A bunch of apples
A teacher must divide 221 apples evenly among 403 students. What is the minimal number of pieces into which she must cut the apples? (A whole uncut apple is one piece.)
This problem is not original.
The answer is 611.
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1 solution
@Guiseppi Butel I have added your solution. If you subscribe to this thread, you will receive notifications when people comment.
Great solution, Guiseppi; I'm glad that Calvin was able to get it posted for you. I ended up using the same process once I realized that the pieces didn't all have to be the same size. I'm not sure if there is any way of rigorously proving that this is indeed the minimum, but at this point I've spent enough time on this question that I'm not going to worry about it anymore. This is such an efficient process I feel that it might already have a name attached to it, but for now I'll just call it Guiseppi's Algorithm. :)
I'm comfortable now to reshare the question; I think over time it may move up to a level 5 problem.
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Yes there is to prove this, even in the general case.
As a hint, the answer is 4 0 3 + 2 2 1 − g cd ( 4 0 3 , 2 2 1 ) .
Think about what it means for each student to receive 4 0 3 2 2 1 of an apple.
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Brilliant formula, Calvin. It works for this problem and the new one that I posted, however, give 7 apples to 18 people a try.
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@Guiseppi Butel – I find that it works for this scenario too. Divide each apple into slices of sizes 1 8 7 , 1 8 7 , 1 8 4 . Distribute the 1 4 larger slices to 1 4 people, then slice 3 of the size 1 8 4 slices into 1 8 3 , 1 8 1 sizes. Then give 3 people slices of sizes 1 8 4 , 1 8 3 and the remaining person slices of sizes 1 8 4 , 1 8 1 , 1 8 1 , 1 8 1 . .
The total number of pieces is 1 4 + 4 + 3 ∗ 2 = 2 4 , just as the formula would indicate. Did you find a more efficient approach for the 7/18 scenario?
I'm still working on a proof of Calvin's formula.
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@Brian Charlesworth – Thanks, Charles, for this enlightenment. I was getting ready to show that 7/18 scenario resulted in 26 pieces. I had cut 2 of the 4/18's cut into 2 2/18's each and 1 of the 4/18's cut into 4 1/18's. Thanks for your timely note.
@Brian Charlesworth – There is a 1 line proof.
Hint: Think about lining up everything in a line. Where must cuts occur?
@Guiseppi Butel – CAN you give a link to the problem by you, sir ?
What if we take lcm(221, 403) to get 6851 pieces. So 2 2 1 6 8 5 1 would give us 31 pieces, is it correct??
Sir, could you please give a little explanation to above solution...Didn't get it.:(
[Calvin's edit: This solution only provides a construction for 611 pieces. It does not show that 611 is indeed the minimum.]