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Suppose it is given :
$\quad \, 5$ & $4\to 100$ ,
$\quad 8$ & $5 \to 400$ ,
$\quad 2$ & $19 \to 361$ ,
$\quad A$ & $5 \to 625$ ,
$\quad 12$ & $3 \to 324$ ,
$\quad 7$ & $6 \to 441$
$\quad$;then $A=?$
where we are getting $100$ from $5$ and $4$ via some logic or combinations of operations
similarly, $400$ from $8$ and $5$ via the same logic as used in $100$
and so on ...

My Thoughts:
$100=10^2$ , $400= 20^2$ , $361=19^2$ , $625=25^2$ , $324=18^2$ , $441=21^2$
$\implies$ all RHS are perfect square

For $1st$ relation,
i.e, $\, 5$ & $4\to 100$
LOGIC: $\quad 5^2 * 4 =100$
but the same logic don't apply to 2nd relation,
i.e, $8^2 * 5 =320 \neq 400$
so, how to solve this reasoning problem? Any suggestions please...

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Note that $100=10^2=(\frac{5\cdot 4}2)^2$, $400=20^2=(\frac{8\cdot 5}2)^2$, etc. Given the information, it seems that $a\&b \to (\frac{ab}2)^2$. So $A$ would be $10$ (if only $A>0$ is allowed.)

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    $\begingroup$ Why not $\pi$? I was pretty sure it should of been $\pi$... $\endgroup$ – Jakobian Feb 25 at 7:32
  • $\begingroup$ @Jakobian I agree with you! The problem is not correct. $\endgroup$ – Michael Rozenberg Feb 25 at 7:35
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    $\begingroup$ @Song Because we generate numbers by numbers. $\pi$ is a number and we are done! $\endgroup$ – Michael Rozenberg Feb 25 at 8:45
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    $\begingroup$ @Suresh Almost surely, this is not what you want, but in principle $A$ can take any values because one can just define $\pi\&25:= 625$ for example. $\endgroup$ – Song Feb 25 at 11:23
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    $\begingroup$ @Suresh It is just defining it, with no regard to the other relations... (this is what I meant by 'in principle'.) Here is a similar kind of example: $1,2,3,4,\color{red}{\pi}$ given by $a_n = n+\frac{\pi-5}{24}(n-1)(n-2)(n-3)(n-4)$. Most would expect $a_5=5$, but it is not necessarily. $\endgroup$ – Song Feb 25 at 12:20

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