For instance, 8 is even because 8 / 4 = 2. In other words, its even because its evenly divisible by something.

2 and 5 are the easiest to test because we are working with a base 10 system.

2 and 5 are divisors of 10.

Whatever system your in, the divisors of that system will be the easiest to work with.

In an expression a + b + c, if a and b are multiples of 5, then in order for the result to be a multiple of 5, c must also be divisible by 5.

For example, 10 + 15 + 20 can be written as 5 * 2 + 5 * 3 + 5 * 4 = 5(2 + 3 + 4)

Another example, a number 473 = 4 * 100 + 7 * 10 + 3

Example 1: 3 3120 since 3 3 + 1 + 3 + 0 The symbol “ ” means “divides into” Example 2: 713 is not divisible by 3 since 7 + 1 + 3 = 11 and 3 does not 11.

Example: 347 = 3(100) + 4(10) + 7 = 3(99 + 1) + 4(9 + 1) + 7 = 3 * 99 + 4 * 9 + (3 + 4 + 7) Notice the 99 and the 9

Breaking 24 into it’s prime number factorizations: 24 = 2 * 3 * 2 * 2 24/6 = 2 * 3 n/6 = P1 * P2 * P3… / 2 * 3 Example: n = 3120 Since 2 | n and 3 | n then we know 6 | n

**For the following number determine all possible ways to fill in the missing**
**ones digit so that the number is divisible by 6.**

4,32_

The ones digit must be either 0, 2, 4, 6 or 8.

3 | 4320? Yes, 4 + 3 + 2 == 9 and 3 | 9

3 | 4 + 3 + 2 and 3 does not go into 4 + 3 + 2 + 2

…

Ex 1: 1784 = (1 * 1000) + (7 * 100) + (8 * 10) + 4

Ex 2: Does 4 divide 342?

No, because 4 does not divide 42.

Ex: 8 3120 since 8 120, i.e. 120 = 8 * 15.

Interesting therem:

For 2^1 check the last digit

For 2 squared check the the last two

For 2 cubed chek the last three

- (Altername means alternate signs between + and -)

Ex: Let n = 6314

6 - 3 + 1 - 4 = 0

So, since 11 | 0 it must be that 11 | 6314

Ex: Let n = 3120

3 - 1 + 2 - 0 = 4

Since 11 does not | 4, then 11 does not | 3120.

What if we computed - 3 + 1 - 2 + 0 = -4?

So since 11 does not | -4 (and in general, if 11 does not | a then 11 | -a)

then the divisibility rule isunaffected.

- Use the following iterative process:
- Step 1: Take the last digit and remove it from the rest of the number.
- Last digit: d = n % 10, remove last digit: n = n/10
- Step 2: Subtract twice the last digit from the number formed by the remaining digits
- n = n - 2 * d
- Repeat steps one and two as needed.
- The origional number n is divisible by 7 if and only if the number obtained by this iterative process is divisible by 7.

Ex: n = 392

Step 1: d = 2, n = 39

Step 2: n - 2d = 39 - 4 = 35

True since 7 | 35, then 7 | 392

Ex: n = 1384

Step 1: d = 4, n = 138

Step 2: 138 - 2d = 130

Repeat:

Step 1: d = 0, n = 13

Step 2: 13 - 2(0) = 13

Since 7 does not | 13, then 7 does not divide 1384.