Question

in a three digit number unit's digit is 2 less than hundred's digit and hundred's digit is three less than tens digit. if sum of the original number and number obtained by reversing the digits is 968. find the number

Answer 1

let tens digit =x
hundred place digit = x-3
ones place digit =x-5
100(x-3)+10(x)+x-5= original no.

100x-300+10x+x-5=111x-305(eq.1)

100(x-5)+10(x-3)+×=after reversing
100x-500+10x-30+x=111x-530 ( eq3)

100(x)+10(x-5)+x-3=reversing again
100x+10x-50+x-3=111x-53 (eq2)

their sum= 111x-305+111x-530 +111x-53=968
= 333x-888=968
=3x=968+888
=3x=1856
=x=1856/3
x=618

Answer 2

Answer:100(x-3)+10x+x-5 is the number obtained

after reversing 100(x-5)+10x+x-3

sum of both=200x-800+20x+2x-8=968

solve for x

222x=968+808

222x=1776

x=8

number is 583

Step-by-step explanation: