Difference between revisions of "2011 AMC 8 Problems/Problem 24"
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However, <math>9999</math> is clearly divisible by <math>3</math> | However, <math>9999</math> is clearly divisible by <math>3</math> | ||
so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> | so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=23|num-a=25}} | {{AMC8 box|year=2011|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:12, 18 November 2014
In how many ways can be written as the sum of two primes?
Solution
For the sum of two numbers to be odd, one must be odd and the other must be even, because All odd numbers are of the form where n is an integer, and all even numbers are of the form where m is an integer. and is an integer because and are both integers. The only even prime number is so our only combination could be and However, is clearly divisible by so the number of ways can be written as the sum of two primes is
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
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