well, the only possibilities for 3 kids, whose age's product is 72 is:

72*1*1
36*2*1
24*3*1
18*4*1
18*2*2
12*6*1
12*3*2
9*4*2
9*8*1
8*3*3
6*6*2
6*4*3


but then the bartender says to check the address. the guy does, but it's still inconclusive. this means that there are two possible combinations with the same sum - if not, the guy could've figured it out.

thus, the remaining combos are

8*3*3
6*6*2

both with a sum of 14


but then he says his YOUNGEST loves ice cream. obviously, the youngest does not have a twin. the 8*3*3 combo is not possible.

you're left with 6*6*2



it's still confusing, how come you don't pick
9*4*2
9*8*1
6*4*3

 

yeah, what about that?

 

I got a good hard brain teaser, I'll post it in a couple minutes

 

(If I can remember it )

 

george, the reason the combos below don't work

9*4*2 (15)
9*8*1 (18)
6*4*3 (13)

is that they have different sums.

here's the deal... here are the possible combos, and their respective sum of the digits:

72*1*1 (74)
36*2*1 (39)
24*3*1 (28)
18*4*1 (23)
18*2*2 (22)
12*6*1 (19)
12*3*2 (17)
9*4*2 (15)
9*8*1 (18)
8*3*3 (14)
6*6*2 (14)
6*4*3 (13)

now, when the guy goes out to look at the address of the bar, he still can NOT decide which age combo it is. the only reason would be if there are two possible combos (if there are 2 combos with identical sums). it's obvious the address is 14. if the address were 15, for example, it'd obviously be the 9*4*2 combo, because that's the only combo with the sum of 15.

but because the guy could still not decide after seeing the address, we have to assume that there were two possibilities. that in and of itself disqualifies all other combos, except 8*3*3 and 6*6*2.



from there, the 8*3*3 combo is disqualified because he said the youngest likes the ice cream. youngest is singular, and thus can not be a twin. thus, you're left with the youngest being the 2-year-old in the 6*6*2 combo

 

OHHH

 

(8*3)^2/3