Question

The digits in 5512 are rearranged to form all possible 4-digit positive integers. Find the last three digits of the sum of all such positive integers.

Answer 1

• 329

Explanation:

There are possible:

• 4*4 = 16 4-digit numbers and since 5 is repeated, the number becomes:
• 4*3 = 12

The numbers are:

1. 1255
2. 1525
3. 1552
4. 2155
5. 2515
6. 2551
7. 5125
8. 5152
9. 5215
10. 5251
11. 5512
12. 5521

The sum is:

• 43329

The last 3 digits of the sum is 329