HW11 - HW11 Solutions Dilip Raghavan December 2, 2008 1...

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HW11 Solutions Dilip Raghavan December 2, 2008 1 Appendix A.4 Problem 8. Let us arrange the numbers in strictly increasing order (we can do this since they are all distinct). So we have 1 k 1 < k 2 < · · · < k n + 1 2 n . Now if the di erence between k i + 1 and k i is always at least 2, then the di erence between k 1 and k n + 1 is at least 2 n . But the di erence between two numbers laying between 1 and 2 n is at most 2 n - 1. So the di erence between some k i and k i + 1 must be 1, which means they are consecutive. Problem 11. For each number from 1 to 14 consider the highest power of 2 dividing it. The part that is left over must be an odd number between 1 and 14, of which there are only 7. That is, suppose that the 8 numbers given to us are n 1 , . . . , n 8 . For each i between 1 and 8 write n i = 2 m i k i , where 2 m i is the highest power of 2 dividing n i , and k i is an odd number. Notice that k i is between 1 and 14. There are only 7 odd numbers between 1 and 14. So there are i , j such that k i = k j . Now the one with the smaller m divides the
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HW11 - HW11 Solutions Dilip Raghavan December 2, 2008 1...

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