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 4 of 8.

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.

Durga said:   1 decade ago
But how we will know 37 is the H.C.F first only.

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.

Ramesh said:   1 decade ago
Given HCF = 37.

Product of two numbers is = 4107.

As HCF is greatest divisor so the number will be 4107/37 = 111.

The greatest number is 111.

Sandip said:   1 decade ago
We know that HCF*LCM = Product of two number.

So the other no is = 4107/37 = 111.

Prakruthi said:   1 decade ago
a*b = HCF*LCM;
4107 = 37*LCM;

LCM = 4107/37;
LCM = 111;

Arif said:   1 decade ago
But in case of smaller number?

Nikita said:   1 decade ago
Please explain the given evaluation properly. By co-prime method only.

MJani said:   1 decade ago
Explain any short method to solve this problem? Its too long of 4107/1369.

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.
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