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78 = 2*3*13

66 = 2*3*11

Since both are factors of x then x will have 2*3*11*13.

Now we need to calculate positive factors of x so look at it this way. We can either pick a number or drop a number. So we have 2 options for each of these. So we have 2*2*2*2 factors possible and that's 16.
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Bunuel
If 78 and 66 are both factors of the positive integer x, what is the smallest number of positive factors x could have in total?

A. 12
B. 16
C. 18
D. 24
E. 36

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\(LCM ( 78, 66) = 858\)

Now, \(858 = 2^1 3^1 11^1 13^1\)

So, No of factors of 858 = (1+1)(1+1)(1+1)(1+1) => 16 , Answer must be (B)
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If 78 and 66 are both factors of the positive integer x, what is the smallest number of positive factors x could have in total?

A. 12
B. 16 --> correct: the smallest number x that is divisible by 78 & 66 i.e the smallest number x = LCM (78, 66) = LCM (2*3*13, 2*3*11) = \(2^1*3^1*11^1*13^1\); so number of positive factors x = (1+1)*(1+1)*(1+1)*(1+1) = 2^4 = 16
C. 18
D. 24
E. 36
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