Question 996686: What is the sum of the digits of the product of 23 ones and 32 tens?
Answer by KMST(5345)   (Show Source):
You can put this solution on YOUR website! MULTIPLYING:
You get  , and
adding the digits you get  .
WITHOUT MULTIPLYING:
 ,
so the product  is  tens,
with the same digits as with a  added to the right.
We need to find the sum of the digits of .
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  ,
when we divide  by  , the remainder is  , and
when we divide  by , the remainder is .
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 by 9 is
the remainder of dividing  by 9.
which is  .
However, is number ending in  , and
Is a 3-digit number between  and  .
So, the sum of the first and last digits of that product is greater than  , and
the sum of all 3 digits of must be less than  .
Then, the sum of all 3 digits of must be a 2-digit number between  and  ,
The only such number that divided by 9 has a remainder of  is
 ,
because  .
|
|
|
