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14 Dec 2022, 10:16
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:11) correct
43%
(02:16)
wrong
based on 234
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If the LCM of x and 36 is 756, then how many values of x are possible?
A. 1
B. 2
C. 3
D. 6
E. 12
A. 1
B. 2
C. 3
D. 6
E. 12
15 Dec 2022, 07:24
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gmatophobia
756=\(36*21=2^2*3^3*7\)
Now, \(36=2^2*3^2\)
So x has to have the highest powers of primes that are not there in 36.
That is minimum value of x = \(3^3*7\)
Maximum value of x will be 756 itself, that is \(2^2*3^3*7\)
The total possible values will be \(3^3*7\) or 189 multiplied by factors of \(2^2\) or 1, 2 and 4. Hence, THREE values.
=> 189*1, 189*2 and 189*4.
C
15 Dec 2022, 10:49
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gmatophobia
Solution:
- We are given \(LCM(x,36)=756\)
- Or \(LCM(x, 2^2\times 3^2)=2^2\times 3^3\times 7\)
- From this, we can infer that the minimum value of x is \(x_{min}=3^3\times 7\)
- Because 36 doesn't have \(3^3\) and \(7\) which we find in 756 i.e., the LCM
- Other values of x include \(2\times 3^3\times 7\) and \(2^2\times 3^3\times 7\)
- In total 3 values
Hence the right answer is Option C
