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:
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.
Rakesh said:
1 decade ago
How you got ab=3?
(1)
Pannu said:
1 decade ago
37a*37b=4107
1369ab=4107
ab=4107/1369
ab=3
1369ab=4107
ab=4107/1369
ab=3
(1)
Raj said:
1 decade ago
Simple method:
LCM=product of co primes/HCF
=>4107/37=111
LCM=product of co primes/HCF
=>4107/37=111
(1)
Jaz said:
1 decade ago
Thanks raj.
(1)
Sanjivani said:
1 decade ago
Co prime means not get co-prime (1, 3).
(1)
PVNARASIMHARAO said:
1 decade ago
L.C.M * H.C.F=FIRSTNUMEBR * SECONDNUMBER
IF THEY ARE COPRIMES
IF THEY ARE COPRIMES
Mehar said:
1 decade ago
Easy way to get ans:divide the product with H.C.F
-->4107/37=111;
-->4107/37=111;
Jyoti said:
1 decade ago
Simply divide the product with H.C.F
i.e 4107/37=111
i.e 4107/37=111
Anand said:
1 decade ago
Simply multiply the value 37*111=4107
Rahul said:
1 decade ago
Thanks Pannu.
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