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, 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.

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1984 AIME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3 (bash)

See also