Question 487277: Пlease help me with those questions!!!
1. Find all triples (x; y; z) of positive integers such that x < y < z and
(1/x)-(1/xy)-(1/xyz)=19/9
2. Show that from any five integers, one can always choose three of these integers such that their sum is divisible by 3.
Thank you very much!!!
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! 1. There are none. Since x < y < z, the maximum value of the LHS is about 1 (set x = 1, y,z to be large numbers). The LHS can never get larger than 1, so there are no solutions (x,y,z) of positive integers.
2. Out of any five integers, there will either exist an integer for each residue modulo 3 (in which we're done), or all five integers will occupy two different residues mod 3. If this is the case, then by a simple Pigeonhole argument, three of them have the same residue mod 3, in which we're also done (since the sum of these three is 0 mod 3).
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