whats the best way to figure out divisble numbers when factoring?

Question

its something that I've always struggled with. for example factor the number 105 is there an easier way than just guessing numbers? really need help

Answer 1

Yes.

- A number is divisible by 2 if the last digit is divisible by 2
- A number is divisible by 3 if the sum of the digits is divisible by 3
- A number is divisible by 4 if the last two digits are divisible by 4
- A number is divisible by 5 if the last digit is a 0 or 5
- A number is divisible by 6 if it is divisible by 2 and 3
- There is no easy rule for 7, but below I will show you a method.
- For small numbers, there is no easy rule for 8 (use the same method that I show below for 7). For larger numbers, a number is divisible by 8 if its last 3 digits are divisible by 8.
- A number is divisible by 9 if the sum of its digits is divisible by 9.
- A number is divisible by 10 if its last digit is 0.

So for your example:
105
Last digit is 5 so it is not divisible by 2 (and thus not divisible by 6 either) or 10, but it is divisible by 5.
The sum of the digits is 6 so it is divisible by 3 but it is not divisible by 9.
The last two digits are 05 which is not divisible by 4, so 105 is not divisible by 4.

To figure out if it is divisible by 7 or 8:
(this logic can be applied to prime numbers greater than 10 such as finding if a number is divisible by 11, 13, etc.)
You know 70 is divisible by 7 so:
105 - 70 = 35
If 35 is divisible by 7, then so is 105. And 35 is divisible by 7.

You know 80 is divisible by 8 so:
105 - 80 = 25
If 25 is divisible by 8, then so is 105. Since 25 is not divisible by 8, neither is 105.


So to recap, 105 is divisible by 3, 5, and 7

Now, since it is divisible by 3,5, and 7 it is also divisible by these numbers multiplied by each other. So it is divisible by:
3 * 5 = 15
3 * 7 = 21
5 * 7 = 35

So 105 is divisible by 3, 5, 7, 15, 21, and 35

Answer 2

An even number is divisible by 2.
A number ending in 5 is divisible by 5
A number ending in 0 is divisible by 10
If the sum of digirts is divisible by 3, the number is divisible by 3.
In your case 105/5= 21 and 21= 7*3 so 105 = 3*5*7

Answer 3

All I know are the quick rote memory systems. 105 is obviously divisible by 5. It is also easily divisible by 3 ( 1+0+5 = 6 which is divisible by 3). Therefore, 15 is a pretty good size number to be divisible by...i.e. (15*7=105)

Answer 4

Many factors have simple rules:

If its divisable by 2 its even
If its divisable by 5 it ends in 0 or 5
If its divisable by 3, sum of its digits is divisable by 3

105 (your example)

Ends in 5 so it is divisable by 5
Digits add up to 6, 6 is divisable by 3, so 105 is divisable by 3

After using some of these types of rules, you can figure the other factors.