Problem:
What is the largest even integer which cannot be written as the sum of two odd composite numbers? (Recall that a positive integer is said to be composite if it is divisible by at least one positive integer other than 1 and itself.)
Solution:
We will show that if k is an even integer and if k≥40, then k is expressible as the sum of two composites. This leaves 38 as the candidate for the largest even integer not expressible in such manner; it is easy to check that 38 indeed satisfies this requirement. The proof of our claim for k≥40 hinges on the fact that if n is odd and greater than 1, then 5n is an odd composite ending in 5. So, to express k as desired, it suffices to find small odd composites ending in 5, 7, 9,1 and 3, and to add these to numbers of the form 5n. Indeed, 15, 27,9,21 and 33 will satisfy the above condition, and in each of the following cases one can find an odd integer n, n>1, such that
if k ends in 0 (i.e. 40,50,…), then k=15+5n,if k ends in 2 (i.e. 42,52,…), then k=27+5n,if k ends in 4 (i.e. 44,54,…), then k=9+5n,if k ends in 6 (i.e. 46,56,…), then k=21+5n,if k ends in 8 (i.e. 48,58,…), then k=33+5n,​
OR
First observe that 6n+9 is an odd composite number for n=0,1,2,…. Now partition the set of even positive integers into three residue classes, modulo 6, and note that in each of the following cases the indicated decompositions are satisfied by some nonnegative integer, n:
if k≡0(mod6) and k≥18, then k=9+(6n+9), if k≡2(mod6) and k≥44, then k=35+(6n+9), if k≡4(mod6) and k≥34, then k=25+(6n+9). ​
This takes care of all even integers greater than 38. Checking again shows that 38​ is the answer to the problem.
The problems on this page are the property of the MAA's American Mathematics Competitions