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Oct 29, 2019
edited by sinclairdragon428  Nov 20, 2019

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Camy made a list of every possible distinct five-digit positive integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list?

I assume, we count the numbers of all permutations of 13459.

The sum is:

$$\begin{array}{|rcll|} \hline && \dfrac{5!}{5} \times (1+3+4+5+9)\times (10^0+10^1+10^2+10^3+10^4) \quad | \quad 5!=120 \\ &=& \dfrac{120}{5} \times 22\times 11111 \\ &=& 24 \times 22\times 11111 \\ &=&\mathbf{ 5 866 608 } \\ \hline \end{array}$$

The sum of the integers on Camy's list is 5866608

Oct 29, 2019
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thank you!

sinclairdragon428  Oct 29, 2019