Number theory using modulo 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
 A: 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.
A: 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 $38$.
A: 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:
$$ 2, 4, 6, 8, 10, 12, 14, 16, 20, 22, 26, 28, 32, 38.$$
