# HCF and LCM problem

• August 12th 2009, 03:55 AM
louis12345
HCF and LCM problem
i have a question about HCF and LCM, which I have really forgot about.

Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of
a and c is 3.

I am always weak at HCF and LCM, i only know their definition. Hope someone explain me

and if it is possible, i hope someone can provide me some difficult examples on stuff of LCM and HCF, thanks all!!!
• August 12th 2009, 04:24 AM
bruxism
HCF stands for highest common factor.

If you factorise a number, it looks like this.

Example
12= 1 x 12 = 2 x 6 = 3 x 4

So the factors of 12 are 1, 2, 3, 4, 6, 12

Lets factorise another number

18 = 1 x 18 = 2 x 9 = 3 x 6

Factors are 1, 2, 3, 6, 9, 18

the COMMON factors between 12 and 18 are , 1, 2, 3 and 6

The HIGHEST COMMON factor between 12 and 18 is 6.

Now with a hint that A = 6, can you find the values for B and C?
• August 12th 2009, 04:46 AM
Amer
Quote:

Originally Posted by louis12345
i have a question about HCF and LCM, which I have really forgot about.

Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of
a and c is 3.

I am always weak at HCF and LCM, i only know their definition. Hope someone explain me

and if it is possible, i hope someone can provide me some difficult examples on stuff of LCM and HCF, thanks all!!!

look
HCF is the biggest number you can factor from two numbers Example what is the HCF for 15 , 25 it is 5 since it is the biggest factor you can take

Examples:-
$1) 144,204$ here you need to simplify it

$142=2\times 2\times 2\times 2\times 3\times 3$

$204=2\times 2\times 51$

how I can explain it ok

the similar numbers in the two numbers is "2" just and in 204 you have two "2" and in 142 you have four "2" so take two "2" not four two so 2x2=4 4 is the HCF for 142,204

$2)168,380$

$168=2\times 2\times 2\times 3\times 7$

$380=2\times 2\times 3\times 3\times 5$

you have two digits here "2" and "3" in the two numbers in 168 you have three "2" and one "3" in 380 you have two "2" and two "3" so you take the smaller numbers in each one

two "2" and one "3" so HCF 2x2x3=12

$3)2800,2940$

$2800=2\times 2\times 2\times 2\times 5\times 5\times 7$

$2940=2\times 2\times 3\times 5\times 7\times 7$

similar digits "2" and "5" and "7"
2800 have four "2" and two "5" and one "7"
2940 have two "2" and one"5" two "7" so

HCF two "2" and one"5" and one "7"
$HCF=2\times 2\times 5\times 7 = 4(5)(7)=140$

I think it is clear
• August 12th 2009, 05:17 AM
Amer
Quote:

Originally Posted by louis12345
i have a question about HCF and LCM, which I have really forgot about.

Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of
a and c is 3.

I am always weak at HCF and LCM, i only know their definition. Hope someone explain me

and if it is possible, i hope someone can provide me some difficult examples on stuff of LCM and HCF, thanks all!!!

LHM

Examples:-

$420,700$

simplify

$420=2\times 2\times 3\times 5\times 7$

$700=2\times 2\times 5\times 5\times 7$

the difference between the two numbers is 420 have same factor except there is "3" and 700 have same factors except there is another "5" so LCM multiply 420 with the "5" and 700 with "3" so the two numbers will be same and this number is LCM and it equal 2100

$560,528$ simplify

$560=2\times 2\times 2\times 2\times 5\times 7$

$528=2\times 2\times 2\times 2\times 3\times 11$

here 560 have "5" and "7" not the same and 628 have "3" and "11" so multiply 560 with "3" and "11" and 528 with "5" and "7" you will have the same number and that is the LCM and it equal 18480

some note HCF for

1)11,17 equal 1 since there is no factor except 1

2)13,13 equal 13 since it is the biggest factor you can take

LCM for

1)13,20 is 13(20)

2)20,27 is (20)(27)

LCM and HCF

.................................................. .................................................. ............................
• August 12th 2009, 06:52 AM
louis12345
thanks above for the explanation...

"Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of
a and c is 3."

I still don't quite understand how to solve it

people can try to explain me more deeply, I understand what is
HCF and LCM and short division. But I am always weak at "Algebra" while good at Calculus. In fact I am a new college teacher to help students with Maths, but I really forget what is LCM and HCF, because the terms are really history to me, Thanks alll.
• August 12th 2009, 07:14 AM
Amer
Quote:

Originally Posted by louis12345
thanks above for the explanation...

"Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of
a and c is 3."

