The digits of a three-digit number taken in an order are in geometric progression. If one is added to the middle digit, they form an arithmetic progression. If 594 is subtracted from the number, then a new number with the same digits in reverse order is formed. The original number is divisible by

Step-by-step solution

Let the digits be a, ar, ar^2 in GP. The number is 100a + 10ar + ar^2 . AP condition ( a, b+1, c in AP): 2(ar+1) = a + ar^2 2ar + 2 = a(1+r^2) ... (1) Reverse condition: (100a + 10ar + ar^2) - 594 = 100ar^2 + 10ar + a 99a - 99ar^2 = 594 a(1-r^2) = 6 ... (2) From (1): a - ar^2 = 2 - 2ar . Substituting into (2): 2 - 2ar = 6 ar = -2 Since digits must be positive and a > c (as 594 is subtracted), try r = 1 2 , a = 8 : b = 8 1 2 = 4 , c = 8 1 4 = 2 Verification: 8, 4, 2 are in GP. 8, 5, 2 are in AP. 842 - 594 = 248 (rev