1984 AIME Problems/Problem 14

Problem

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

Solution 1

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, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

Solution 2

Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$ , then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$ , which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$ , yielding a maximal answer of 38. Since $38-25=13$ , which is prime, the answer is $\boxed{038}$ .

Solution 3 (bash)

Let $2n$ be an even integer. Using the Chicken McNugget Theorem on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as $n+n$ . We bash each case until we find one that works.

Solution 5

Claim: The answer is $\boxed{038}$ .

Proof: It is fairly easy to show 38 can't be split into 2 odd composites.

Assume there exists an even integer $m > 38$ that m can't be split into 2 odd composites.

Then, we can consider m modulo 5.

If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since $m > 20$ , $k > 1$ so $k \geq 2$ . Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well.

If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since $m > 26$ , $k > 1$ so $k \geq 2$ . Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well.

If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since $m > 32$ , $k > 1$ so $k \geq 2$ . Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well.

If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since $m > 38$ , $k > 1$ so $k \geq 2$ . Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well.

If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since $m > 14$ , $k > 1$ so $k \geq 2$ . Thus, 5k is composite. Since 9 is odd and composite, m is even, 5k should be odd as well.

Thus, in all cases we can split m into 2 odd composites, and we get at a contradiction. $\blacksquare$

-Alexlikemath

Solution 6

All numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \cdot 25 - 9 - 25 = 225-34 = 191$ . We then bash from top to bottom:

1. $178 = 89 \cdot 2 => 87 + 91$ - refuted

2. $166 = 83 \cdot 2 => 81 + 85$ - refuted

3. $158 = 79 \cdot 2 => 77 + 81$ - refuted

4. $146 = 73 \cdot 2 => 69 + 77$ - refuted

5. $142 = 71 \cdot 2 => 65 + 77$ - refuted

6. $134 = 67 \cdot 2 => 65 + 69$ - refuted

7. $122 = 61 \cdot 2 => 57 + 65$ - refuted

8. $118 = 59 \cdot 2 => 55 + 63$ - refuted

9. $106 = 53 \cdot 2 => 51 + 55$ - refuted

10. $94 = 47 \cdot 2 => 45 + 49$ - refuted

11. $86 = 43 \cdot 2 => 35 + 51$ - refuted

12. $82 = 41 \cdot 2 => 33 + 49$ - refuted

13. $74 = 37 \cdot 2 => 35 + 39$ - refuted

14. $62 = 31 \cdot 2 => 27 + 35$ - refuted

15. $58 = 29 \cdot 2 => 25 + 33$ - refuted

16. $46 = 23 \cdot 2 => 21 + 25$ - refuted

17. $38 = 19 \cdot 2 => = 19 + 19$ - it works!

Because we did a very systematic bash as shown, we are confident the answer is $\boxed {038}$

~Arcticturn

Solution 7 (casework)

As stated above, all numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \cdot p$ can be expressed as ( $p$ + $q$ )+( $p$ - $q$ ) for some positive, even $q$ less then $p$ .

If $p$ = $1 \bmod{30}$ , p±4 would both be composite

If $p$ = $7 \bmod{30}$ , p±2 would both be composite

If $p$ = $11 \bmod{30}$ , p±14 would both be composite

If $p$ = $13 \bmod{30}$ , p±8 would both be composite

If $p$ = $17 \bmod{30}$ , p±8 would both be composite

If $p$ = $19 \bmod{30}$ , p±14 would both be composite

If $p$ = $23 \bmod{30}$ , p±2 would both be composite

If $p$ = $29 \bmod{30}$ , p±4 would both be composite

So $p$ < 30

From here, just try all possible p and find the answer is $\boxed {038}$

~Mathophobia

Video Solution

1984 AIME #14

MathProblemSolvingSkills.com

Video Solution using Bashing

https://www.youtube.com/watch?v=n98zEG1-Hrs ~North America Math Contest Go Go Go

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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