Let the least number of six digits which when divided by 4, 6, 10, 15 leaves…

To find the sum of the digits of N, the least six-digit number that leaves a remainder of 2 when divided by 4, 6, 10, and 15, we follow these steps:

Step-by-Step Calculation
1. Find the Least Common Multiple (LCM):
First, find the LCM of the divisors 4, 6, 10, and 15.

Prime factorization:

4 = 2^2

6 = 2 * 3

10 = 2 * 5

15 = 3 * 5

LCM = 2^2 * 3 * 5 = 60.

2. Find the smallest six-digit number divisible by the LCM:
The smallest six-digit number is 100,000.

Divide 100,000 by 60: 100,000 / 60 = 1666 with a remainder of 40.

To find the smallest six-digit multiple of 60, add the difference between the LCM and the remainder to the smallest six-digit number:

100,000 + (60 - 40) = 100,020.

3. Add the remainder to satisfy the condition:
The problem states the number must leave a remainder of 2 in each case.

N = 100,020 + 2 = 100,022.

4. Find the sum of the digits of N:

Sum = 1 + 0 + 0 + 0 + 2 + 2 = 5.