The L.C.M of any 2 or more numbers is the lowest number that can be divisible by any of the numbers without a remainder.
Ex: find the L.C.M of 12 15 and 21
Solution:
|
2
|
12
|
15
|
21
|
|
2
|
6
|
15
|
21
|
|
3
|
3
|
15
|
21
|
|
5
|
1
|
5
|
7
|
|
7
|
1
|
1
|
7
|
|
|
1
|
1
|
1
|
22*3*5*7=420
Test of divisibility
This involves determining whether a number is divisible by 2,3,4,5,6,8,9,10,12 and 25
- A number is divisible by 2 if it’s last digit is even or 0
- A number is divisible by 3 if the sum of its digits is divisible by 3
- A number is divisible by 4 if it’s last 2 digits is divisible by 4
- A number is divisible by 5 it it’s last digit is 5 or 0
- A number is divisible by 8 if it’s last 3 digits is 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 it’s last digit is 0
- A number is divisible by 12 if it is divisible by 3 and 4
- A number is divisible by 25 if it’s last 2 digits is divisible by 25
