2002 AMC 12A Problems/Problem 17

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Problem

Several sets of prime numbers, such as $\{7,83,421,659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?

$\text{(A) }193 \qquad \text{(B) }207 \qquad \text{(C) }225 \qquad \text{(D) }252 \qquad \text{(E) }447$

Solution

Neither of the digits $4$, $6$, and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$.

We can indeed create a set of primes with this sum, for example the following set works: $\{ 41, 67, 89, 2, 3, 5 \}$.

Thus the answer is $\boxed{207}$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions
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