6. How many ways are there to choose integers $a, b$ and $c$ with $a<b<c$ from the list $1,5,8,21,22,27,30,33,37,39,46,50$ so that the product abc is a multiple of 12 ?
6. For the product $a b c$ to be a multiple of 12 , the integers $a, b$ and $c$ must include between them at least 1 factor of 3 and 2 factors of 2 .
Among the integers in the list, there are five categories with respect to divisibility by 2 and 3 :
- (A): 3 are not divisible by 2 or 3 (these are $1,5,37$ )
- (B): 3 are divisible by 2 but not by 4 or 3 (these are $22,46,50$ )
- (C): 1 is divisible by 4 but not by 3 (this is 8 )
- (D): 4 are divisible by 3 but not by 2 (these are $21,27,33,39$ )
- (E): 1 is divisible by 2 and by 3 (this is 30 )
We work through two possibilities: either 30 is chosen or 30 is not chosen. We do not need to worry about whether 30 is $a, b$ or $c$, since the three chosen integers can be arranged in increasing order after they are chosen.
Case 1: 30 is chosen
In this case, at least one of the remaining integers is even, since 30 includes only 1 factor of 2 .
If both of the remaining integers are even, then 2 of the 4 integers from (B) and (C) are chosen. There are 6 ways to do this: 22 and $46 ; 22$ and $50 ; 22$ and $8 ; 46$ and $50 ; 46$ and $8 ; 50$ and 8 . Therefore, there are 6 ways to choose in this sub-case.
If only one of the remaining integers is even, this even integer can come from either (B) or (C). There are 4 such integers.
The third integer chosen must be odd, and so is from (A) or (D). There is a total of 7 integers in these.
Thus, in this sub-case there are $4 \times 7=28$ ways to choose since each of the 4 even integers can be paired with each of the 7 odd integers.
Case 2: 30 is not chosen
In this case, one or two of the three integers chosen must be from (D), since $a b c$ must include at least 1 factor of 3 . (If all three were from (D), there would be no factors of 2.)
If two of the integers are from (D), there are 6 ways of choosing these 2 integers (21 and 27;21 and $33 ; 21$ and $39 ; 27$ and $33 ; 27$ and $39 ; 33$ and 39 ).
The third integer chosen must then be a multiple of 4 in order for $a b c$ to have at least 2 factors of 2 . Thus, the third integer is 8 .
Thus, in this sub-case there are 6 ways to choose.
If one of the integers is from (D), there are 4 ways of choosing that integer.
The remaining two integers chosen then need to include a multiple of 4 and another even integer, or a multiple of 4 and an odd integer not divisible by 3 , or two even integers that are
not multiples of 4. (Can you see why these are all of the possible cases?)
If these integers are a multiple of 4 and another even integer, the multiple of 4 must be 8 and the even integer comes from the 3 integers in (B).
Thus, there are $4 \times 3=12$ ways to choose in this sub-case, since each of the 4 integers from (D) can be paired with each of the 3 integers from (B).

If the two integers are a multiple of 4 and an odd integer not divisible by 3 , the multiple of 4 must be 8 and the odd integer comes from the 3 integers in (A).
Thus, there are $4 \times 3=12$ ways to choose in this sub-case.
If the two integers are even and not multiples of 4 , we choose 2 of the 3 integers from (B) and there are 3 ways to do this, as we saw earlier.
Thus, there are $4 \times 3=12$ ways to choose in this sub-case.
In total, there are $6+28+6+12+12+12=76$ ways to choose the three integers.
ANSWER: 76