2011 AMC 8 Problems/Problem 24

Revision as of 01:27, 22 July 2023 by Huangjialan (talk | contribs) (Solution 3)

Problem

In how many ways can $10001$ be written as the sum of two primes?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

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 $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers. The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$, so the number of ways $10001$ can be written as the sum of two primes is $\boxed{\textbf{(A)}\ 0}$

Solution 2 (Sort of)

One interesting way to do this is to think of $10001$ as if it's binary. Converting it to base $10$ would result in the number $17$. Since $17$ cannot be written as the sum of two primes, the answer is $\boxed{\textbf{(A)} 0}$.

Note: This is not a valid way to do problems like this. For example, the number $1000$ can be written as the sum of two primes in $28$ ways, but if we convert $1000$ to base ten, we would get $16$ which obviously cannot be written as the sum of two primes in $28$ ways.

Solution 3(simple)

First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is 0

Video Solution

https://youtu.be/qJuoLucUn9o by David

Video Solution 2

https://youtu.be/GqTHx0tOB4o

~savannahsolver

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
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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