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

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== Problem ==
 
== Problem ==
  
Several sets of prime numbers, such as <math>\{7,83,421,659\}</math> use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Several sets of prime numbers, such as <math>\{7,83,421,659\}</math> use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math>
 
<math>
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Neither of the digits <math>4</math>, <math>6</math>, and <math>8</math> can be a units digit of a prime. Therefore the sum of the set is at least <math>40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207</math>.
 
Neither of the digits <math>4</math>, <math>6</math>, and <math>8</math> can be a units digit of a prime. Therefore the sum of the set is at least <math>40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207</math>.
  
We can indeed create a set of primes with this sum, for example the following set works: <math>\{ 41, 67, 89, 2, 3, 5 \}</math>.
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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>.
  
Thus the answer is <math>\boxed{(B)207}</math>.
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Thus the answer is <math>207\implies \boxed{\mathrm{(B)}}</math>.
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== Video Solution ==
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https://www.youtube.com/watch?v=V6z7GiitBUM
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 15:01, 8 September 2022

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 $207\implies \boxed{\mathrm{(B)}}$.

Video Solution

https://www.youtube.com/watch?v=V6z7GiitBUM

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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