The Sum of all four(4) digit numbers formed with the digits [closed]

The Sum of all four(4) digit numbers formed with the digits one(1),three(3),three(3), and zero(0) Without repeating them

closed as off-topic by Henrik, Scientifica, GNUSupporter 8964民主女神 地下教會, projectilemotion, LeucippusMar 13 '17 at 1:33

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• Do you consider $0133$ four-digit? Have you tried just listing them all and adding them? – vrugtehagel Mar 12 '17 at 21:45

First, we list out all numbers. This is relatively easy to do, especially if it is completed systematically as below. You know you have all of the numbers if they only have the digits {1, 3, 3, 0}, each number only once, exactly 12 numbers (12 is found by taking (number of digits, 4)!/(number of times 3 is repeated, twice!).

===

All numbers for 1 in the thousands place:

1330, 1303, 1033

All numbers for 3 in the thousands place:

3310, 3301, 3130, 3103, 3013, 3031

All numbers for 0 in the thousands place:

0133, 0313, 0331

===

Since we followed the criteria listed above, we know we have the correct numbers. Now all you have to do is find the sum. It's 23331.

Suppose $\{a_1,...,a_4\}$ are the digits that we want to consider. Every $a_i$ appears the same number of times in every position, say $n$. So we have $$S=\sum_i n(a_i \times 1+a_i \times 10+a_i \times 100+a_i \times 1000)=\sum_i 1111 a_i n=1111 n \sum_i a_i$$ It is easy to see that $n=6$, and as we have that two digits are the same, the result is $S/2$