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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.

User Sangil
by
8.2k points

1 Answer

6 votes

Answer:

  • 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

User Pavel Sher
by
7.7k points
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