# Math Help - HCF

1. ## HCF

Has anyone got a quick way to find the Highest Common Factor (HCF) of these numbers?

1). 6, 15
2). 12, 16
3). 12, 30
4). 21, 14
5). 36, 16
6). 50, 35
7). 45, 27
8). 64, 88
9). 35, 14
10). 20, 40
11). 28, 84
12). 48, 84
13). 99, 77
14). 90, 36
15). 60, 48
16). 88, 66
17). 96, 144
18). 140, 252
19). 175, 250
20). 396, 252
21). 27, 18, 99
22). 42, 56, 98
23). 108, 54, 90
24). 96, 192, 144

2. Originally Posted by Natasha1
Has anyone got a quick way to find the Highest Common Factor (HCF) of these numbers?

1). 6, 15
2). 12, 16
3). 12, 30
4). 21, 14
5). 36, 16
6). 50, 35
7). 45, 27
8). 64, 88
9). 35, 14
10). 20, 40
11). 28, 84
12). 48, 84
13). 99, 77
14). 90, 36
15). 60, 48
16). 88, 66
17). 96, 144
18). 140, 252
19). 175, 250
20). 396, 252
21). 27, 18, 99
22). 42, 56, 98
23). 108, 54, 90
24). 96, 192, 144
Edit: You're asking far too many questions here, nobody will do all 24 for you. Especially if there is no evidence that you'd tried them yourself

Split them into their prime factors and multiply the highest power of the factors which appear in both

Eg 1:

$6 = 2 \times 3$
$15 = 3 \times 5$
$\text{HCF} = 3$

Eg 17

$96 = 2^5 \times 3$
$144 = 2^4 \times 3^2$
$\text{HCF} = 2^4 \times 3 = 48$

Note that I only multiplied by $2^4$ because that is a factor of both

3. Hello Natasha1
Originally Posted by Natasha1
Has anyone got a quick way to find the Highest Common Factor (HCF) of these numbers?

1). 6, 15
2). 12, 16
3). 12, 30
4). 21, 14
5). 36, 16
6). 50, 35
7). 45, 27
8). 64, 88
9). 35, 14
10). 20, 40
11). 28, 84
12). 48, 84
13). 99, 77
14). 90, 36
15). 60, 48
16). 88, 66
17). 96, 144
18). 140, 252
19). 175, 250
20). 396, 252
21). 27, 18, 99
22). 42, 56, 98
23). 108, 54, 90
24). 96, 192, 144
This is easier to do than to explain, but it's something like this:
1. Find the prime factors of each number.

2. For each prime factor that they have in common, choose the lower of their two exponents (powers).

3. Multiply these common factors together.
For instance, here's question 14, with the numbers $90$ and $36$:
$90 = 2^1\times3^2\times 5^1$

$36 = 2^2\times 3^2$
The common prime factors are $2$ and $3$; the lower of their exponents then are:
$2^1$ and $3^2$
So the HCF is $2^1\times 3^2 = 18$.

Did you follow that?

Grandad

4. Thanks for this but it's going to take me agesssssssssssssssssss :-(

Can anyone do this on a computer and give me the answers please

5. Originally Posted by Natasha1
Thanks for this but it's going to take me agesssssssssssssssssss :-(

Can anyone do this on a computer and give me the answers please
I sure can, but I'm not going to. Surely if you wanted the answers you could google them or use wolfram...

6. oh man! Harsh!!!

What is Wolfram

7. Originally Posted by Natasha1
oh man! Harsh!!!

What is Wolfram
Perhaps, and my apologies if I offended you. I stand by what I said though, I understand these problems and I've got plenty to do.

Wolfram: Wolfram|Alpha although it doesn't seem to like prime factord

8. Originally Posted by e^(i*pi)
Perhaps, and my apologies if I offended you. I stand by what I said though, I understand these problems and I've got plenty to do.

Wolfram: Wolfram|Alpha although it doesn't seem to like prime factord
Great site thanks for the tip :-)