AMC 10 · 2014 · #20
Grade 6 number-theoryThe product (8)(888…8), where the second factor has k digits, is an integer whose digits have a sum of 1000. What is k?
Pick an answer.
AMC 10 2014 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 $8$ by a number made of $k$ eights (like $88\dots8$). The resulting integer has digits that add up to $1000$. Find how many eights $k$ the second factor has.
Givens: The first factor is $8$; The second factor is $\underbrace{88\dots8}_{k}$, a string of $k$ eights; The digits of the product add up to $1000$; Five answer choices: (A) $901$, (B) $911$, (C) $919$, (D) $991$, (E) $999$
Unknowns: The number of digits $k$ in the second factor
Understand
Restated: Multiply $8$ by a number made of $k$ eights (like $88\dots8$). The resulting integer has digits that add up to $1000$. Find how many eights $k$ the second factor has.
Givens: The first factor is $8$; The second factor is $\underbrace{88\dots8}_{k}$, a string of $k$ eights; The digits of the product add up to $1000$; Five answer choices: (A) $901$, (B) $911$, (C) $919$, (D) $991$, (E) $999$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #4 Introduce a Variable, #13 Convert to Algebra
A number with $k$ eights is far too big to multiply directly for the size of $k$ we expect. So compute a few small cases and hunt for a pattern (Tool #5): the products $8\cdot88$, $8\cdot888$, and so on. Once the digit shape of the product is clear, name the count of eights with a variable $k$ (Tool #4) to write the digit sum as a formula in $k$, then turn "digit sum $=1000$" into a simple equation and solve it (Tool #13).
Execute — Answer: D
5.NBT.B.5 Step 1 Multiply the first few cases
- Start small and actually do the multiplication for a few sizes.
- Work out $8$ times a string of $2$, $3$, $4$, and $5$ eights.
- Keep the full products so you can compare their digits side by side.
💡 Doing a few concrete products by hand lets the shape of the answer show itself.
4.OA.C.5 Step 2 Read off the digit pattern
- Line the products up: $704$, $7104$, $71104$, $711104$.
- Each one starts with a $7$, ends with $04$, and has a block of $1$s in the middle.
- Counting, the second factor with $k$ eights produces a $7$, then $(k-2)$ ones, then $0$ and $4$.
- So the product looks like $7\underbrace{1\dots1}_{k-2}04$.
💡 The middle just gains one more $1$ each time you add an $8$, so the rule is easy to extend.
4.OA.C.5 Step 3 Write the digit sum in terms of k
- Now add the digits of that product.
- There is one $7$, then $(k-2)$ copies of $1$, then a $0$ and a $4$.
- Adding these gives $7+(k-2)+0+4$.
- Combine the constants: $7-2+4=9$, so the digit sum is $k+9$.
💡 Every extra $8$ adds exactly one more $1$ to the middle, so each step raises the digit sum by just $1$.
6.EE.B.7 Step 4 Solve for k
- The problem says the digit sum equals $1000$.
- Set the formula equal to $1000$ and solve: $k+9=1000$, so $k=1000-9=991$.
- That matches choice (D).
💡 Once the digit sum is a tidy formula $k+9$, finding $k$ is just undoing the $+9$.
5.NBT.B.5 Start small and actually do the multiplication for a few sizes. Work out $8$ tim 4.OA.C.5 Line the products up: $704$, $7104$, $71104$, $711104$. Each one starts with a $ 4.OA.C.5 Now add the digits of that product. There is one $7$, then $(k-2)$ copies of $1$ 6.EE.B.7 The problem says the digit sum equals $1000$. Set the formula equal to $1000$ an Review
Reasonableness: Test the formula on a case we computed: with $k=3$ the product is $7104$, whose digits sum to $7+1+0+4=12$, and the formula gives $k+9=3+9=12$. It matches, so trusting $k+9=1000$ gives $k=991$. Choice (D) also sits sensibly just below $999$, which fits a digit sum barely over $k$.
Alternative: Skip the digit shape and track only how the digit sum grows (Tool #5 on the sums themselves): the sums are $11,12,13,\dots$ for $k=2,3,4,\dots$, so the sum is always $9$ more than $k$. Setting $k+9=1000$ again gives $k=991$.
CCSS standards used (min grade 6)
5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the sample products $8\cdot88=704$, $8\cdot888=7104$, and so on.)4.OA.C.5Generate a number or shape pattern following a given rule (Spotting the $7\,1\dots1\,04$ digit pattern and turning it into the digit-sum rule $k+9$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form x + p = q (Solving $k+9=1000$ to get $k=991$.)
⭐ Multiply a few small cases, catch the digit pattern, and the giant problem shrinks to $k+9=1000$.
⭐ Multiply a few small cases, catch the digit pattern, and the giant problem shrinks to $k+9=1000$.
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