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December 18, 2014

December 18, 2014

Posted by **risa** on Monday, November 7, 2011 at 5:04pm.

- math -
**Steve**, Monday, November 7, 2011 at 7:27pmw=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

- math -
**Nic**, Monday, September 10, 2012 at 11:13pmGood 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.

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