Difference between revisions of "2002 AMC 10A Problems/Problem 15"
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(You could also guess the numbers, if you have a lot of spare time.) | (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== | ||
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+ | https://youtu.be/4dr5Mk4-hOI | ||
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+ | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== |
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 |
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