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  1. Teen kids are in the forest picking mushrooms. No one came home empty-handed, together they collected 54 mushrooms. Show that at least two children picked the same amount of mushrooms.

  2. Paul has a rectangular patio which is covered with concrete slabs. Some of the plates in the form of 2x2, other form of 1x4.One of the plates has been broken, but as a possible replacement Paul only have a single plate of the other form. Can Paul replace the broken plate with the one he has, or possibly rearrange the tiles?

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    $\begingroup$ A few things: ask one question per post, please share your own thoughts on the problem, and don't expect us to do your homework for you. $\endgroup$ – Sean Roberson Jul 11 '17 at 19:37
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    $\begingroup$ Look at the related questions on the side (->). Can you tell the difference between them based on their titles? If no, you see how the title you've chosen is terrible. $\endgroup$ – Trevor Gunn Jul 11 '17 at 19:40
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    $\begingroup$ See How to ask a good question?. $\endgroup$ – Trevor Gunn Jul 11 '17 at 19:41
  • $\begingroup$ No one came home empty handed. 11 to 19 pickers. That leaves 54-19=35 to 54-11 =43 mushrooms left after each had found one. Now to avoid that each had had the same start filling up more on each. How much would you need to "fill up" from everyones 1 to get a unique number for each? $\endgroup$ – mathreadler Jul 11 '17 at 19:43
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    $\begingroup$ Do you mean "ten" instead of "teen"? $\endgroup$ – user84413 Jul 11 '17 at 20:21
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If two of them did not pick the same amount of mushrooms, then they must have at least collected 1,2,3...,10 mushrooms, with one mushroom coresponding to one child. But that gives a total of at least 55 mushrooms, a contradiction.

For the second one, it cannot be done. This is because, suppose that the 2×2 tile has been broken. Then replacing it with one of the 1×4 tiles changes the parity of the dimension across which the 1 width tile is placed. In that case, we cannot compensate for this by rearrangement. Likewise vice versa.

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  • $\begingroup$ I think first needs some more explaining for most, but you are right. $\endgroup$ – mathreadler Jul 11 '17 at 19:45
  • $\begingroup$ Thank you . Actually, i have more doubts on the second although I am sure it is right. $\endgroup$ – астон вілла олоф мэллбэрг Jul 11 '17 at 19:49

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