In a 3-digit number unit's digit is 2 less than hundred's digit and hundred's digit is 3 less than the ten's digit. If the sum of the original number and number obtained by reversing the digits is 968, find the number.
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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
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