AMC 8 · 2017 · #4
Easy mode Grade 5Problem
Picture multiplying these two numbers together: 0.000315 and 7,928,564.
The first number is a small decimal. The second number is close to 8 million.
You don't need the exact answer. You just need to know which of the choices is closest to the real product.
Which one of the choices below is closest?
Pick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Multiply the tiny decimal $0.000315$ by the large whole number $7{,}928{,}564$ and decide which of the five answer choices the product is closest to. Because the choices ($210, 240, 2100, 2400, 24000$) are spread across very different orders of magnitude, an exact long multiplication is unnecessary — a careful estimate will pick out the right one.
Givens: First factor: $0.000315$ (a small decimal between $0$ and $1$); Second factor: $7{,}928{,}564$ (a $7$-digit whole number, close to $8$ million); Answer choices: (A) $210$, (B) $240$, (C) $2100$, (D) $2400$, (E) $24000$
Unknowns: Which answer choice is closest to the product $0.000315 \times 7{,}928{,}564$
Understand
Restated: Multiply the tiny decimal $0.000315$ by the large whole number $7{,}928{,}564$ and decide which of the five answer choices the product is closest to. Because the choices ($210, 240, 2100, 2400, 24000$) are spread across very different orders of magnitude, an exact long multiplication is unnecessary — a careful estimate will pick out the right one.
Givens: First factor: $0.000315$ (a small decimal between $0$ and $1$); Second factor: $7{,}928{,}564$ (a $7$-digit whole number, close to $8$ million); Answer choices: (A) $210$, (B) $240$, (C) $2100$, (D) $2400$, (E) $24000$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #3 Eliminate Possibilities
The exact product $0.000315 \times 7{,}928{,}564$ 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 — $0.000315 \to 0.0003$ and $7{,}928{,}564 \to 8{,}000{,}000$ — 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.
Execute — Answer: D
5.NBT.A.4 Step 1 - Replace each factor with a nearby number that has only one non-zero digit.
- $0.000315$ is very close to $0.0003$ (we drop the last two digits, $15$, which barely change the value).
- $7{,}928{,}564$ is very close to $8{,}000{,}000$ (we round up by about $1\%$).
💡 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.2 Step 2 - Multiply the easy version by separating the leading digits from the zeros.
- $0.0003 = 3 \times \tfrac{1}{10{,}000}$ and $8{,}000{,}000 = 8 \times 1{,}000{,}000$.
- So the product is $(3 \times 8) \times \tfrac{1{,}000{,}000}{10{,}000}$.
💡 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 $3 \times 8 = 24$ with a $\times 100$ rescue.
4.NBT.A.2 Step 3 - Match the estimate $2400$ against the five choices and eliminate the ones that are in the wrong order of magnitude.
- (A) $210$ and (B) $240$ are roughly $10$ times too small; (E) $24{,}000$ is $10$ times too big; (C) $2100$ is in the right ballpark but our estimate of $2400$ is closer to (D) than to (C).
- The honest rounding (one factor up by $\sim 1\%$, the other down by $\sim 5\%$) is small enough that (D) is the answer.
💡 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.
5.NBT.A.4 Replace each factor with a nearby number that has only one non-zero digit. $0.00 5.NBT.A.2 Multiply the easy version by separating the leading digits from the zeros. $0.00 4.NBT.A.2 Match the estimate $2400$ against the five choices and eliminate the ones that a Review
Reasonableness: A quick sanity check: $3 \times 8 = 24$, and the place-value bookkeeping gives $10^{-4} \times 10^{6} = 10^{2} = 100$, so the product is around $24 \times 100 = 2400$. The exact product is $0.000315 \times 7{,}928{,}564 \approx 2497.5$, which rounds to $2500$ — much closer to (D) $2400$ than to (C) $2100$ or (E) $24{,}000$. The answer (D) is consistent with both the rough estimate and the true value.
Alternative: Tool #8 (Analyze the Units) via scientific notation: write $0.000315 = 3.15 \times 10^{-4}$ and $7{,}928{,}564 = 7.93 \times 10^{6}$. Then the product is $(3.15 \times 7.93) \times 10^{-4+6} = (\text{about } 25) \times 10^{2} \approx 2500$. Same answer (D), reached by tracking exponents instead of swapping in easier numbers.
CCSS standards used (min grade 5)
5.NBT.A.4Round decimals to any place (Rounding $0.000315$ down to $0.0003$ and $7{,}928{,}564$ up to $8{,}000{,}000$ — the place-value rounding move that turns the messy product into a one-digit-by-one-digit estimate.)5.NBT.A.2Explain patterns in number of zeros and placement of decimal point (Reducing $0.0003 \times 8{,}000{,}000$ to $3 \times 8 = 24$ followed by a $\times 100$ shift, using the rule that multiplying or dividing by powers of $10$ just moves the decimal point.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing the estimate $2400$ against the five answer choices to eliminate those that are off by a factor of $10$ and pick (D).)
⭐ This AMC 8 problem only needs Grade 5 place-value rounding and the powers-of-$10$ shortcut you already know!
⭐ This AMC 8 problem only needs Grade 5 place-value rounding and the powers-of-$10$ shortcut you already know!
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