# If $\, 5$ & $4\to 100$ , $8$ & $5 \to 400$ , $2$ & $19 \to 361$ , $A$ & $5 \to 625$ , $12$ & $3 \to 324$ , $7$ & $6 \to 441$ ; then $A=?$

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...

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.)
• Why not $\pi$? I was pretty sure it should of been $\pi$... – Jakobian Feb 25 at 7:32
• @Song Because we generate numbers by numbers. $\pi$ is a number and we are done! – Michael Rozenberg Feb 25 at 8:45
• @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. – Song Feb 25 at 11:23
• @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. – Song Feb 25 at 12:20