Aspire Faculty ID #16484 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Using the digits 1,5,2,8 all possible four digit numbers are formed and the sum of all such numbers is between

Solution

Digits = 1,5,2,8
Number of arrangements = 4! = 24

Each digit appears 6 times in each place (unit, tens, hundreds, thousands).

Sum contributed by one digit = 6 × (1000 + 100 + 10 + 1) × digit
= 6 × 1111 × digit = 6666 × digit

Sum of digits = 1 + 5 + 2 + 8 = 16
Total sum = 6666 × 16 = 106656

This lies between 50000 and 100000.
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