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:
101
107
111
185
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 2 of 8.

Mouni said:   4 years ago
HCF=37
The product of two no's=4107
37 * x = 4107, where x=greatest no
x = 4107%37 = 111.
(4)

Frio said:   1 year ago
Thanks all for the explanation.
(3)

Nilay said:   4 years ago
Thank you @Shilpa.
(1)

Pannu said:   1 decade ago
37a*37b=4107
1369ab=4107
ab=4107/1369
ab=3
(1)

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.
(1)

Sangeetha said:   9 years ago
How will we decide 111 is the largest number?
(1)

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.
(1)

Rakesh said:   1 decade ago
How you got ab=3?
(1)

Raj said:   1 decade ago
Simple method:
LCM=product of co primes/HCF
=>4107/37=111
(1)

Jaz said:   1 decade ago
Thanks raj.
(1)


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