Question

the digits of a positive number of three digits are in A.P. and their sum is 15 .the number obtained by reversing the digits is 594 less than the original number. find the number.

Answer 1

Let the numbers be (a-d), a, (a+d)
(a+d)×100+a×10+(a−d)=111a+99d(a+d)×100+a×10+(a−d)=111a+99dOn reversing(a−d)×100+a×10+(a+d)=111a -99d
a+ d+a+a-d=15
a=15
d=3
Therefore the number is 111× 5+99 ×3=852

Answer 2

Answer:

Let the numbers be (a-d), a, (a+d)

(a+d)×100+a×10+(a−d)=111a+99d(a+d)×100+a×10+(a−d)=111a+99dOn reversing(a−d)×100+a×10+(a+d)=111a -99d

a+ d+a+a-d=15

a=15

d=3

Therefore the number is 111× 5+99 ×3=852