Let x be the least number which when divided by 8, 12, 20, 28, 35 leaves a…

Concept: When a number greater than every one of the given divisors leaves the same non-zero remainder r on division by each of them, it must equal a common multiple of the divisors plus r (the degenerate case x = r itself involves no real division, so it is excluded); the least such genuine solution is LCM(divisors) + r.

Application:

1. Find the prime factorisation of each divisor: 8 = 23, 12 = 22 × 3, 20 = 22 × 5, 28 = 22 × 7, 35 = 5 × 7.

2. Take the highest power of every prime that appears across the divisors: 23 (needed for 8), 31 (needed for 12), 51 (needed for 20 and 35), 71 (needed for 28 and 35).

3. Multiply these highest powers to get the LCM: LCM = 23 × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840.

4. The least such number that leaves remainder 5 on division by every one of 8, 12, 20, 28, 35 is x = LCM + 5 = 840 + 5 = 845.

5. Sum of the digits of x = 8 + 4 + 5 = 17.

Cross-check: Check by dividing 845 by each divisor and confirming the remainder is 5 every time: 845 = 8 × 105 + 5 = 12 × 70 + 5 = 20 × 42 + 5 = 28 × 30 + 5 = 35 × 24 + 5.

Hence, the sum of the digits of x is 17.