Difference between revisions of "2002 AMC 10A Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is <math>20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)}</math>. | + | Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is <math>20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)= 190}</math>. |
(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.) | (Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.) | ||
Revision as of 20:58, 11 September 2016
Problem
Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
Solution
Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is 
.
(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is 
, 
, 
, and 
.)
See Also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
