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:
77 comments Page 3 of 8.

Jaz said:   2 decades ago
Thanks raj.
(1)

Sanjivani said:   2 decades ago
Co prime means not get co-prime (1, 3).
(1)

Sai vamsi said:   6 years ago
x*y = 4107.
Hcf of two nums is 37. So these two nums are divisible by 37.
37x*37y = 4107.
37(x*y) = 4107.
x*y = 4107/37 = 111.

Asif said:   9 years ago
How we come to know that in this question we have to find LCM, in the question we have to find the greater number? please.

Harwani Payal said:   7 years ago
Assume Two numbers p and q.
Multiply them with 37.
37p*37p.

Now, 37p*37p = 4107.
1369pq = 4107.
pq = 4107/1369.
pq = 3.

Nidhu said:   7 years ago
We know that;

H.C.F * L.C.M = product of two numbers.
37* x = 4107,
x=4107/37,
x = 111.

Manikanta said:   7 years ago
For a*b =3. So.

1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).

Then how you are deviating to 1, 2 and 2, 1? Please tell me.

Manikanta said:   7 years ago
For a*b =3. So.

1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).

Then how you are deviating to 1, 2 and 2, 1? Please tell me.

Abitha said:   8 years ago
Here for ab=3, its possibilities are (1, 2) & (2, 1) Only right?

Please, anyone, help me.

Vani said:   1 decade ago
Can LCM always be the greatest no.?


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