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If $p, q, r, s$ are distinct integers such that:

$f (p, q, r, s) = \text{ max } (p, q, r, s)$

$g (p, q, r, s) = \text{ min } (p, q, r, s)$

$h (p, q, r, s) = \text {remainder of } \frac{(p \times q)} {(r \times s)} \text{ if } (p \times q) > (r \times s)$
$\text{ or remainder of } \frac {(r \times s)}{(p \times q)} \text{ if } (r \times s) > (p \times q)$

Also a function $fgh (p, q, r, s) = f(p, q, r, s) \times g(p, q, r, s) \times h (p, q, r, s)$
Also the same operations are valid with two variable functions of the form $f(p, q)$
What is the value of $fg \left(h \left(2, 5, 7, 3\right), 4, 6, 8\right)$?

whats the point of just posting the answer?

@Lakshman Patel RJIT @GO Classes @sk Shouldn’t this question approach be like:
fg(h(2,5,7,3),4,6,8)=f(h(2,5,7,3),4,6,8)*g(h(2,5,7,3),4,6,8)*g(h(2,5,7,3),4,6,8)

h(2,5,7,3)=(7*3)mod (2*5)=1, now the function reduces to f(1,4,6,8)*g(1,4,6,8)*h(1,4,6,8)

h(1,4,6,8)=(6*8)mod(1*4)=0

now function becomes f(1,4,6,8)*g(1,4,6,8)*0=0

It is given that $h(p, q, r, s) =$ remainder of $\frac{(p \times q)}{(r \times s)}$ if $(p \times q) > (r \times s)$ or remainder of $\frac{(r \times s)}{(p \times q)}$ if $(r \times s) > (p \times q)$.

$h(2,5,7,3) =$ remainder of $\frac{(7 \times 3)} {(2 \times 5)}, \;\; \because (7 \times 3) > (2 \times 5)$

Thus, $h(2,5,7,3) = 1$

Again, it is given that $fg(p, q, r, s) = f(p, q, r, s) \times g(p, q, r, s)$

Also, $f(p, q, r, s) = \max(p, q, r, s)$, and $g(p, q, r, s) = \min(p, q, r, s)$

So we have:

$fg(1,4,6,8) = 8 \times 1, \;\; \because \max(1,4,6,8) = 8 \;\&\; \min(1,4,6,8) = 1$

Thus, $fg(1,4,6,8) = 8$

Answer: $8$

@Pragy I believe that @Arjun meant that the particular extension can't be assumed. For beyond this, let's wait till March 12, 2015.
It was not a typo :P
I meant it can be assumed if we are not strictly formal.

GATE should give Marks to All. But it doesn't always happen.
so its not a composite function right ?

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