I still don't quite understand how to solve it

people can try to explain me more deeply, I understand what is
HCF and LCM and short division. But I am always weak at "Algebra" while good at Calculus. In fact I am a new college teacher to help students with Maths, but I really forget what is LCM and HCF, because the terms are really history to me, Thanks alll.

2 is factor of a,b ...(1)
4 is factor of b,c ...(2)
3 is factor of a,c ...(3)

$a=2\times 3\times \frac{a}{6}$

$b=2\times 2 \times \frac{b}{4}$

$c=2\times 2\times 3\times \frac{c}{12}$

I think there is a mistake in the question since "2" is a factor of "a" and a factor of "c" this from (1) and (2) but "3" is also a factor of "a" and "c" so HCF for "a" and "c" should be at least 6
• August 12th 2009, 07:21 AM
bruxism
Quote:

"Suggest three numbers a,b,c, such that the HCF of a and b is 2, HCF of b and c is 4 and HCF of a and c is 3."
When you hit a problem like this, try writing down what you know about each number. You know that

A = 2 x ? = 3 x ?
B = 2 X ? = 4 x ?
C = 4 x ? = 3 x ?

Ok so we know that ABC must satisfy these things. A B and C all have factors of either 2 or 4. This means they are all even. That's another clue for you. A also has a factor of 3.

All possible A's must be even and be multiples of 3.

A could be 6, 12, 18, 24 etc.

Now you can test A=6 and see if the other two are satisfied for this A.

Now C must have a HCF with A of 3 and must also be a multiple of 4. 12 satisfies both of these conditions.

A = 6
C = 12

Now b's highest common factor with A is 2, and with C is 4. B = 8 satisfies these conditions.

So
A=6
b=8
c=12

Listing all factors and highlighting the highest common factors we get

6 (1,2,3,6)
8 (1,2,4,8)
12 (1,2,3,4,12)

To be honest this is all very long winded. Perhaps work at finding highest common multiples of two numbers and then move to doing problems like this after you are familiar with that.

Try questions like
Two different numbers, A and B have HCF's of 5, while A has a factor of 10. What are the lowest numbers that satisfy these conditions?
• August 12th 2009, 07:24 AM
Amer
Quote:

Originally Posted by bruxism
When you hit a problem like this, try writing down what you know about each number. You know that

A = 2 x ? = 3 x ?
B = 2 X ? = 4 x ?
C = 4 x ? = 3 x ?

Ok so we know that ABC must satisfy these things. A B and C all have factors of either 2 or 4. This means they are all even. That's another clue for you. A also has a factor of 3.

All possible A's must be even and be multiples of 3.

A could be 6, 12, 18, 24 etc.

Now you can test A=6 and see if the other two are satisfied for this A.

Now C must have a HCF with A of 3 and must also be a multiple of 4. 12 satisfies both of these conditions.

A = 6
C = 12

Now b's highest common factor with A is 2, and with C is 4. B = 8 satisfies these conditions.

So
A=6
b=8
c=12

Listing all factors and highlighting the highest common factors we get

6 (1,2,3,6)
8 (1,2,4,8)
12 (1,2,3,4,6,12) you forgot 6

To be honest this is all very long winded. Perhaps work at finding highest common multiples of two numbers and then move to doing problems like this after you are familiar with that.

Try questions like
Two different numbers, A and B have HCF's of 5, while A has a factor of 10. What are the lowest numbers that satisfy these conditions?

....
• August 12th 2009, 07:33 AM
bruxism
yep, definitely forgot 6. I can't see there being a solution to this problem unless we disregard the factors 1 x ?.

I think when dealing with HCF, using

6 = 1 x 6 is pointless as it has no practical application in simplifying algebraic equations. So I believe my answer is correct, while my list of factors should look more like....

A = 6 (2,3)
B = 8 (2,4)
C = 12 (2,3,4,6)

What do you think Amer??
• August 12th 2009, 07:37 AM
Amer
Quote:

Originally Posted by bruxism
yep, definitely forgot 6. I can't see there being a solution to this problem unless we disregard the factors 1 x ?.

I think when dealing with HCF, using

6 = 1 x 6 is pointless as it has no practical application in simplifying algebraic equations. So I believe my answer is correct, while my list of factors should look more like....

A = 6 (2,3)
B = 8 (2,4)
C = 12 (2,3,4,6)

What do you think Amer??

I think the question is wrong since 2 is factor of "a" and "c" and 3 also "3" is factor of "a" and "c" so HCF for "a" and "c" can't be 3 or 2 as I said before at least 6
• August 12th 2009, 07:46 AM
bruxism
i think if you narrow the definition of factor to any factor except for 1 and the number itself, then the HCF of A and C is 3, and the question makes sense.