Question

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?

Answer 1

Step 1: Describe the symmetry in the specified list of numbers.

If every possible five-digit positive integer formed using the five distinct digits $1,3,4,5$ and $9$ is listed, then, all digits will appear at all possible place values an equal number of times ($4!$ times, to be specific). This fact is verified in the following example.

Consider the number of times the particular digit $1$, appears in one's place. In other words, the number of ways to form the five-digit integer of the format $1$. The four empty boxes are to be filled with the remaining 4 distinct digits, which has $4!$ possible arrangements.

Thus, in the list of all five-digit numbers formed using the five distinct digits $1,3,4,5$ and $9$, the digit $1$ appears at the one's place $4!$ times. Note that the above result is unchanged if any other digit among $1,3,4,5$ and $9$ is considered. The result is also unchanged if we consider any other place in the five-digit number.

Step 2: Determine the sum of digits with a particular place value.

The sum of one's digits of all numbers in the list is $4!\times 1+4!\times 3+4!\times 4+4!\times 5+4!\times 9=4!\times (1+3+4+5+9)=4!\times 22$.

This will remain the same for the digits in tens, hundreds, thousands, and ten-thousands place also.

Step 3: Determine the contribution of digits with a particular place value to the sum of numbers in the list.

Also, the contribution of the sum of digits at a particular place depends on its place value. For example, a $4$ at the thousandth place will contribute $4\times 1000=4000$ to the sum.

Thus, the contribution to the sum of numbers from the digits at a particular place in the numbers is $4!\times 22\times \text{place value of the digits}$. There are 5 places in five-digit numbers having place values $1,10,100,1000\text{and}10000$.

Step 4: Determine the sum of integers.

Thus, the sum of the list of integers can be written as $4!\times 22\times (1+10+100+1000+10000)$.

Evaluate $4!\times 22\times (1+10+100+1000+10000)$:

$\begin{array}{rcl}4!\times 22\times (1+10+100+1000+10000)& =& 4!\times 22\times 11111\\ & =& 24\times 22\times 11111\\ & =& 528\times 11111\\ & =& 5866608\end{array}$

Hence, the sum of the integers on Camy's list is $\mathbf{5866608}$.