A is the smallest three-digit number which when divided by 3, 4 and 5 leaves remainders 1, 2 and 3 respectively. What is the sum of the digits of A?

Here the LCM of 3, 4 and 5 is 60.

We need the smallest three-digit number. so we can multiply 60 by 2. So  $$60\times\ 2$$ = 120.

 The number is divided by 3, 4 and 5 respectively, and leaves 1, 2, and 3 as a remainder.

So 3-1 = 2, 4-2 = 2, 5-3 = 2

Hence 120 will be subtracted by 2.

So the smallest three-digit number = A = 120-2 = 118

Sum of the digits of A = 1+1+8 = 10

We can also obtain the value 'A' by checking the random number starting from the smallest one. In this (A-3) will be the multiple of five. The numbers can be 103, 108, 113, 118 etc.. Now 113 and 108, when subtracted by 1, then it will not be divisible by 3. Now 103 and 118 are remaining. 103 when subtracted by 2, then it will not be divisible by 4. Hence we are left with 118 which is following all the given conditions.