Question

The digits of positive integer, having three digits are in A.P and ter sum is 15. The number obtained by reversing the digits is 594 less than a original number. Find a number.

Answer 1

Let the number 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)a+d+a+a−d⟹=111a−99d=15a=5 and d = 3(a−d)×100+a×10+(a+d)=111a−99da+d+a+a−d=15⟹a=5 and d = 3Hence the no is 111×5+99×3=852