Difference between revisions of "2002 AMC 10A Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | + | Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. 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>.) | ||
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| + | (You could also guess the numbers, if you have a lot of spare time.) | ||
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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/4dr5Mk4-hOI | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:56, 1 September 2024
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
Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. 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.)
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
