2002 AMC 10A Problems/Problem 15: Difference between revisions
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(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>.) | ||
(you could also guess the numbers, if you have a lot of spare time) | |||
==See Also== | ==See Also== | ||
Revision as of 20:11, 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 
.)
(you could also guess the numbers, if you have a lot of spare time)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
