A number consists of three digits in geometric progression. The sum of the right hand and left hand digits exceeds twice the middle digit by 1 and the sum of left hand and middle digits is two third of the sum of the middle and right hand digits. Then the sum of digits of number is

Step-by-step solution

Let the three digits of the number be a/r , a , and ar , where r is the common ratio. Since these are digits, a/r, a, ar 0, 1, 2, ..., 9 . The first condition states that the sum of the right hand and left hand digits exceeds twice the middle digit by 1: a r + ar = 2a + 1 The second condition states that the sum of the left hand and middle digits is two-thirds of the sum of the middle and right hand digits: a r + a = 2 3 (a + ar) From the second equation, dividing by a (assuming a 0 ): 1 r + 1 = 2 3 (1 + r) 1+r r =