2014 AMC 10A Problem 20

Compute the first few products:
$$8 \cdot 8 = 64,\quad 8 \cdot 88 = 704,\quad 8 \cdot 888 = 7104,\quad 8 \cdot 8888 = 71104,\quad 8 \cdot 88888 = 711104.$$

For $k \ge 2$ the product is a $7$, followed by $k - 2$ ones, followed by $04$. To see why the pattern continues, let $P_k = 8 \cdot \underbrace{88\dots8}_{k}$. Since $\underbrace{88\dots8}_{k} = 10 \cdot \underbrace{88\dots8}_{k-1} + 8$,
$$P_k = 10 P_{k-1} + 64.$$
Appending a $0$ to $71\dots104$ gives $71\dots1040$, and adding $64$ turns the final $040$ into $104$: one more $1$ appears and the ending $04$ is preserved.

The digit sum of $P_k$ is therefore
$$7 + (k - 2)\cdot 1 + 0 + 4 = k + 9.$$
Setting $k + 9 = 1000$ gives $k = 991$.

The answer is $\boxed{\textbf{(D)}\ 991}$.

Products of repdigits have rigid digit patterns because every column receives the same partial product, and the carries stabilize after the first couple of digits. Computing three or four small cases reveals the pattern; the recurrence $P_k = 10P_{k-1} + 64$ is the one-line justification that it never breaks. The problem is really about reading digits, not about the size of the number.

Write $\underbrace{88\dots8}_{k} = \frac{8(10^k - 1)}{9}$, so the product is $64 \cdot \underbrace{11\dots1}_{k}$. Multiplying a string of $k$ ones by $64$ column by column gives $7\,\underbrace{1\dots1}_{k-2}\,04$ once the carries settle, the same pattern.

• Assuming each of the k eights contributes 6 + 4 = 10 to the digit sum, ignoring the carries that turn most digits into 1.
• Miscounting the number of ones as $k - 1$ or $k - 3$, giving $k = 990$ or $992$; check against $8 \cdot 888 = 7104$, which has exactly one $1$ for $k = 3$.