Difference between revisions of "2002 AMC 12A Problems/Problem 17"

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We can indeed create a set of primes with this sum, for example the following sets work: <math>\{ 41, 67, 89, 2, 3, 5 \}</math> or <math>\{ 43, 61, 89, 2, 5, 7 \}</math>.
 
We can indeed create a set of primes with this sum, for example the following sets work: <math>\{ 41, 67, 89, 2, 3, 5 \}</math> or <math>\{ 43, 61, 89, 2, 5, 7 \}</math>.
  
Thus the answer is <math>\boxed{(B)207}</math>.
+
Thus the answer is <math>207\implies \boxed{(B)}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 07:23, 8 November 2019

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 sets work: $\{ 41, 67, 89, 2, 3, 5 \}$ or $\{ 43, 61, 89, 2, 5, 7 \}$.

Thus the answer is $207\implies \boxed{(B)}$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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