Question 996686
MULTIPLYING:
You get {{{23*320=7360}}} , and
adding the digits you get {{{7+3+6+0=highlight(16)}}} .


WITHOUT MULTIPLYING:
{{{23*320=23*32*10}}} ,
so the product {{{23*320}}} is {{{23*32}}} tens,
with the same digits as {{{23*32}}} with a {{{0}}} added to the right.
We need to find the sum of the digits of {{{23*32}}} .
If you add the digits of a number, and add the digits of the sum, and so on, until you get a 1-digit number, the final result is the remainder of dividing the original number by 9.
Since {{{2+3=5}}} ,
when we divide {{{23}}} by {{{9}}} , the remainder is {{{5}}} , and
when we divide {{{32}}} by {{{9}}} , the remainder is {{{5}}} .
When you multiply numbers, the remainder from dividing the product by 9
is the the remainder of dividing the product of the remainders of dividing the factors by 9.
So, the remainder from dividing the product {{{23*32}}} by 9 is
the remainder of dividing {{{5*5=25}}} by 9.
which is {{{2+5=7}}} .
However, {{{23*32}}} is number ending in {{{2*3=6}}} , and
Is a 3-digit number between {{{20*30=600}}} and {{{25*32=800}}} .
So, the sum of the first and last digits of that product is greater than {{{6+6=12}}} , and
the sum of all 3 digits of {{{23*32}}} must be less than {{{7+9+6=22}}} .
Then, the sum of all 3 digits of {{{23*32}}} must be a 2-digit number between {{{12}}} and {{{22}}} ,
The only such number that divided by 9 has a remainder of {{{7}}} is
{{{9+7=highlight(16)}}} ,
because {{{2*9+7=18+7=25}}} .