Cool

The answer is perl has built a hash table in your while statement but when it comes to if it see's $p.. which is a new variable accordint to perl.. So there comes the problem. This is why it states the error in a wrong place


Note:
Use Strict and warnings

Update:
OOps sorry i mis understood the problem. Ok The problem lies in this line
if (($n < $p[$i]) && ($x > $p[-1])) # error here

just remove the previous code and use this u will not see the warning.
if (($n < $p[$i]))

This is becasue the lt operator should work by finding the modulus and then determine the solution.(I am not sure with that but for this error i hope this could be the reason. i will update on you again after 15 minutes

Sorry i was not able to find it out....

Thanks
Sasi Kumar

In reply to Re^3: Modulus zero error by sasikumar
in thread Modulus zero error by Anonymous Monk

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