kinjiGC
BunuelI solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?As 220 < 225, so maximum prime I need to check for divisibility is 13.kinjiGCThanks,
Gaurav

hi
GauravSolanky,
the logic behind "the rule" of divisiblity by \(\sqrt{x}\) as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..
Thanks and kudos to both of you.