Difference between revisions of "2002 AMC 10A Problems/Problem 15"
5849206328x (talk | contribs) m (→Solution: Typo) |
|||
(5 intermediate revisions by 4 users not shown) | |||
Line 5: | Line 5: | ||
==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>.) | ||
+ | |||
+ | (You could also guess the numbers, if you have a lot of spare time.) | ||
==See Also== | ==See Also== | ||
Line 13: | Line 15: | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 01:50, 26 January 2018
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.)
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.