AMC 10 · 2021 · #8
Grade 8 arithmeticPick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only five candidate values {15, 30, 45, 60, 75} — Tool #6 (Guess and Check) plugs each into the repeating-vs-terminating gap and tests whether 66 × (gap) = 0.5. Tool #3 (Eliminate) discards every N that misses 0.5. Tool #11 (Work Backwards) is the verification: starting from the required gap = , run the gap formula in reverse to recover N in one step — this confirms the unique answer without sweeping all five choices.
Let N be the two-digit block. The terminating 1.ab is 1 + ; the repeating 1.ab is 1 + .
Two-place terminating → /100; two-place repeating → /99. Standard Grade 8 conversion.
8.NS.A.1Work BackwardsSubtract; the 1s cancel and - = over the common denominator 9900.
The two fractions differ by exactly — a small number because the missed repeat is tiny.
5.NF.A.1Work BackwardsMultiply the gap by 66: 66· = , since 9900 = 150 × 66.
9900 = 150 × 66, so collapses to .
6.NS.B.3Work BackwardsThe problem's gap is 0.5, so = 0.5 and N = 75.
Running the formula backwards: multiply both sides by 150.
6.EE.B.7Work BackwardsCheck N=75: correct 66·(1+) minus wrong 66·1.75, gap = = 0.5 ✓.
Plug the candidate back into the original word relation — must hit 0.5 exactly.
7.NS.A.3Guess And CheckN = 75 matches choice (E).
Read the two-digit value off the five options.
4.NBT.A.2Eliminate PossibilitiesThis AMC 10 problem only needs Grade 8 repeating-decimal-to-fraction conversion you already know — turn 1.ab = 1 + and 1.ab = 1 + , find their gap , multiply by 66 to get = 0.5, and N = 75.