AMC 8 · 2017 · #5
Easy mode Grade 4Problem
Look at this fraction:
1+2+3+4+5+6+7+81⋅2⋅3⋅4⋅5⋅6⋅7⋅8
The top is the numbers from 1 to 8 all multiplied together. The bottom is the same numbers all added together.
Work out the top. Then work out the bottom. Then do the division.
What is the value of the fraction?
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: Evaluate the fraction whose numerator is the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$ and whose denominator is the sum $1+2+3+4+5+6+7+8$, then pick the matching answer choice.
Givens: Numerator $= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$ (the product of the first $8$ positive integers); Denominator $= 1+2+3+4+5+6+7+8$ (the sum of the first $8$ positive integers); Answer choices: (A) $1020$, (B) $1120$, (C) $1220$, (D) $2240$, (E) $3360$
Unknowns: The numeric value of $\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$, matched to a letter (A)-(E)
Understand
Restated: Evaluate the fraction whose numerator is the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$ and whose denominator is the sum $1+2+3+4+5+6+7+8$, then pick the matching answer choice.
Givens: Numerator $= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$ (the product of the first $8$ positive integers); Denominator $= 1+2+3+4+5+6+7+8$ (the sum of the first $8$ positive integers); Answer choices: (A) $1020$, (B) $1120$, (C) $1220$, (D) $2240$, (E) $3360$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern
The expression splits cleanly into two independent subproblems: the denominator (a sum) and the numerator (a product), so Tool #7 says 'solve each piece, then combine'. Tool #5 helps with the sum step: the familiar pattern $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ gives the denominator instantly as $\dfrac{8 \cdot 9}{2} = 36$. Once the denominator is in hand, we treat the big fraction as a third subproblem — simplify by spotting $36 = 4 \times 9$ inside the numerator and canceling — instead of multiplying out $8! = 40320$ and dividing.
Execute — Answer: B
2.NBT.B.5 Step 1 - First subproblem: add the integers from $1$ to $8$.
- Notice the pattern that the first $n$ counting numbers sum to $\dfrac{n(n+1)}{2}$ (pair the ends: $1+8=9$, $2+7=9$, $3+6=9$, $4+5=9$ — four pairs of $9$).
- With $n=8$ this gives $\dfrac{8 \cdot 9}{2} = 36$.
💡 Adding small whole numbers up to $100$ is a Grade 2 fluency skill; the pairing pattern is just a faster way to do that same addition.
4.OA.A.3 Step 2 Second subproblem: rewrite the original fraction with the denominator we just found, so the question becomes $\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36}$.
💡 Substituting the computed sum back into the fraction is a multi-step Grade 4 word-problem move — finish one piece, then use it.
4.OA.B.4 Step 3 - Third subproblem: factor the denominator into pieces that already live inside the numerator.
- Find factor pairs of $36$: $36 = 4 \times 9 = 4 \times 3 \times 3$.
- The numerator already contains a $4$, a $3$, and a $6 = 2 \times 3$ — so all the factors of $36$ are hidden inside it.
💡 Listing factor pairs of $36$ and checking which factors are 'already there' is exactly the Grade 4 factor-pair / divisibility skill.
4.NF.A.1 Step 4 - Cancel the matching factors.
- The $4$ in the numerator cancels the $4$ in $36$; the $3$ in the numerator cancels one of the $3$s; and the $6 = 2 \cdot 3$ in the numerator cancels the other $3$ while leaving a $2$ behind.
- Canceling matched factors gives an equivalent fraction, which is the same idea as reducing $\tfrac{6}{8} = \tfrac{3}{4}$ — just with more factors.
💡 Dividing the top and the bottom by the same number to get an equivalent fraction is the Grade 4 equivalent-fractions rule applied piece by piece.
3.OA.C.7 Step 5 Multiply what is left, grouping $2 \times 5 = 10$ first to make the arithmetic painless: $1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8 = (2 \cdot 5) \cdot (2 \cdot 7) \cdot 8 = 10 \cdot 14 \cdot 8 = 10 \cdot 112 = 1120$, which is choice (B).
💡 Each remaining multiplication ($2 \cdot 5$, $2 \cdot 7$, $14 \cdot 8$, $\times 10$) is a within-$100$ basic fact — Grade 3 multiplication fluency.
2.NBT.B.5 First subproblem: add the integers from $1$ to $8$. Notice the pattern that the 4.OA.A.3 Second subproblem: rewrite the original fraction with the denominator we just fo 4.OA.B.4 Third subproblem: factor the denominator into pieces that already live inside th 4.NF.A.1 Cancel the matching factors. The $4$ in the numerator cancels the $4$ in $36$; t 3.OA.C.7 Multiply what is left, grouping $2 \times 5 = 10$ first to make the arithmetic p Review
Reasonableness: Sanity-check by computing $8!$ the long way: $8! = 40320$, and $40320 \div 36 = 1120$, matching choice (B). Magnitude makes sense too — $40320 / 36$ should land near $40000 / 40 = 1000$, so a four-digit answer just above $1000$ is exactly the right size, ruling out (D) $2240$ and (E) $3360$ on size alone.
Alternative: Tool #3 (Eliminate Possibilities). The denominator $36$ has $9$ as a factor, so the answer must be divisible by $1$ (trivial), and the original numerator $8!$ is divisible by $9$ (since $3 \cdot 6$ contributes $9$). More usefully, $8!$ is divisible by $7$ (it contains a $7$), so the quotient $8!/36$ must also be divisible by $7$. Testing the choices: $1020/7$ is not whole, $1120/7 = 160$ is, $1220/7$ is not, $2240/7 = 320$ is, $3360/7 = 480$ is. Now use divisibility by $8$: $8!/36 = 8!/36$ contains the factor $8$ untouched, so the answer must be divisible by $8$. $1120/8 = 140$ ✓, $2240/8 = 280$ ✓, $3360/8 = 420$ ✓. Finally, the answer must be exactly $40320/36 = 1120$, and only (B) gives the magnitude that matches a rough estimate of $40000/40 = 1000$.
CCSS standards used (min grade 4)
2.NBT.B.5Fluently add and subtract within 100 (Adding the eight whole numbers $1+2+\cdots+8 = 36$ to compute the denominator.)3.OA.C.7Fluently multiply and divide within 100 (Carrying out the small multiplications after cancellation ($2 \times 5$, $2 \times 7 = 14$, $14 \times 8 = 112$) and the final $\times 10$ to reach $1120$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Sequencing the calculation: first find the sum, then substitute, then simplify, then multiply — a multi-step plan combining addition, multiplication, and division.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing $36 = 4 \times 9 = 4 \times 3 \times 3$ and matching those factors to the $4$, $3$, and $6$ already sitting in the numerator.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Canceling common factors from the numerator and the denominator to produce an equivalent — but much simpler — fraction before multiplying.)
⭐ This AMC 8 problem only needs Grade 4 factor pairs and equivalent fractions you already know — no calculator, just smart cancellation!
⭐ This AMC 8 problem only needs Grade 4 factor pairs and equivalent fractions you already know — no calculator, just smart cancellation!
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