Question
Quantitative Aptitude Question on LCM and HCF
If the LCM of three natural numbers 2412, 1816 and a is 3618, then how many values of a are possible?
A
37
B
39
C
38
D
36
Answer
37
Explanation
Solution
The correct option is (A): 37.
2412 = (23 x 3)12 = 236 x 312
1816 = (2 x 32)16 = 216 x 332
3618 = (22 x 32)18 = 236 x 336
The power of 2 in a can be any value from 0 to 36 because the LCM has 236 which is already present in 2412. So, a may or may not contain powers of 2.
Thus, there are 37 possible cases.