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Find the number of non-negative integer solutions of $a^2$ + $2b$ + $c^3$ + $4ac$ + $d$ = 144 where a,b,c,d $\geqslant$ $0$

I tried by applying $C((n+r-1),(r-1))$.

But by this method it is becoming too lengthy with so many cases. And the time given to solve this question is only 3 mins.

Shaun
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jame samajoe
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  • $a\in [0,12]$, $c\in [0,5]$, $ac\in [0,36]$ it shouldn't be hard.

    The number of solutions of $2b+d=x$ is known $(x=144)$ etc...

     – Toni Mhax Sep 06 '18 at 15:39
  • This is a terribly difficult question to do in $3$ minutes, especially because $c$, which seems to be the most influential variable in terms of the left hand side, can still go up to $5$, so we need to do the cases $c = 0 \to 5$ separately, and this gives (as I saw for checking three(at maximum four is the limit, I would say) cases in SIX minutes) pandemonium. $d+2b$ is not a problem, but the question involves a lot of cases, and is better worth being left out in a competitive exam where students are short on time. – Sarvesh Ravichandran Iyer Sep 06 '18 at 15:40
  • @ToniMhax i too reached till here but main problem comes after this step,taking each cases and combining all the ways possible to form a proper combination – jame samajoe Sep 06 '18 at 15:53
  • again need formula for $2b+d=x$ once you have that, start $a=0$ for $c\in [0,5]$ you add up the count.

    $a=1$ same $a=2$ same til $a=7$ for $a=8$, $c\in [0,4]$ respecting $ac$ bound and so on. You should find the formula for the number of non-negative integer solutions to $ax+by=n$, $a$ and $b$ coprime, i could give you one (they are not unique).

     – Toni Mhax Sep 06 '18 at 16:07

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