Question

A digits of a three digit natural number are in A.P their sum is $15$. the obtained no. by reversing the digits is $396$ less than the original number. Find the number.

Answer 1

Let the numbers be,

$a-d,a,a+d$

Given,

$a-d+a+a+d=15$

$3a=15$

$\therefore a=5$

$(a-d)100+10a+(a+d)=(a+d)100+10a+(a-d)+396$

$100a-100d+10a+a+d=100a+100d+10a+a-d+396$

$-198d=396$

$\therefore d=-2$

Therefore the digits are $7,5,3$

Therefore the number is $753$