Let m = 2^k - 1. To check divisibility of n by m, 1. If n is zero, return true. 2. If n is negative, replace n with -n. 3. While n > m, replace n with (n >> k) + (n & m). 4. Return (n == m).
This is equivalent to the "casting out nines" algorithm to determine if a number is a multiple of 9. Dave On Jun 22, 3:37 pm, divya <[email protected]> wrote: > u are given any binary no...... u have to check its divisbility by > 3,7,15, > 31......(any no. of the form 2^x-1) > .u cant use any division > and modulo operator for that..... -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
