ddo wrote:Hi,
Which would be the fastest way to find all the divisors of a nbr (192 for example)?
I do the prime factorization 192=2sq6x3, but then which method is the fastest to find all divisors?
Thanks!
You have 192 = 2^6 * 3. First, focus on one of the primes, say 2^6. This has the following divisors:
1, 2, 2^2, 2^3, 2^4, 2^5, 2^6
Now we get more divisors by using the 3; just multiply all of the above divisors by 3:
3, 3*2, 3*2^2, 3*2^3, 3*2^4, 3*2^5, 3*2^6.
That's it - we've listed every possible combination we can make by selecting primes from the six 2's and one 3 that divide 192. So 192 has 14 divisors in total. If we had instead been looking at 2^6 * 3^2 = 576, we would have had another set of seven divisors, which you'd get by multiplying the original set above by 3^2. So 576 has 21 divisors in total.
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