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Divisibility Tests

A number is…

divisible by 2 if and only if it’s ones digit is even (i.e. 0, 2, 4, 6, 8).

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

divisible by 5 if and only if its ones digit it 0 or 5.

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.

divisible by 10 if and only if its ones digit is 0.

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

divisible by 3 if and only if the sum of its digits is divisible by 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.

divisible by 9 if and only if the sum of its digits is divisible by 9.

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

divisible by 6 if and only if and only if it is divisible by both 2 and 3.

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
…

divisible by 4 if and only if the number represented by its lats two digits is divisible by 4.

Ex 1: 1784 = (1 * 1000) + (7 * 100) + (8 * 10) + 4
Ex 2: Does 4 divide 342?
No, because 4 does not divide 42.

divisible by 8 if and only if the number represented by its last 3 digits is divisible by 8.

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

divisible by 11 if and only if the alternating sum of its digits is divisible by 11.

  • (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 is unaffected.

divisible by 7 if and only if

  • 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.