9/24/2023

Unfortunately, today I didn't get a lot of time to do math/physics :(

In the time I did get though, I did a few AMC problems from the last five. I found the following problem cool:

Problem:

The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

${ \textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}}\ 991\qquad\textbf{(E)}\ 999 $

Solution:

Note that$$8 \cdot (888 \dots 8) = 64 \cdot (111 \dots 1)$$$$=(60+4) \cdot (111 \dots 1)$$$$=\underbrace{666 \dots 60}_{k+1} + \underbrace{444 \dots 4}_k$$$$=7\underbrace{111 \dots 1}_{k-2} 04$$The sum of digits is$$7+1(k-2)+0+4$$, which should be $1000$.

Solving, we get $k=\boxed{991}$.