2002 AMC 10A Problem 15

We don't actually need to find the four primes. We only need to know which digits are in the tens place and which are in the ones place.

1. Last digits. A two-digit number ending in $2, 4, 6$ is even, and one ending in $5$ is a multiple of $5$, so none of those are prime. A two-digit prime must end in $1, 3, 7$ or $9$.

1. Our digits include exactly four of these: $1, 3, 7, 9$. Four primes need four last digits, so these four are the last digits.

1. First digits. The remaining digits, $2, 4, 5, 6$, must be the tens digits.

4. Add by place value. Each tens digit is worth ten times itself:
$$10 \times (2 + 4 + 5 + 6) + (1 + 3 + 7 + 9) = 10 \times 17 + 20 = 190.$$
This does not depend on how the digits are paired.

1. Make sure it is possible. For example $23$, $41$, $59$, $67$ are all prime and use each digit once, and indeed $23 + 41 + 59 + 67 = 190$.

The answer is $\boxed{\textbf{(E)}\ 190}$.

The sum of some two-digit numbers depends only on which digits are tens digits and which are ones digits, not on how they are matched up. Rules for divisibility by $2$ and $5$ decide the ones digits completely. When a problem asks only for a total, look for what stays the same no matter how the unknown choices are made.

A two-digit prime is odd and not a multiple of $5$, so its units digit must be $1$, $3$, $7$ or $9$. The eight given digits contain exactly four of these, namely $1, 3, 7, 9$, and four primes need four units digits, so these are the units digits. The other four digits, $2, 4, 5, 6$, must all be tens digits.

Whatever the pairing, the sum of the four numbers is
$$10\,(2 + 4 + 5 + 6) + (1 + 3 + 7 + 9) = 170 + 20 = 190.$$

(Such a pairing does exist, for example $23, 41, 59, 67$, so the problem is consistent.)

The answer is $\boxed{\textbf{(E)}\ 190}$.