Difference between revisions of "2002 AMC 12A Problems/Problem 17"
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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 | + | 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>\boxed{(B)207}</math>. | ||
Revision as of 11:15, 25 December 2018
Problem
Several sets of prime numbers, such as 
use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
Solution
Neither of the digits 
, 
, and 
can be a units digit of a prime. Therefore the sum of the set is at least 
.
We can indeed create a set of primes with this sum, for example the following sets work: 
or 
.
Thus the answer is 
.
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
