Ice cream math problem

Hmm. One equation, 4 variables - there's obviously some sort of trick here. My instinct is that partial fractions need to be used here, but it's been so long since I've done basic calc I can't really work out how they would apply. I would also assume, intuitively, that at least one of the ice creams cost less than $1.

Wait, it's 3 equations - ugh, I need to not try to do math just after getting up 4 hours after the GSI finishes.

2 equations 4 variables, that means that there are infinite solutions

Since the values cannot have precision smaller than 0.01, there aren't that many answers.

My train of thought is to break down the number into it's factors and see if something can be done from there to get an answer.

The prime factors of 711 are 3, 3, and 79. However, it's possible that we're working with a number one magnitude higher, like 7110, in which case the factors are 2,3,3,5,79.

1.2
3.16
1.25
1.5

since we are using cents, we have to have solutions such that the numbers are some-what close to each other.

the cents make it so that we need factors of 300, 300, 79.

thus this gives us 2 2 2 2 3 3 5 5 5 5 79 to work with. thus, we combine these so that adding them up gives us 1 at the end. the hardest to work with is .79. we need a number that ends with .2 which is impossible here.
so we try to multiply it to make it bigger. we try 2* .79 which ends in a 8 which needs a 3 at the end, which is also impossible
thus we try 3*.79 which ends in a 7 which needs a 3 at the end which is impossible. thus we try 4*.79 which ends in a 6 which implies we need a 5 which is possible given our number mix.

the rest of the numbers follows easily. (u take out the 2 2 from the list and recompose the numbers to get dollar numbers)

I didn't get that part. Could you explain it?

I made two equations:
x+y+z+v = 711
xyzv = 711000000
factorizing 711000000 gives the following numbers:
2, 2, 2, 2, 2, 2, 3, 3, 5, 5, 5, 5, 5, 5, 79
I was too lazy to work out how to combine them so the sum fits. But somehow your way seemed to make that a little easier. Or did you just use that they must be somewhat evenly distributed among the four variables? But how did you get to 300?
K, I guess apart from that it would have been easy to find the right combinations. Damn laziness. :D