Ways in which 38 can be divided into 3 positive parts such that the first is divisible by 8, the second by 7 and the third part by 3? I am stuck with the question, i have tried a couple of random approaches but none of them is correct. The answer is 2. Please help if you know how to solve this question.

Assuming that each part must be an integer, this is equivalent to the number of non negative integral solutions of
$$8x+7y+3z=38$$
We have $x,y,z\ge1$. So, let $a=x-1,b=y-1,c=z-1$ such that $a,b,c\ge0$.
Then,
$$8a-8+7b-7+3c-3=38$$
$$8a+7b+3c=56$$
Let $X=8a,Y=7b,Z=3c$.
$$X+Y+Z=56$$
Allowed values of $X=0,8,16,24,32,40,48,56$.
Allowed values of $Y=0,7,14,21,28,35,42,49,56$.
Allowed values of $Z=0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54$.
Number of solutions=Coefficient of $x^{56}$ in the expansion of
$$(1+x^8+x^{16}+\cdots+x^{56})(1+x^7+x^{14}+\cdots+x^{56})(1+x^{3}+x^6+\cdots+x^{54})$$
that is coefficient of $x^{56}$ in
$$\left(\frac{1-x^{64}}{1-x^8}\right)\left(\frac{1-x^{63}}{1-x^7}\right)\left(\frac{1-x^{57}}{1-x^3}\right)$$
which can be further simplified using binomial expansions.
Using this, you can find all possible solutions in general. In this case, however, trial and error may be more convenient.
