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I know how to find the sum of all numbers formed by digits 1,2,3,4,5 taken all at a time without repetitions. How can I find the sum when digit repetitions are allowed?

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    $\begingroup$ Is the length limited? Because otherwise you could created arbitrarily large numbers ... $\endgroup$
    – Matti P.
    Jan 30, 2020 at 9:23

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Let us focus on the last digit in the ten-thousands place. Every digit there has to be multiplied by $10^4$. Each number $1,2,3,4,5$ can occur $4^5$ times. This is because each other $4$ digits places can be occupied by any of the $5$ numbers. Hence multipliying all this gives:

$$S=10^4\times 4^5\times (1+2+3+4+5)$$ $$S=10^4\times 4^5\times 15$$

Now we simply extend this for every digit's place. The final sum will be $$S= 4^5\times 15\times 11111$$

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