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:
74 comments Page 1 of 8.

Bijip said:   2 months ago
Why we will take co-prime here? Please explain me.

Vaibhavi said:   1 year 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.
(17)

MANJULADEVI K said:   1 year ago
The product of 2 number is 4107.
The HCF of these number is 37.
4107= 37 * a
4107/37 = a
Answer is 111.
(49)

Omprakash said:   2 years ago
Thanks, everyone.
(3)

Srivani said:   2 years ago
How you get ab=3? please explain.
(6)

Mayur said:   3 years ago
Simple;

HCF*LCM = PRODUCT.

37 * 111 = 4107.
(19)

Gopi said:   3 years ago
37 * x = 4107.
x = 111.
(4)

Mouni said:   3 years ago
HCF=37
The product of two no's=4107
37 * x = 4107, where x=greatest no
x = 4107%37 = 111.
(3)

Nilay said:   3 years ago
Thank you @Shilpa.
(1)

Hemanth reddy said:   3 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.
(11)


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