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 1 of 8.

MANJULADEVI K said:   3 years ago
The product of 2 number is 4107.
The HCF of these number is 37.
4107= 37 * a
4107/37 = a
Answer is 111.
(82)

Mayur said:   4 years ago
Simple;

HCF*LCM = PRODUCT.

37 * 111 = 4107.
(38)

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.
(26)

Sonu Gurjar said:   4 months ago
Formula:

4107 = 37 * lcm,
Lcm = 4107/37 = 111.
(13)

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.
(13)

Hemanth reddy said:   4 years ago
AB = hcf * lcm.
4107 = 37 * x.
X = 4107/37.
Lcm = 111.

We know that lcm is always greater than HCF.
So the greatest number is 111.
In case the smallest number is 37.
(13)

Srivani said:   4 years ago
How you get ab=3? please explain.
(10)

Bijip said:   2 years ago
Why we will take co-prime here? Please explain me.
(10)

Gopi said:   4 years ago
37 * x = 4107.
x = 111.
(7)

Omprakash said:   3 years ago
Thanks, everyone.
(6)


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