Another solution:
Simply keep on substracting 15 from x until result becomes 0 or negative. If result is zero, x is divisible by 15. if negative then not divisible :)
If the '%' operator is allowed....simply check the remainder of number%15. If it is 0, it is divisible, else not....however % uses / internally..... :)
use bitwise operator
Another solution:
Simply keep on substracting 15 from x until result becomes 0 or negative. If result is zero, x is divisible by 15. if negative then not divisible :)
use bitwise operator
If the '%' operator is allowed....simply check the remainder of number%15. If it is 0, it is divisible, else not....however % uses / internally..... :)
use bitwise operator
n % k is equivalent to n & (k-1) as long as k is a power of 2..
E.g. 32 % 16 == 0
so 32 & 15 is also 0
This is true only when k is a power of 2 as I mentioned.
use bitwise operator
Somehow i dont agree with geekgod! Say 30 is divisible by 15 but 31=(30 + 1) is not divisible by 16. So ur statement isnt true! Am i right!
[quote="geekgod"]If a number n is divisible by 15.. then n+1 is divisible by 16
to check if n+1 is divisible by 16 you can either use
(n+1) % 16 == 0 OR
(n+1) & 15 == 0[/quote][color=red][/color][size=18][/size][size=12][/size]
use bitwise operator
n % k is equivalent to n & (k-1) as long as k is a power of 2..
use bitwise operators..(avoid /,%)
(n+1) & 15 == 0 ---->will this work for any input????
it will not work..
use bitwise operator
If a number n is divisible by 15.. then n+1 is divisible by 16
to check if n+1 is divisible by 16 you can either use
(n+1) % 16 == 0 OR
(n+1) & 15 == 0