Question

3]the digits of a three-digit number are in A.P. their sum is 18.the number obtained by reversing the digits is 594 less than the original number. find the original number? Q.4. Attempt​

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.