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- Divisibility Rules (Tests) Easily test if one number can be evenly divided by another

Friday, 15 May 2015

## Divisibility Rules (Tests) Easily test if one number can be evenly divided by another

Divisibility is a good way to factor large numbers. Learn to test for divisibility by 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Although there are divisibility tests for numbers in any radix, and they are all different, this article presents rules and examples only for decimal numbers.

**#within 2 Seconds | Divisibility rules | #Learn Now**

This video covers the divisibility rules of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. After watching this quick video you are gonna answer seeing the number at a glance whether or not this number is divisible by the numbers ranging between 1 - 13.

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Thanks for watching the video. We at Team MAST (Maven Scientists) just want to help our community with all that is possible.

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__Probliem 1:__

**Determine whether 7,168 is divisible by 2, 3, 4, 5, 6, 8, 9 and 10.**

7,168 is divisible by 2 since the last digit is 8.

7,168 is not divisible by 3 since the sum of the digits is 22, and 22 is not divisible by 3.

7,168 is divisible by 4 since 168 is divisible by 4.

7,168 is not divisible by 5 since the last digit is not 0 or 5.

7,168 is not divisible by 6 since it is not divisible by both 2 and 3.

7,168 is divisible by 8 since the last 3 digits are 168, and 168 is divisible by 8.

7,168 is not divisible by 9 since the sum of the digits is 22, and 22 is not divisible by 9.

7,168 is not divisible by 10 since the last digit is not 0 or 5.

Solution: 7,168 is divisible by 2, 4 and 8.

__Problem 2:__

91 is not divisible by 2 since the last digit is not 0, 2, 4, 6 or 8.

91 is not divisible by 3 since the sum of the digits (9+1=10) is not divisible by 3.

91 is not evenly divisible by 4 (remainder is 3).

91 is not divisible by 5 since the last digit is not 0 or 5.

91 is not divisible by 6 since it is not divisible by both 2 and 3.

91 divided by 7 is 13.

Solution: The number 91 is divisible by 1, 7, 13 and 91. Therefore 91 is composite since it has more than two factors.

There are many tricks to teach children divisibility in mathematics. Some tricks that I used to use in my classroom are listed here. If you know of some that I may have missed, drop into the forum and let everyone know. I'll add them to this list as I see them.