A positive integer N is divisable by both 6 and 4. What is the minimum number of positive divsors N must have?
Since N is divisable by both 6 and 4, its factorization must include both 3 and 4; i.e., it must be a multiple of at least 12. 12 has factors (1, 2, 3, 4, 6, 12).
My question is, why must the factorization of N include both 3 and 4?
The prime factors of 6 are 2, 3.
The prime factors of 4 are 2, 2.
Why do we multiply 3 and 4 to get 12 to get 6 factors?
When the number is divisible by 6, it should be divisible by 2x3.
Similarly, when the number is divisible by 4, it should be divisible by 2x2.
Thus, the number should be divisible by 2x2 and 3x2. Even if one of the 2s is common, it should still be divisible by 2x2x3. This is 3x4. This also follows from the fact that the LCM of 6 and 4 is 12, which is 3x4. (You may also recall that to calculate the LCM we cross out common factors).
This problem is easily tackled by taking an example, LCM, being 12.
Whatever should hold true for 12 would be the minimum
number of factors.
Hope that helps.
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