Hey there,
What you could do is use some of the properties of numbers. For instance:
- if it is an even number, it's always divisible by 2
- if the sum of the digits is divisible by 3, then the number will be divisible by 3 (example: 192 - 1+9+2 = 12, which is divisible by 3. That means that 192 will also be divisible by 3)
- if the last two digits of your number make a number that is divisible by 4, then the number itself is divisible by 4 (example: 144 - 44 is divisible by 4, so 144 will be divisible by 4, it's actuallly 12^2). This rule can be extended to all of the powers of 2: if the last x digits of the number in question make a number that is divisible by 2^x, then said number is divisible by 2^x
- same goes for the powers of 3: if the sum of the digits of a number divisible by 3^x, then the number itself is divisible by 3^x
- numbers that end in either 0 or 5 are divisible by 5
I also use prime factorization most of the time. besides, I've noticed that the numbers CATs "come up" with are usually reasonable, no more than three digits long...
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You have 192 = 2^6 * 3. First, focus on one of the primes, say 2^6. This has the following divisors: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!
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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