AMC 8 · 2017 · #5

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorsfraction-arithmeticsequences-arithmeticfactorial identify-subproblems ↑ Prerequisites: multi-digit-arithmeticfraction-arithmetic

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$ ?

Pick an answer.

(A)

1020

(B)

1120

(C)

1220

(D)

2240

(E)

3360

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.

View mode:

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

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

#5 Look for a Pattern 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 $\dfrac{8 \cdot 9}{2} = 36$.

$$1+2+3+4+5+6+7+8 = \dfrac{8 \cdot 9}{2} = 36$$

💡 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.

#7 Identify Subproblems 4.OA.A.3 Step 2

Second subproblem: rewrite the original fraction with the denominator we just found, so the question becomes $\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36}$.

$$\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8} = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36}$$

💡 Substituting the computed sum back into the fraction is a multi-step Grade 4 word-problem move — finish one piece, then use it.

#7 Identify Subproblems 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.

$$36 = 4 \times 9 = 4 \times 3 \times 3$$

💡 Listing factor pairs of $36$ and checking which factors are 'already there' is exactly the Grade 4 factor-pair / divisibility skill.

#7 Identify Subproblems 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.

$$\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \dfrac{1 \cdot 2 \cdot \cancel{3} \cdot \cancel{4} \cdot 5 \cdot (2 \cdot \cancel{3}) \cdot 7 \cdot 8}{\cancel{4} \cdot \cancel{3} \cdot \cancel{3}} = 1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8$$

💡 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.

#7 Identify Subproblems 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).

$$(2 \cdot 5) \cdot (2 \cdot 7) \cdot 8 = 10 \cdot 14 \cdot 8 = 1120 \;\Rightarrow\; \textbf{(B)}$$

💡 Each remaining multiplication ($2 \cdot 5$, $2 \cdot 7$, $14 \cdot 8$, $\times 10$) is a within-$100$ basic fact — Grade 3 multiplication fluency.

[1] #5 2.NBT.B.5 First subproblem: add the integers from $1$ to $8$. Notice the pattern that the

[2] #7 4.OA.A.3 Second subproblem: rewrite the original fraction with the denominator we just fo

[3] #7 4.OA.B.4 Third subproblem: factor the denominator into pieces that already live inside th

[4] #7 4.NF.A.1 Cancel the matching factors. The $4$ in the numerator cancels the $4$ in $36$; t

[5] #7 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.5 Fluently add and subtract within 100 (Adding the eight whole numbers $1+2+\cdots+8 = 36$ to compute the denominator.)
3.OA.C.7 Fluently 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.3 Solve 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.4 Find 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.1 Explain 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

More like this