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Difference between revisions of "1984 AIME Problems/Problem 14"

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== Solution ==
 
== Solution ==
  
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Let the desired integer be <math>2n</math> for some positive integer <math>n</math>. Notice that we must have <math>2n-9</math>, <math>2n-15</math>, <math>2n-21</math>, <math>2n-25</math>, ..., <math>2n-k</math> all prime for every odd composite number <math>k</math> less than <math>2n</math>. Therefore <math>n</math> must be small. Also, we find that <math>n</math> is not divisible by 3, 5, 7, and so on. Clearly, <math>n</math> must be a prime. We can just check small primes and guess that <math>n=19</math> gives us our maximum value of <math>38</math>.
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Take an even positive integer <math>x</math>. <math>x</math> is either <math>0 \bmod{6}</math>, <math>2 \bmod{6}</math>, or <math>4 \bmod{6}</math>. Notice that the numbers <math>9</math>, <math>15</math>, <math>21</math>, ... , and in general <math>9 + 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases:
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If <math>x \ge 18</math> and is <math>0 \bmod{6}</math>, <math>x</math> can be expressed as <math>9 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>9</math> and <math>9+6n</math> are both odd composites.  
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If <math>x\ge 44</math> and is <math>2 \bmod{6}</math>, <math>x</math> can be expressed as <math>35 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>35</math> and <math>9+6n</math> are both odd composites.
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If <math>x\ge 34</math> and is <math>4 \bmod{6}</math>, <math>x</math> can be expressed as <math>25 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>25</math> and <math>9+6n</math> are both odd composites.
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Clearly, if <math>x \ge 44</math>, it can be expressed as a sum of 2 odd composites. However, if <math>x = 42</math>, it can also be expressed using case 1, and if <math>x = 40</math>, using case 3. <math>38</math> is the largest even integer that our cases do not cover. If we examine the possible ways of splitting <math>38</math> into two addends, we see that no pair of odd composites add to <math>38</math>. Therefore, <math>38</math> is the largest possible number that is not expressible as the sum of two odd composite numbers.
  
 
== See also ==
 
== See also ==

Revision as of 14:00, 5 February 2011

Problem

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution

Take an even positive integer $x$. $x$ is either $0 \bmod{6}$, $2 \bmod{6}$, or $4 \bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:

If $x \ge 18$ and is $0 \bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.

If $x\ge 44$ and is $2 \bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.

If $x\ge 34$ and is $4 \bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.


Clearly, if $x \ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $38$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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