Sum of a two-digit number and the number obtained by reversing the digits is always divisible by?
Given :
The given statement is 'The sum of a two-digit number and the number obtained by reversing the digits'.
To do :
We have to find the sum is always divisible by which number.
Solution :
Let the two digit number be $10x+y$.
The number obtained by reversing the digits is $10y+x$.
The sum of the two digit number and the number obtained by reversing the digits is,
$(10x+y)+(10y+x)=10x+x+10y+y$
$=11x+11y$
$=11(x+y)$
$11(x+y) = 11 \times (x+y)$
Therefore, the sum of the two-digit number and the number obtained by reversing the digits is always divisible by 11.
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