## Sum of Digits

Find the sum of the digits in the expression $333333333\times 777777777$.
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
If we multiply the numbers by hand, we have the following addition
$.\;\;\;\;\;\;\;\;\;2333333331$
$.\;\;\;\;\;\;\;\;2333333331$
$.\;\;\;\;\;\;\;2333333331$
$.\;\;\;\;\;\;2333333331$
$.\;\;\;\;\;2333333331$
$.\;\;\;\;2333333331$
$.\;\;\;2333333331$
$.\;2333333331$
$2333333331$
———————————-
$259259258740740741$
The sum of the digits equals $81$.

Answer: $81$

Alternative solution
Let $p=333333333$ and $q=777777777$.
$pq=(3/3)pq$
$=(3p)(q/3)$
$=(999999999)(259259259)$
$=(10^9-1)(259259259)$
$=259259259\cdot 10^9-259259259$
We have the following subtraction
$259259259000000000$
$-\;\;\;\;\;\;\;\;\;\;\;\;\;259259259$
———————————
$259259258740740741$
If we add the digit $2$ in the first column to the digit $7$ in the tenth column and so on, we have
$2+7\!=\!9,5+4\!=\!9,9+0\!=\!9,2+7\!=\!9,5+4\!=\!9,9+0\!=\!9,2+7\!=\!9,5+4\!=\!9,8+1\!=\!9$
$9(9)=81$
This is true for any random 9-digit number, for example $187971192$
$187971192000000000$
$-\;\;\;\;\;\;\;\;\;\;\;\;\;187971192$
———————————
$187971191812028808$
$1+8\!=\!9,8+1\!=\!9,7+2\!=\!9,9+0\!=\!9,7+2\!=\!9,1+8\!=\!9,1+8\!=\!9,9+0\!=\!9,1+8\!=\!9$
$9(9)=81$
In fact, the number $777777777$ is not necessary in this problem; any 9-digit number that is a multiple of $3$ will do.