"PQR is a three-digit number such that it can be

Solution

Let PQR be (100p + 10q + r) where 'p', 'q' and 'r' are integers between (0 - 9). ATQ; 100p + 10q + r = (10p + q) + (10q + r) + (10r + p) Or, 100p + 10q + r = 11p + 11q + 11r So, 89p - q = 10r And, values of 'p', 'q' and 'r' must lie between 0 and 9. At p = 0, -q = 10r (Since, 'q' cannot be negative, this is never possible) At p = 1, 89 - q = 10c Maximum value of q = 9, So, 89 - 9 = 10r Or, r = 8 So, one possible pair of (p, q, and r) is (1, 9 and 8) Now at p = 1, the value of 10r will never be a multiple of 10, as the value of 'q' cannot exceed 9. So, only one pair would exist at p = 1. At p = 2, 178 - q = 10r Now, for any value of 'q', the value of '10r' would exceed 100, meaning the value of 'r' would exceed 9. So, 'p' cannot be more than 1. So, only pair of 'p', 'q' and 'r' that would satisfy the equation is 1, 9 and 8. So, required value = 1 + (6 × 9) + (4 × 8) = 1 + 54 + 32 = 87