AMC 8 · 2017 · #4
Grade 5 arithmeticPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The exact product 0.000315 × 7,928,564 is painful to compute by hand, but the problem only asks which choice it is closest to. Tool #9 (Easier Related Problem) says: swap the ugly numbers for friendly nearby ones — 0.000315 → 0.0003 and 7,928,564 → 8,000,000 — and solve that easier version instead. Tool #3 (Eliminate Possibilities) then takes our estimate and matches it against the five widely-spaced choices; because (A)/(B) are around 200, (C)/(D) around 2000, and (E) around 24,000, a single-digit estimate is enough to land on exactly one choice.
Swap each factor for the nearest one-nonzero-digit number: 0.0003 and 8,000,000.
Rounding to a single non-zero digit by place value is exactly the Grade 5 "round decimals to any place" idea, applied to both the tiny decimal and the big whole number.
5.NBT.A.4Solve An Easier Related ProblemSplit off the leading digits: 3 × 8 = 24, and 10⁻⁴ × 10⁶ = 10², so the estimate is 2400.
Multiplying by powers of 10 just shifts the decimal point — the Grade 5 "patterns in zeros and the decimal point" standard turns the calculation into 3 × 8 = 24 with a × 100 rescue.
5.NBT.A.2Solve An Easier Related ProblemOnly 2400 matches the order of magnitude — (A)/(B) are 10× too small, (E) is 10× too big, and (C) 2100 is farther.
Comparing multi-digit numbers by place value — Grade 4 — is all that is needed to throw out the choices that are off by a factor of 10 and pick the one that matches the estimate.
4.NBT.A.2Eliminate PossibilitiesThis AMC 8 problem only needs Grade 5 place-value rounding and the powers-of-10 shortcut you already know!