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If the least common multiple of a positive integer m and n is 120 and m:n is 3:4 what is the greatest common factor of m and n

3
5
6
10
12

Please provide detailed explanations on how to solve this

many thanks

D for me!
x is the least common factor of m and n
m*n=3x*4x=120*x, so x=10, because x can not be 0!
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D for me!
x is the least common factor of m and n
m*n=3x*4x
I don't get this (where does it come from ?)

sondenso
3x*4x=120*x
This is just false

sondenso
=120*x, so x=10, because x can not be 0!
I don't get this either.


Can you explain ?
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walker
D

fist way:

LCM=120=3*2^3*5

1. LCM contains prime number 5, so m or n or both also contain 5.
2. if m:n=3:4 then only both m and m contain 5. Therefore, GCD is 5 or 10.
3. LCM contains 2^3, but in ratio m:n we have only 4=2^2. So, both m and m contain 2. GCD=10.
I like this. Thanks !


walker
second way:

m*n=LCM*GCD - it is a formula.
m*n=3x*4x=120*GCD ---> GCD=x^2/10 ---> only 10 works.
Thanks for the refresh on the formula, I did not remember.

But why is m*n = 3x*4x ?
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Oski
But why is m*n = 3x*4x ?

m:n=3:4 --> m=3x, n=4x where x is an integer (m/n=3x/4x=3/4)
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Oski
But why is m*n = 3x*4x ?

m:n=3:4 --> m=3x, n=4x where x is an integer (m/n=3x/4x=3/4)
Yes, sure, but why is this x necessarily the GCD ?

Edit : Okay, I got it. This is because there is no common divisors in 3 and 4... (I guess this should be part of the demonstration ^^)
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vd
If the least common multiple of a positive integer m and n is 120 and m:n is 3:4 what is the greatest common factor of m and n

3
5
6
10
12

Please provide detailed explanations on how to solve this

many thanks

I just plugged in numbers for m & n. If the ratio is 3:4, then using 3 & 4 for their values works fine, since we're trying to find the GCF, and not the actual values of m&n,

so

m*n=LCM*GCF

3*4=120*GCF
12=120*GCF
10=GCF

D.
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m:n = 3:4

\(m = \frac{120}{4} = 30\)

\(n = \frac{120}{3} = 40\)

GCD of 30 & 40 = 10

Answer = D
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120=5*2*3*2*2

3/4 means 3 and 2*2 in two numbers exist, so we should give another 2*5 to both numbers to get the same ratio

3*5*2/2*2*2*5=30/40

GCF=10

D
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vd
If the least common multiple of positive integer m and n is 120, and m:n=3:4, what is the greatest common factor of m and n?

(A) 3
(B) 5
(C) 6
(D) 10
(E) 12
Supposing that m and n are 3*x and 4*x; LCM of m and n = 12*x= 120=>x=10
Since 3 and 4 are co-prime numbers 10 must be the gcf.
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Nice Question.
Here is what i did =>
As m/n=3/4
Let m=3x
n=4x

Now GCD = Common elements => x
LCM=common elements *leftovers => x*3*4=> 12x

Hence !2x=120
so x=10
Hence GCD=10
Hence D
Additionally we can say that numbers m and n must be 30 and 40
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vd
If the least common multiple of positive integer m and n is 120, and m:n=3:4, what is the greatest common factor of m and n?

(A) 3
(B) 5
(C) 6
(D) 10
(E) 12

We can let m = 3x and n = 4x for some positive integer x. We see that since 3 and 4 have no common factor (other than 1), the least common multiple (LCM) of m and n must be (3)(4)(x). Since it’s given that the LCM is 120, we have:

(3)(4)(x) = 120

12x = 120

x = 10

So m = 30 and n = 40, and we see that 10 is the greatest common factor of m and n.

Answer: D
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vd
If the least common multiple of positive integer m and n is 120, and m:n=3:4, what is the greatest common factor of m and n?

(A) 3
(B) 5
(C) 6
(D) 10
(E) 12

We can let m = 3x and n = 4x for some positive integer x. We see that since 3 and 4 have no common factor (other than 1), the least common multiple (LCM) of m and n must be (3)(4)(x). Since it’s given that the LCM is 120, we have:

(3)(4)(x) = 120

12x = 120

x = 10

So m = 30 and n = 40, and we see that 10 is the greatest common factor of m and n.

Answer: D

wd it be correct to say that 10 is the LOWEST gcf of m,n?

in other words, is it correct to assume that given the restrictions on m,n, there cd be other values of m,n that can give an lcm of 120 and have 10 as the gcf?
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Mansoor50


wd it be correct to say that 10 is the LOWEST gcf of m,n?

in other words, is it correct to assume that given the restrictions on m,n, there cd be other values of m,n that can give an lcm of 120 and have 10 as the gcf?

The short answer is no. There's only one pair of m,n which are in the ratio of 3 to 4 and which have a LCM of 120. Thus, 10 is The greatest common factor of m and n.

Any other pair m,n will either not have a ratio of 3 to 4 or will not have a LCM of 120. Since the ratio of m to n is 3:4, different values of m = 3x and n = 4x will be given by different values of x. Since the LCM of m and n is 120, both 120/3x = 40/x and 120/4x = 30/x are integers. Thus, x can only equal 1, 2, 5 or

10. Hence, the only possibilities for the pair m, n are 3 and 4, 6 and 8, 15 and 20, 30 and 40. Of these pairs, only 30 and 40 have a LCM of 120.
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why is LCM product of x and 'leftovers'? is this always true?
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vd
If the least common multiple of positive integer m and n is 120, and m:n=3:4, what is the greatest common factor of m and n?

(A) 3
(B) 5
(C) 6
(D) 10
(E) 12

Asked: If the least common multiple of positive integer m and n is 120, and m:n=3:4, what is the greatest common factor of m and n?

LCM (m,n) = 120 ; m = 3k , n= 4k
m*n = 3k * 4k = 12k^2 = 120 * GCF(m,n)
k = GCF (m,n)
12k^2 = 120 * k
k = 10 = GCF (m,n)

IMO D
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Hi,

Got it wrong at first then here is how I tacked it:

LCM 120 = 12*10= 4*3*5*2 = 2^3*3*5

M:N = 3x:4x
My 3 must come from M and I have two 2's that must come from N therefore my third 2 must be in x, and because I have it in x it must be in M too.
The remaining integer, 5, is also in x so we have x= 5*2=10

M=30 and N=40, Therefore the GCF is 10.

Answer D)
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For any two numbers a and b,
LCM*HCF = a*b


Let the positive integer m be 3x.
Thus, n = 4x

Thus, the HCF of the numbers will be x.

So,
120*x = 3x*4x
120 = 12x
x = 12.

Thus, the correct option is E.
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