If the sum of digits of any number between 100 and 1000 be subtracted from the number, by what number is the result always divisible?
Given :
sum of digits of any number between 100 and 1000 be subtracted from the number
To find :
We have to find by what number the result is always divisible
Solution :
A number lying between 100 and 1000 can be represented as $100a+10b+c$.
Sum of the digits of such number = $a+b+c$
The difference between the number and the sum of its digits = $100a+10b+c-(a+b+c)=(100-1)a+(10-1)b+c-c=99a+9b$$=9(11a+b)$
This implies,
$9(11a+b)$ is divisible by 9.
A number that is divisible by 9 is also divisible by 3.
Every number is divisible by 1.
Therefore,
The difference between the number lying between 100-1000 and the sum of its
digits is divisible by 1,3 and 9.
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