Question

The digits of a positive int teger, having 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. The number is
8​

Answer 1

final number would be:-

Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is

(a+d)×100+a×10+(a−d)=111a+99d

The number obtained by reversing the digits is

(a−d)×+a×10+(a+d)=111a−99d

It is given that the sum of the digits is 15.

(a−d)+a+(a+d)=15 ...(i)

Also it is given that the number obtained by reversing the digits is 594 less than the original number.

∴111a−99d=111a+99d−594 ...(ii)

⟹3a=15 and 198d=594

⟹a=5 and d=3

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So, the number is 111×5+99×3=852