Difference between revisions of "2011 AMC 8 Problems/Problem 24"

Latest revision as of 19:52, 25 October 2024

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 $9999.$ But, $9999$ 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(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 also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is $\boxed{\textbf{(A)} 0}$

Solution 3 (assumed previous knowledge)

It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of $2$ primes - and also, that the ${\textbf{odd numbers}\ }$ that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form $2+p$ . Hence, the odd numbers that are the sum of two primes are exactly $2$ more than a prime. Relating to the problem, $10001$ is not $2$ more than a prime, because $10001-2=9999$ and we can easily see that $9999$ is divisible by $3$ . Therefore, $10001$ cannot be written as the sum of two primes, and the answer is $\boxed{\textbf{(A)}\ 0}$ ~mk

i do not recommend this at all why just why

Video Solution

https://youtu.be/qJuoLucUn9o by David

Video Solution 2

https://youtu.be/GqTHx0tOB4o

~savannahsolver

@@ Line 9: / Line 9: @@
 <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>
-However, <math>9999</math> is clearly divisible by <math>3</math>,
+But, <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>.
-==Solution 2 (Sort of)==
+==Solution 2(Simple)==
-One interesting way to do this is to think of <math>10001</math> as if it's binary. Converting it to base <math>10</math> would result in the number <math>17</math>. Since <math>17</math> cannot be written as the sum of two primes, the answer is <math>\boxed{\textbf{(A)} 0}</math>.
+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 also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is <math>\boxed{\textbf{(A)} 0}</math>
-Note: This is not a valid way to do problems like this. For example, the number <math>1000</math> can be written as the sum of two primes in <math>28</math> ways, but if we convert <math>1000</math> to base ten, we would get <math>16</math> which obviously cannot be written as the sum of two primes in <math>28</math> ways.
+==Solution 3 (assumed previous knowledge)==
+It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of <math>2</math> primes - and also, that the <math>{\textbf{odd numbers}\ }</math> that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form <math>2+p</math>. Hence, the odd numbers that are the sum of two primes are exactly <math>2</math> more than a prime. Relating to the problem, <math>10001</math> is not <math>2</math> more than a prime, because <math>10001-2=9999</math> and we can easily see that <math>9999</math> is divisible by <math>3</math>. Therefore, <math>10001</math> cannot be written as the sum of two primes, and the answer is <math>\boxed{\textbf{(A)}\ 0}</math>
+~mk
+i do not recommend this at all why just why
 ==Video Solution==
-https://youtu.be/qJuoLucUn9o
+https://youtu.be/qJuoLucUn9o   by David
+==Video Solution 2==
+https://youtu.be/GqTHx0tOB4o
+~savannahsolver
 ==See Also==
 {{AMC8 box|year=2011|num-b=23|num-a=25}}
 {{MAA Notice}}

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
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