Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 6)
6.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Answer: Option
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Discussion:
76 comments Page 1 of 8.
Ramkumar LS said:
1 decade ago
Simplest method in few secs.
Formula:
Product of two numbers = HCF X LCM.
Given:
Product of two numbers = 4107, HCF = 37 and LCM = unknown.
Now, simply substitute the given values.
4107 = (37 X unknown).
Lets keep unknown on one side and bring 37 below.
Unknown = 4107/37.
Therefore, Unknown= 111 //simply divide 4107 by 37.
Note: When you do this on paper this will only take 2 secs to find.
Formula:
Product of two numbers = HCF X LCM.
Given:
Product of two numbers = 4107, HCF = 37 and LCM = unknown.
Now, simply substitute the given values.
4107 = (37 X unknown).
Lets keep unknown on one side and bring 37 below.
Unknown = 4107/37.
Therefore, Unknown= 111 //simply divide 4107 by 37.
Note: When you do this on paper this will only take 2 secs to find.
Shilpa said:
4 years ago
Co prime means the only common factor of two numbers is 1.
For this problem, let the numbers be 37a and 37b as we know both numbers is multiple of 37.
So, 37a*37b = 4107.
37ab = 111,
ab=3.
Here there's only one way to split 3. Ie 1*3 or 3*1.
So let a=1 and b=3.
Numbers are 37a and 37b.
Substitute the value for a and b.
37*1=37 and 37*3=111.
Here the highest number is 111.
Hence the answer is 111.
For this problem, let the numbers be 37a and 37b as we know both numbers is multiple of 37.
So, 37a*37b = 4107.
37ab = 111,
ab=3.
Here there's only one way to split 3. Ie 1*3 or 3*1.
So let a=1 and b=3.
Numbers are 37a and 37b.
Substitute the value for a and b.
37*1=37 and 37*3=111.
Here the highest number is 111.
Hence the answer is 111.
(13)
Sugandh said:
1 decade ago
Let we have two numbers 2 and 6...
now product of these two numbers=12
hcf(2 and 6)=2
lcm=6
now we hav 2 find d greater numbr that is 6..how it comes..it is simply d lcm...
it means we hav to find out the lcm....so somply divide the product f two numbers by their hcf to get the lcm;,,,,
12/2=6 =greatest number
now product of these two numbers=12
hcf(2 and 6)=2
lcm=6
now we hav 2 find d greater numbr that is 6..how it comes..it is simply d lcm...
it means we hav to find out the lcm....so somply divide the product f two numbers by their hcf to get the lcm;,,,,
12/2=6 =greatest number
Bipin said:
1 decade ago
Simply as we know:
LCM x HCF = product of co-prime or a x b.
Given that,
Product of two nos = 4107
HCF = 37
LCM = ?
Finally we can put the value on above formula,
LCM*HCF = product.
LCM = 4107/37.
LCM = 111 Answer.
LCM x HCF = product of co-prime or a x b.
Given that,
Product of two nos = 4107
HCF = 37
LCM = ?
Finally we can put the value on above formula,
LCM*HCF = product.
LCM = 4107/37.
LCM = 111 Answer.
DHEERAJ SINGH said:
1 decade ago
HCF = 37.
Product of two numbers are = 4107.
HCFxLCM = Product of two nos.
37xLCM = 4107.
LCM = 111.
Since LCM is the least no which will be divided by the number.
So the number will be a multiple of 111, so from above options 111 is the only multiple of 111. So answer is 111.
Product of two numbers are = 4107.
HCFxLCM = Product of two nos.
37xLCM = 4107.
LCM = 111.
Since LCM is the least no which will be divided by the number.
So the number will be a multiple of 111, so from above options 111 is the only multiple of 111. So answer is 111.
Vinay gudipati said:
7 years ago
Consider the equation be like x*y=4107.
Here x and why are the taken when product them we get 4107.
And also given that HCF of the two numbers is 37 so.
37*x*37*y=4107.
x*y= (4107/ (37*37) ) =3.
xy=3.
x*y=3*1.
x=3.
y=1.
Therefore 37*3=111.
37*1=37 so the answer is 111.
Here x and why are the taken when product them we get 4107.
And also given that HCF of the two numbers is 37 so.
37*x*37*y=4107.
x*y= (4107/ (37*37) ) =3.
xy=3.
x*y=3*1.
x=3.
y=1.
Therefore 37*3=111.
37*1=37 so the answer is 111.
(1)
Ravi said:
1 decade ago
@ vani
Its false statement that LCM is the the greater no.
We know as a*b = lcm*hcf
Here 4107 = lcm*37
lcm = 111 or (3*37)
so a*b = 37*3*37
or a*b = (37*3)*37
or a*b = 37*(3*37)
So greater no. is 3*37=111
Its false statement that LCM is the the greater no.
We know as a*b = lcm*hcf
Here 4107 = lcm*37
lcm = 111 or (3*37)
so a*b = 37*3*37
or a*b = (37*3)*37
or a*b = 37*(3*37)
So greater no. is 3*37=111
Vaibhavi said:
2 years ago
Let the numbers be 37a × 37b
Then, 37a × 37b = 4107.
37ab = 4107/37,
37ab = 111,
ab = 111/37.
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
The Greater number = 111.
Then, 37a × 37b = 4107.
37ab = 4107/37,
37ab = 111,
ab = 111/37.
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
The Greater number = 111.
(26)
Prasad said:
1 decade ago
Generally ((hcf*lcm)/(n1*n2)) = 1.
So here the hcf is=37;
Let lcm is =x;
The product of two numbers = 4107.
So,
Now ((hcf*lcm)/(n1*n2)) = 1.
(37*x)/(4107) = 1.
By solving above expression we get x value as 111.
So here the hcf is=37;
Let lcm is =x;
The product of two numbers = 4107.
So,
Now ((hcf*lcm)/(n1*n2)) = 1.
(37*x)/(4107) = 1.
By solving above expression we get x value as 111.
Riya said:
1 decade ago
Look it is really easy:.
As we know the formula.
Product of two numbers = HCF*LCM.
And that 4107 = 37*LCM.
So, by a simple method which is,
LCM = 4107/37 = 111.
This method always works try for yourself.
As we know the formula.
Product of two numbers = HCF*LCM.
And that 4107 = 37*LCM.
So, by a simple method which is,
LCM = 4107/37 = 111.
This method always works try for yourself.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers