
Darlene N. answered 06/14/14
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Experienced Math Teacher and Doctoral Candidate in Math Education
If I'm reading this correctly and the guys are throwing out one of the four parts they divided the chocolates into, the answer is 32 chocolates.
For the thieves to divide them equally, there had to be 4 at the end.
When the fourth guy woke, there were 4. Since the third guy had divided the previous amount into fourths, ate one fourth and threw one fourth out, what remained was half of what he started with. So the third guy started with 8.
When the third guy woke, there were 8. Since the second guy had divided the previous amount into fourths, ate one fourth and threw one fourth out, what remained was half of what he started with. So the second guy started with 16.
When the second guy woke, there were 16. Since the first guy had divided the previous amount into fourths, ate one fourth and threw one fourth out, what remained was half of what he started with. So the first guy started with 32.
Working it backward to verify...
- When the first thief woke, there were 32 chocolates. He split it into four parts (8 in each part), ate one part (-8) and threw one part out (-8), leaving 16 for the second guy.
- Second guy woke and broke the remaining into four parts (4 per part), ate one part (-4) and threw one part out (-4), leaving 8 for the third guy.
- Third guy woke and broke the remaining into four parts (2 per parts), ate one part (-2) and threw one part out (-2), leaving 4 for the fourth guy.


Darlene N.
You're right about that assumption...I did assume whole chocolates. I also assumed they want the lowest possible answer using whole chocolates. Good points.
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06/14/14
Talha K.
Can you tell me What is lowest possibility?
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06/14/14

Darlene N.
Tahla, assuming they keep the chocolates whole, it is 32. Mathematically, the formula is 2n+1, where n is the number of thieves.
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06/14/14
Bob A.
06/14/14