math
posted by risa on .
Mia is approaching her goal of 119 points for the video game. She wants to put her initials on the computerized list of Top Ten Point Scorers forr the game, She moves Freddie the frog all over the terminal screen and at last she does it. Freddie the Frog eats a total of 22 critters: worm ( worth 6 points each), bugs ( worth 7 points each). and butterfies ( worth 4 points each). if Mia earned the total of 119 points, how many critters of each type did her frog eat?

w=worms, g=bugs, f=butterflies
count critters: w+g+f = 22
count points: 6w+7g+4f = 119
6w+7g+4f = 4(w+g+f) + 2w + 5f
4(22) + 2w + 5g = 119
2w + 5g = 11988 = 31
Now, we don't know how many w,g we have, but since 2w is even, that means 5g must be odd, which means that g must be odd.
Take a look at possibilities:
g=1: 2w+5 = 31 ==> w=13 f=8
g=3: 2w+15 = 31 ==> w=8 f=11
g=5: 2w+25 = 31 ==> w=3 f=14
Those are the only possible solutions 
Good solution, Steve, but w+b+f=6+7+4=17.
worms (w)  6
bugs (b)  7
butterflies (f) 4
critters: w+b+f=22
points: 6w+7b+4f=119 (the only way to get an odd number is if b is odd)
points from having one of each of the critters: w+b+f > 6+7+4=17
so, using the points now:
6w+7b+4f=119
Break apart each set of critters into 4+something:
4w+2w+4b+3b+4f=119
Which means 4w+4b+4f+2w+3b=119
But 4w+4b+4f=4(w+b+f)=4(22) – (this is when she has equal amounts of each critter)
So 4(22)+2w+3b=119
2w+3b=11988=31
2w+3b=31 (which can only be odd if b is odd, so try values for b)
b=1, 2w+3=31, w=14, f=7 (f=22114) > w=14, b=1, f=7
b=3, 2w+9=31, w=11, f=8 > w=11, b=3, f=8
b=5, 2w+15=31, w=8, f=9 > w=8, b=5, f=9
b=7, 2w+21=31, w=5, f=10 > w=5, b=7, f=10
b=9, 2w+27=31, w=2, f=11 > w=2, b=9, f=11
If b is larger than 9, 2w+3b will be larger than 31, so no more solutions. 
Nic, your solution is a bit confusing as it contains 6+4+7. instead, try this:
wormsw9
bugsb10
butterfliesf3
11b+1w
or...
11b+3f
11b=110
1w=9
3f=9
110+9=119
SOLVED!