Let p be the least number, which when divided by 5, 6,

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Let p be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of p is

1. 21
2. 22
3. 18
4. 24

Correct Option: C

We find LCM of 5, 6, 7 and 8 = 840

∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
∴ Required number = 840p + 3 which is divisible by 9 for a certain least value of p.
Now, 840p + 3 = 93p × 9 + 3p + 3
3p + 3 , is divisible by 9 for p = 2
∴ Required number = 840 × 2 + 3 = 1680 + 3 = 1683
∴ Sum of digits = 1 + 6 + 8 + 3 = 18