This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

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A **prime number** is a number that is only divisible by 1 and itself. Below is a list of the first 15 prime numbers:

\[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\]

Every number can be written as a product of prime numbers. For example,

\[6=2 \times 3\]

\[8 = 2 \times 2 \times 2 \text{ or } 2^3\]

\[21 = 3\times 7\]

In the above examples,

- \(2 \times 3\) is the prime factorisation of \(6\),
- \(2\times 2\times 2\) is the prime factorisation of \(8\),
- \(3\times 7\) is the prime factorisation of \(21\).

There may be times when you are asked to find the prime factorisation of a number, but the number is very large. See the video below for an example of finding the prime factorisation of a large number.

The **Least Common Multiple **of a group of numbers is the smallest number that is divisible by all of the numbers in the group.

For example, let's try to find the least common multiple of the numbers \(2,3\) and \(4\) - i.e., we want to find the smallest number that is divisible by \(2,3\) and \(4\).

Let's look at the multiples of each of the numbers and identify the first one that is common to all three:

- Multiples of \(2\): \(2,4,6,8,10,\)\(12\),\(14,16,18,20\)
- Multiples of \(3\): \(3,6,9,\)\(12\),\(15,18,21\)
- Multiples of \(4\): \(4,8,\)\(12\),\(16,20,24\)

The smallest number that is in all of the lists is the least common multiple - \(12\).

Writing out the list of multiples and comparing lists is a valid way to find the least common multiple, but this can be difficult if the numbers get large or the lists get long. There is also a method to finding the least common multiple using the prime factorisation of each of the numbers.

- Write down the prime factorisation of each of the numbers.
- Multiply each factor together the greatest number of times it appears in each prime factorisation.

**Example: **Find the least common multiple of \(8,9\) and \(12\).

Let's start by writing the prime factorisations of each:

- \(8 = 2\times 2\times 2\)
- \(9 = 3\times 3\)
- \(12 = 2\times 2\times 3\)

The only prime numbers present in the prime factorisations are \(2\) and \(3\). The greatest number of times \(2\) appears in the above factorisations is three times (in \(8\)) and the greatest number of times \(3\) appears is twice (in \(9\)). Therefore, the least common multiple is

\[2\times 2\times 2\times 3\times 3 = 72\]

so \(72\) is the least common multiple of \(8,9\) and \(12\). As an exercise, verify this by writing out all the multiples of \(8,9\) and \(12\) and finding the smallest number in each of the lists - i.e., complete the lists below:

- Multiples of \(8\): \(8,16,24,\) ...
- Multiples of \(9\): \(9,18,27,\) ...
- Multiples of \(12\): \(12,24,36,\) ...

The product of two numbers is the number you get when you multiply the numbers. For example, the product of \(4\) and \(5\) is \(20\).**Product:**A factor of a number is a number that divides in evenly. For example, factors of \(20\) are \(1\), \(2\), \(4\), \(5\), \(10\) and \(20\) because all of those numbers divide in to \(20\), with no remainder.**Factor:**A factorisation of a number is a is what you get when you break a number up into a product of other numbers. For example, \(4 \times 5\) is a factorisation of \(20\). So is \(2 \times 10\).**Factorisation:**A number \(A\) is divisible by another number \(B\) if \(B\) divides into \(A\) without a remainder. For example, \(20\) is divisible by \(4\) because \(4\) divides into \(20\) \(5\) times with no remainder.**Divisible:**A multiple of a number is that number multiplied by a whole number. For example, \(20\) is a multiple of \(5\) because it is \(5 \times 4\). To check if a number \(A\) is a multiple of \(B\), divide \(B\) into \(A\) and see if there is a remainder. If there is no remainder, then it is a multiple.**Multiple:**

- Last Updated: May 1, 2024 10:36 AM
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