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Show that among any seven distinct positive integers not greater than 126 one can find two of them, say x and y, satisfying the inequalities $1 < \dfrac y x \le 2.$

This is what I have done

Notice that $ 126/7 = 18$, therefore the maximum spread of all the numbers is $1, 19,37,55,73,91,109$. Therefore we are trying to solve the inequality $1<(x+18)/x<2$

therefore $x>18$ Hence whenever $x>18$ the inequality holds, however when $x<18$ the maximum range between the numbers is $18/7$ which is approximately equal to $3$.

Therefore we are solving the inequality $1<(x+3)/x<2$. Trivially $x>3$ and if the gap between numbers is greater than three than there is at least one pair of numbers with a small enough range to satisfy the initial inequality.

This is what i came up with, if I am in the right direction please suggest improvements if not please indicate where I have gone wrong thanks!

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up vote 4 down vote accepted

I think you should work with constant ratios rather than constant increments.

Note that if the numbers are arranged in increasing order, to not satisfy the desired condition each $a_{n+1} > 2\cdot a_n$. If you carry this out seven times the contradiction should follow.

Actually, your problem statement must have a typo, since $1,2,4,8,16,32,64$ are 7 numbers not satisfying your inequality. Either you meant eight numbers, or (I suspect) you should have $1\lt \frac yx \le 2$ . Using the latter change, the smallest numbers at each increment would be $1,3,7,15,31,63,127$ .

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Yes there is an equal to thanks! – fosho Dec 27 '12 at 20:08

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