CEMC Pascal

We label the digits of the unknown number as vwxyz.
Since vwxyz and 71794 have 0 matching digits, then $v \neq 7$ and $w \neq 1$ and $x \neq 7$ and $y \neq 9$ and $z \neq 4$.
Since vwxyz and 71744 have 1 matching digit, then the preceding information tells us that $y=4$.
Since $v w x 4 z$ and 51545 have 2 matching digits and $w \neq 1$, then $v w x y z$ is of one of the following three forms: $5 w x 4 z$ or $v w 54 z$ or $v w x 45$.
Case 1: vwxyz $=5 w x 4 z$
Since $5 w x 4 z$ and 21531 have 1 matching digit and $w \neq 1$, then either $x=5$ or $z=1$.
If $x=5$, then $5 w x 4 z$ and 51545 would have 3 matching digits, which violates the given condition. Thus, $z=1$.
Thus, $v w x y z=5 w x 41$ and we know that $w \neq 1$ and $x \neq 5,7$.
To this point, this form is consistent with the 1st, 2 nd, 3 rd and 7 th rows of the table.
Since $5 w x 41$ and 59135 have 1 matching digit, this is taken care of by the fact that $v=5$ and we note that $w \neq 9$ and $x \neq 1$.
Since $5 w x 41$ and 58342 have 2 matching digits, this is taken care of by the fact that $v=5$ and $y=4$, and we note that $w \neq 8$ and $x \neq 3$.
Since $5 w x 41$ and 37348 have 2 matching digits and $y=4$, then either $w=7$ or $x=3$.
But we already know that $x \neq 3$, and so $w=7$.
Therefore, vwxyz $=57 x 41$ with the restrictions that $x \neq 1,3,5,7$.
We note that the integers $57041,57241,57441,57641,57841,57941$ satisfy the requirements, so are all possibilities for Sam's numbers.
Case 2: vwxyz $=v w 54 z$
Since $v w 54 z$ and 51545 have only 2 matching digits, so $v \neq 5$ and $z \neq 5$.
Since $v w 54 z$ and 21531 have 1 matching digit, then this is taken care of by the fact that $x=5$, and we note that $v \neq 2$ and $z \neq 1$. (We already know that $w \neq 1$.)
Since $v w 54 z$ and 59135 have 1 matching digit, then $v=5$ or $w=9$ or $z=5$.
This means that we must have $w=9$.
Thus, vwxyz $=v 954 z$ and we know that $v \neq 2,7,5$ and $z \neq 1,4,5$.
To this point, this form is consistent with the 1 st, 2 nd, 3 rd , 4 th, and 7 th rows of the table.
Since $v 954 z$ and 58342 have 2 matching digits and $v \neq 5$, then $z=2$.
Since $v 9542$ and 37348 have 2 matching digits, then $v=3$.
In this case, the integer 39542 is the only possibility, and it satisfies all of the requirements.
Case 3: vwxyz $=v w x 45$
Since $v w x 45$ and 21531 have 1 matching digit and we know that $w \neq 1$, then $v=2$ or $x=5$.
But if $x=5$, then $v w 545$ and 51545 would have 3 matching digits, so $x \neq 5$ and $v=2$.
Thus, vwxyz $=2 w x 45$ and we know that $w \neq 1$ and $x \neq 5,7$.
To this point, this form is consistent with the 1st, 2 nd, 3rd and 7 th rows of the table.
Since $2 w x 45$ and 59135 have 1 matching digit, this is taken care of by the fact that $z=5$ and we note that $w \neq 9$ and $x \neq 1$.
Since $2 w x 45$ and 58342 have 2 matching digits, then $w=8$ or $x=3$, but not both.
Since $2 w x 45$ and 37348 have 2 matching digits, then $w=7$ or $x=3$, but not both.
If $w=8$, then we have to have $x \neq 3$, and so neither $w=7$ nor $x=3$ is true.
Thus, it must be the case that $x=3$ and $w \neq 7,8$.
Therefore, vwxyz $=2 w 345$ with the restrictions that $w \neq 1,7,8,9$.
We note that the integers $20345,22345,23345,24345,25345,26345$ satisfy the requirements, so are all possibilities for Sam's numbers.
Thus, there are 13 possibilities for Sam's numbers and the sum of these is 526758.