Difference between revisions of "2011 AMC 8 Problems/Problem 24"
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<cmath> 2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1 </cmath> and <math>m+n</math> is an integer because <math>m</math> and <math>n</math> are both integers. | <cmath> 2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1 </cmath> and <math>m+n</math> is an integer because <math>m</math> and <math>n</math> are both integers. | ||
The only even prime number is <math>2,</math> so our only combination could be <math>2</math> and <math>9999.</math> | The only even prime number is <math>2,</math> so our only combination could be <math>2</math> and <math>9999.</math> | ||
− | 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> | ||
Revision as of 10:03, 5 October 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.