-3
$\begingroup$

If $$\quad 5 * 5 = 10$$ ,
$$\quad 6*6=18$$
$$\quad 7*7=36$$
then $$\quad 7* 8 = ?$$

a) $54 \quad$ b) $51 \quad$ c) $30$

NOTE: Here '$*$' is not simple multiplication
It is some logic or combinations of operations from which we are getting $10$ from $5$ and $5$
Similarly, we are getting the rest of the relations via the same logic or combinations of operations

My Thoughts:
$$10= 5*(5-3)$$ i.e, LOGIC: $\quad x* (x-3) = $ RHS
Similarly $2nd$ relation is satisfied, i.e, $$18=6*(6-3)$$ $$\text{But, } 36 \neq 7 * (7-3)$$ So, how to solve this puzzle? Any Suggestions please...

$\endgroup$

closed as off-topic by Saad, Rhys Steele, Riccardo.Alestra, Vinyl_cape_jawa, Xander Henderson Mar 7 at 13:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Saad, Rhys Steele, Riccardo.Alestra, Vinyl_cape_jawa, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ All of 3 can be correct. $\endgroup$ – Botond Feb 26 at 8:10
  • $\begingroup$ How? Can you please explain... $\endgroup$ – Suresh Feb 26 at 10:09
  • $\begingroup$ The $*$ function is not well defined, so $7*8$ can be anything. $\endgroup$ – Botond Feb 26 at 11:25
  • $\begingroup$ "$*$ is not well defined" , I agreed; but in this problem that is what required to be find. For eg.: '$*$' should be defined in such a way that it satisfies given $3$ relations, and then that definition of '$*$' should be used to evaluate $7*8$ $\endgroup$ – Suresh Feb 26 at 13:33
  • $\begingroup$ But you can't give an unique extension with the given conditions. You can let, for example $5*5=10, 6*6=18,7*7=36$ and $0$ otherwise. Why would any other extension be better or worse than this one? $\endgroup$ – Botond Feb 26 at 14:10
2
$\begingroup$

Suggestion: Let $ab=10x+y$. Then $a*b=f(x,y)$.

$\endgroup$
  • $\begingroup$ Sorry i could not understand, could you please explain how we can relate $10x + y \, $ with $5*5=10$ ? $\endgroup$ – Suresh Feb 27 at 6:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.