Question

the digit of three number are in A.P their sum is 18 the number obtained by reversing the digit is 594 less than the original number find the original number
(write answer step by step )
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Answer 1

Answer:

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

So, the number is 111×5+99×3=852.