Posted by risa on Monday, November 7, 2011 at 5:04pm.
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 = 119-88 = 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=119-88=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=22-1-14) -> 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:
worms-w-9
bugs-b-10
butterflies-f-3
11b+1w
or...
11b+3f
11b=110
1w=9
3f=9
110+9=119
SOLVED!