# Find the largest even number which cannot be expressed as the sum of two odd composite numbers.

Ques - Find the largest even number which cannot be expressed as the sum of two odd composite numbers.

Honestly speaking , I don't even know where to start . It would be great if someone could show me the way forward . Thanking in advance

Let $n$ be an even number, hence $n$ is $0\pmod 6$ or $2\pmod 6$ or $4\pmod 6$.

Any number of the form $9+6k$ is composite for $k\ge 0$ (as it divisible by $3$).

Now, if $n\ge 18$ and $n\equiv 0\pmod 6$ we can write $n=9+(9+6k)$, i.e $n$ is sum of two odd composite numbers.

If $n\ge 44$ and $n\equiv 2\pmod 6$ we can write $n=35+(9+6k)$, i.e $n$ is sum of two odd composite numbers.

If $n\ge 34$ and $n\equiv 4\pmod 6$ we can write $n=25+(9+6k)$, i.e $n$ is sum of two odd composite numbers.

All we need to check are the numbers $38,40,42$. The last two are covered, but $38$ doesn't, thus it is the largest even number which is not a sum of two odd composite numbers.

• Note that Anatoly's answer, which at first sight looks pretty different from this, is actually the same argument in disguise, but combining the cases 2 and 4 mod 6 into a single case mod 5. Nov 25, 2016 at 16:12

Alternative solution: let us call $k$ the largest integer that cannot be written as a sum of two odd composite numbers. By definition, $k-j$, (where $j$ is a composite odd number $<k$) must be prime. Now let us consider the quantities $$k-3 \cdot 3=k-9$$ $$k-3 \cdot 5=k-15$$ $$k-3 \cdot 7=k-21$$ $$k-3 \cdot 9=k-27$$ $$k-3 \cdot 11=k-33$$

If there exists a value of $k>33$, they have all to be prime or equal to $1$. Now we can note that one of these five numbers must necessary be divisible by $5$ (since they are in aritmetic progression with common difference $6$ ). Thus, the only possibility is given by $k-33=5$. This leads to a maximal value of $k$ equal to $38$.

• Can you pls explain your answer in simple words ? I understood why k - j should be prime , but couldn't understand why we are taking those quantities . Nov 25, 2016 at 12:58
• @AnirudhaBrahma What Anatoly's doing is looking at the possible small (odd composite) values of j. Most of these will have 3 as a factor (since otherwise the smallest prime factor would have to be at least 5, and so j itself would have to be $5 \cdot 5$, $5 \cdot 7$ or something larger); conversely, $3 \cdot (2n+1)$ is always odd and composite (for $n > 0$). [Continued below] Nov 25, 2016 at 15:53
• Now, if $j$ is divisible by 3 and so is $k$, then $k - j$ will automatically be a multiple of 3 as well and so, for large enough $j$, it will be a sum of two odd composite numbers. On the other hand, if $k$ is not divisible by 3 but $j$ still is, then $k - j$ cannot be divisible by 3. Since it is still odd, its smallest prime factor must be at least 5. So if we can find a list of small values of $j$ such that at least one of the $k - j$ must be divisible by 5, then we're essentially done. Nov 25, 2016 at 16:04
• Finally, note that any arithmetic sequence of length $p$ (prime), with successive differences not a multiple of $p$, must contain a multiple of $p$ - this is the bit using modular arithmetic. In this case, the 5-term sequence $k - 33, k - 27, k - 21, k - 15, k - 9$ has successive differences of 6 and so must contain at least one multiple of 5. Unless that number is 5 itself, it will be composite; and since $k$ was even and $j$ was odd, $k - j$ will also be odd. Nov 25, 2016 at 16:06
• I read only now the further question on why we are taking those quantities. Thank you very much to Robin for his comments (+1). Nov 25, 2016 at 17:20

Hint. Any even number greater than 38 can be expressed as the sum of two odd composite numbers.

According to OEIS sequence A118081, the even numbers that can't be represented as the sum of two odd composite numbers are: $$2, 4, 6, 8, 10, 12, 14, 16, 20, 22, 26, 28, 32, 38.$$

• Actually, you can't write 38 as a sum of two odd composite numbers. Nov 25, 2016 at 10:43
• @Galc127 Thanks! Nov 25, 2016 at 10:50