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...
Fastest way to find all the divisors
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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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