Q18 of 47 Page 1

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.

Let the three digits be (a - d), a and (a + d). Clearly these three are in AP


Then the no is = 100(a - d) + 10(a) + a + d = 100a - 100d + 10a + a + d = 111a - 99d


Also sum of digits = 15


(a - d) + a + (a + d) = 15


3a = 15


a = 5


Also,


When no is reversed no is = 100(a + d) + 10(a) + a - d = 111a + 99d


Given that


Original No - 594 = reversed No


111a - 99d - 594 = 111a + 99d


198d = - 594


d = - 3


So the original No is = 111a - 99d = 111(5) - 99( - 3) = 555 + 297 = 852


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