Solution

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Proof

Let A be the given subset of the quasigroup G. Let x be an element of G. In the multiplication table of G every element appears exactly once in every row and column. Let f(x) be the number of times that x appears in the table as the product of elements of A. Then, in the subtable consisting of those rows labelled by an element of A and the columns not labelled by elements of A, the element x appears |A| - f(x) times. Hence, in the subtable consisting of those rows not labelled by elements of A and columns not labelled by elements of A, the element x appears |G| - |A| - (|A| - f(x)) = |G| - 2|A| + f(x)≥0 times. Since |G| - 2|A| < 0, it follows that f(x) > 0; so x is the product of two elements of A (perhaps the square of an element of A).

Dear Alexander Bogomolny,

In the problem: "Given any 6 integers from 1 to 10, some two of them have an odd sum" there is a much simpler solution:

To start with, two integer numbers have an odd sum if and only if one of them is odd and the other is even. From 1 to 10, there are 5 odd and 5 even integers. So in 6 numbers taken from 1 to 10 there is at least one odd and one even number, because of the pigeonhole principle. So there is at least one pair of numbers that have an odd sum.

I really enjoy "cut-the-knot"
Yours,
Odysseus Galanis

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