I've been told about this "Inclusion-Exclusion" but I'm too dumb to understand it on my own ;w;.. also it have scary greek letters on it.

Check how many one-colored packets you could make and then pick the lowest number? That seems reasonable to me but I don't know anything and nobody should ever listen to me (unless I'm correct and I know it, then I'd hate you for not listening and I'd be all like "But I tooold you so")

Actually, that's pretty much how I did it; I split it into three. "How many times does 3 go into 12?", "How many times does 2 go into 7?", and "How many times does 4 go into 11?" I then simply took the smallest result and used that.

If each packet MUST contain the amount on top, then you have to stop as soon as you run out of enough of one color to fill the packets.

The one that would run out first is green after two packets.

There are more specific formulas for this, but it's also just a matter of dividing each of the bottom numbers by the top numbers and then rounding all three results DOWN to the closest whole number and choosing the lowest number as your answer.

So if Red is R, Blue is B, and Green is G:

12R/3R = 4 -> 4
7B/2B = 3.5 -> 3
11G/4G = 2.75 -> 2 (answer)

THIS <3 TOO

You're thinking like this.

For each bag I need 3 red candies, 2 blue candies, and 4 green. I have 12 red candies, 7 blue ones, and 11 green ones.

Try it like this.

red 12/3
blue 7/2
green 11/4

Solve each one one at a time and the lowest answer is weight.

THIS

Thanks.

*Taur hugs*

The way I'd do this in code form would be to check all three colours. So how many packets could I make if I only consider red candies, how many only considering blue candies and how many considering green candies. So do the 12/3, 7/2, 11/4. Then I check which of these is the lowest number with a simple block of if statements. So If(red>blue) then (If(blue>green)then green else blue) else (If(red>green) then green else red).

Ooooh, ive not seen these problems in a while. It's limiting factor sets... xD just solve as everyone's already mentioned and ypure golden. sometimes it's a multi-step.process like...

"Each bagof candies is made of three red and five hundred blues. Each case is made of.five bags of candy. If you have sixty reds and 3000 blues, how many cases can you make?"

The answer is 1. First: (3*5=15) and (500*5=2500). Second: (60/15=4) and (3000/2500=1.2). Round down non-integers, and 1.2 becomes 1.

3 if you cheat and put a extra red in the last one to make up for the missing green

yea, thats actualy the best answer

... math = bane of all existence

but... this is simple (In my eyes) I have not clue what the real answer is, or whether this is correct but:

so if each MUST contain their respective numbers, you look at each one, and pull out part of one set. The one with the least amount you can pull out is the MAX amount of sets you can make, in this case, you can only pull 2 sets out of the greens, 11/4 ~ 2. there for, you can only make 2 sets, and have a partial one left over, with remaining blue and red.

alternate answer is 0

just eat them all and make none

3 packs, if you go by the blues

2 packs if by green

4 packs if by red

2 sets.

probably a simple value check, if you have the luxury of it, you can just do the check against it each time a single trade happens, until it no longer can, which would stop at 2 in this case. for an 'all in one go' coding, you could teach the code to handle it as a value in base 2, base3, or base4. and then have it check against the 'tens' value.

Agh no, chemistry is returning to me...!

Calculate the limiting reagent :P

No, no staaaahp! Do not make me calculate the salty tears (x NaCl- + y H20) ;__;

I have come up with the ultimate answer.

...Math is stupid. Go play video games!

huh... I never realised this before...