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

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

Solution:

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

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