Q. In the decimal system of numeration of six digit numbers in which the sum of the digits is divisible by 5 is
Permutations and Combinations
Solution:
First place from left cannot be filled with 0. Next four places can be filled with any of the 10 digits. After filling the first five places, the last place can be filled in following ways:
Sum of digits in first five places
Digit in the unit's place
5k
0 or 5
5K + 1
4 or 9
5K + 2
3 or 8
5K + 3
2 or 7
5K + 4
1 or 6
Thus, in any case the last place can be filled in two ways.
Hence, the required number of numbers is $9 × 10^4 × 4$
| Sum of digits in first five places | Digit in the unit's place |
|---|---|
| 5k | 0 or 5 |
| 5K + 1 | 4 or 9 |
| 5K + 2 | 3 or 8 |
| 5K + 3 | 2 or 7 |
| 5K + 4 | 1 or 6 |