Question

The sum of the three digits of a positive integer is 15 and these digits are in AP. The number

obtained by reversing the digits is 396 less than the original number. Find the number​

Answer 1

$\textbf{Given:}$

$\textsf{Sum of the 3 digits of a number is 15 and these digits are in A.P}$

$\textsf{The number obtained by reversing the digits is 396 less than the original number}$

$\textbf{To find:}$

$\textsf{The number}$

$\textbf{Solution:}$

$\textsf{Let the three digit number be 100x+10y+z}$

$\mathsf{Then,\;x+y+z=15}$----------(1)

$\textsf{Since x,y and z are in A.P,  we write}$

$\mathsf{y=x+d}$

$\mathsf{z=x+2d}$

$\mathsf{(1)\;becomes}$

$\mathsf{x+(x+d)+(x+2d)=15}$

$\mathsf{3x+3d=15}$

$\mathsf{x+d=5}$

$\implies\boxed{\mathsf{y=5}}$

$\mathsf{(1)\implies}$

$\mathsf{x+5+z=15}$

$\mathsf{x+z=10}$--------------(2)

$\textbf{Also,}$

$\mathsf{(100x+10y+z)-(100z+10y+x)=396}$

$\mathsf{99x-99z=396}$

$\mathsf{x-z=4}$----------------(3)

$\textbf{We solve (2) and (3)}$

$\mathsf{x+z=10}$--------------(2)

$\mathsf{x-z=4}$----------------(3)

$\mathsf{Adding,\;we\;get}$

$\mathsf{2x=14}$

$\implies\boxed{\mathsf{x=7}}$

$\mathsf{(2)\;\implies\;7+z=10}$

$\implies\boxed{\mathsf{z=3}}$

$\therefore\textsf{The required number is}$

$\mathsf{100(7)+10(5)+3}$

$\mathsf{700+50+3}$

$\implies\boxed{\mathsf{753}}$