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
ff
erence between
k
i
+
1
and
k
i
is always at least
2, then the di
ff
erence between
k
1
and
k
n
+
1
is at least 2
n
. But the di
ff
erence between two numbers laying
between 1 and 2
n
is at most 2
n

1. So the di
ff
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
other one. That is, if
m
i
≤
m
j
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 Spring '11
 Hernandez/Rag
 Addition, 2mi

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