and the largest, so the two gaps must be equal.
Equal gaps is a one-line rule — once you see it, the rest is just arithmetic checking on each choice.
3.OA.D.8Guess And CheckTest x = 4: largest = 4 × 4 = 16, but gaps 15 - 4 = 11 and 16 - 15 = 1 differ, so (A) fails.
Multiplying 4 × 4 and subtracting are Grade 3 multiplication / word-problem skills.
3.OA.A.3Eliminate PossibilitiesTest x = 5: largest = 4 × 5 = 20, but gaps 15 - 5 = 10 and 20 - 15 = 5 differ, so (B) fails.
Same Grade 3 multiply-and-subtract check, just on the next candidate.
3.OA.A.3Eliminate PossibilitiesTest x = 6: largest = 4 × 6 = 24, and gaps 15 - 6 = 9 and 24 - 15 = 9 match — this works!
When both differences come out the same, the "equally spaced" rule is satisfied — we've found the smallest number.
3.OA.A.3Guess And CheckCheck the rest: x = 7 gives gaps 8 and 13, x = 8 gives 7 and 17 — both unequal, so only (C) 6 survives.
Eliminating the rest of the choices makes sure (C) is the only valid answer — no second contestant slipped through.
3.OA.D.8Eliminate PossibilitiesThis AMC 8 problem only needs Grade 3 multiplication and subtraction you already know — just try each answer choice and see which one makes equal gaps!