AMC 8 · 2018 · #14
Grade 4 number-theoryProblem
Let N be the greatest five-digit number whose digits have a product of 120. What is the sum of the digits of N?
Pick an answer.
AMC 8 2018 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: Among all five-digit numbers whose five digits multiply to $120$, find the greatest one (call it $N$), and report the sum of its digits.
Givens: $N$ has exactly $5$ digits (so $N$ is between $10000$ and $99999$); The product of the five digits of $N$ equals $120$; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $18$, (E) $20$
Unknowns: The sum of the digits of $N$, the greatest such five-digit number
Understand
Restated: Among all five-digit numbers whose five digits multiply to $120$, find the greatest one (call it $N$), and report the sum of its digits.
Givens: $N$ has exactly $5$ digits (so $N$ is between $10000$ and $99999$); The product of the five digits of $N$ equals $120$; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $18$, (E) $20$
Plan
Primary tool: #6 Guess and Check
Secondary: #7 Identify Subproblems
To maximize a multi-digit number, the leftmost (highest place value) digit matters most, then the next, and so on. So we use Tool #6 (Guess and Check) one digit position at a time: try the biggest candidate ($9$, then $8$, then $7$, \ldots) and keep the first one that still lets the remaining digits multiply to the leftover product. Tool #7 (Identify Subproblems) splits the whole task into five smaller "pick one digit" problems: after we choose $d_1$, we just need four digits whose product is $120 / d_1$, and so on. A quick prime factorization, $120 = 2^3 \times 3 \times 5$, makes the divisibility checks instant.
Execute — Answer: D
4.OA.B.4 Step 1 - Factor $120$ into primes so we can quickly tell which single digits are factors.
- $120 = 8 \times 15 = 2^3 \times 3 \times 5$.
- So the only single digits ($1$\text{--}$9$) that divide $120$ are $\{1, 2, 3, 4, 5, 6, 8\}$; note that $7$ and $9$ are NOT factors of $120$.
💡 Listing factor pairs and recognizing prime factors is a Grade 4 skill.
4.OA.B.4 Step 2 - Pick the ten-thousands digit $d_1$ as large as possible.
- Try $9$: $120 \div 9$ is not a whole number, so $9$ fails.
- Try $8$: $120 \div 8 = 15$, a whole number.
- So $d_1 = 8$ and the remaining four digits must multiply to $15$.
💡 Checking which single digit divides $120$ uses Grade 4 factor / multiple reasoning.
3.OA.C.7 Step 3 - Pick the thousands digit $d_2$ as large as possible, given the remaining product is $15$.
- Try $9, 8, 7, 6$: none divides $15$.
- Try $5$: $15 \div 5 = 3$.
- So $d_2 = 5$ and the remaining three digits must multiply to $3$.
💡 Knowing $15 = 3 \times 5$ is basic Grade 3 multiplication and division fluency.
3.OA.C.7 Step 4 - Pick the hundreds digit $d_3$.
- The remaining product is $3$, and the largest single-digit factor of $3$ is $3$ itself.
- So $d_3 = 3$, leaving a remaining product of $3 \div 3 = 1$ for the last two digits.
💡 Dividing $3$ by $3$ is Grade 3 division fluency.
4.NBT.A.2 Step 5 - The last two digits ($d_4, d_5$) must multiply to $1$.
- The only single-digit factorization of $1$ is $1 \times 1$, so $d_4 = 1$ and $d_5 = 1$.
- The five digits are therefore $\{8, 5, 3, 1, 1\}$, and the greatest arrangement (largest digit in the largest place value) is $N = 85311$.
💡 Putting bigger digits in bigger place values to maximize a number is Grade 4 multi-digit place-value comparison.
2.NBT.B.5 Step 6 Add the digits of $N = 85311$ to get the requested sum.
💡 Adding a few one-digit numbers is Grade 2 fluency within $100$.
4.OA.B.4 Factor $120$ into primes so we can quickly tell which single digits are factors. 4.OA.B.4 Pick the ten-thousands digit $d_1$ as large as possible. Try $9$: $120 \div 9$ i 3.OA.C.7 Pick the thousands digit $d_2$ as large as possible, given the remaining product 3.OA.C.7 Pick the hundreds digit $d_3$. The remaining product is $3$, and the largest sin 4.NBT.A.2 The last two digits ($d_4, d_5$) must multiply to $1$. The only single-digit fac 2.NBT.B.5 Add the digits of $N = 85311$ to get the requested sum. Review
Reasonableness: Verify the product first: $8 \times 5 \times 3 \times 1 \times 1 = 40 \times 3 = 120$. Good. Could any larger five-digit number $N' > 85311$ still have digit product $120$? Any such $N'$ would need a ten-thousands digit of $9$, but $9$ does not divide $120$, so the four remaining digits could not multiply to a whole number — impossible. So $85311$ really is the greatest, and the digit sum $8+5+3+1+1 = 18$ matches choice (D).
Alternative: Tool #2 (Systematic List): enumerate all unordered multisets of five digits in $\{1, \ldots, 9\}$ whose product is $120$, e.g. $\{8,5,3,1,1\}$, $\{8,5,3,1,1\}$, $\{6,5,4,1,1\}$, $\{6,5,2,2,1\}$, $\{5,4,3,2,1\}$, $\{5,4,2,3,1\}$ (duplicates), $\{5,3,2,2,2\}$, etc. For each multiset, the greatest number you can form is obtained by sorting digits in descending order. Comparing the resulting numbers ($85311, 65411, 65221, 54321, 53222$) confirms $85311$ is the largest, again giving digit sum $18$.
CCSS standards used (min grade 4)
2.NBT.B.5Fluently add and subtract within 100 (Adding the five digits of $N$ ($8 + 5 + 3 + 1 + 1 = 18$) to get the final answer.)3.OA.C.7Fluently multiply and divide within 100 (Reducing the remaining product after each chosen digit ($120 / 8 = 15$, $15 / 5 = 3$, $3 / 3 = 1$) using basic multiplication and division facts.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Prime-factoring $120 = 2^3 \times 3 \times 5$ and checking which single digits ($1$\text{--}$9$) divide $120$ or $15$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Using place value to argue that putting bigger digits in higher places (and thus arranging $\{8,5,3,1,1\}$ as $85311$) produces the greatest five-digit number.)
⭐ This AMC 8 problem only needs Grade 4 factor pairs and place value you already know!
⭐ This AMC 8 problem only needs Grade 4 factor pairs and place value you already know!
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