Download Math 6b HW 1 Solutions

Math 6b HW 1 Solutions
Winter 2010
Problem 1.
Given a set S of n distinct integers, there exists a non-empty subset
A of S such that the sum of the integers in A is divisible by n.
Solution. Consider ak =
k
X
sj . If one of the ak ’s is divisible by n, then
j=1
we are done. If no ak is divisible by n, then by Pigeon-hole Principle,
there exist k < l such that ak = al . By considering al − ak , we are also
done.
Problem 2.
Show that there is a positive term of the Fibonacci sequence that is
divisible by 10000.
Solution. Let fn be the n-th term in the Fibonacci sequence, f0 = 0,
f1 = 1, and let f¯n be the minimum non-negative integer satisfying
fn ≡ f¯n (mod 10000). Note that there are at most 100002 choices
for (f¯n , f¯n+1 ). By Pigeon-hold Principle, there exist n < m such that
(f¯n , f¯n+1 ) = (f¯m , f¯m+1 ). Then as we can work from f¯n and f¯n+1 back to
f¯0 = 0, we can also work from f¯m and f¯m+1 back to f¯m−n = 0, meaning
that fm−n is divisible by 10000.
Problem 3.
(Problem 11, Chapter 4) Suppose n > srp then prove that any sequence of n real numbers must contain a strictly increasing subsequence
of length s + 1, or strictly decreasing subsequence of length r + 1 or a
constant subsequence of length p + 1.
Solution. Assume that there does not exist a constant subsequence of
length p + 1. Then if we partition the n numbers into minimum number of constant subsequences and pick one representative from each of
them, we have a subsequence of m distinct real numbers, where m > sr.
We can form a strictly increasing subsequence by taking the first entry,
then going down the sequence and appending a number to the subsequence if it is larger than the last entry of our subsequence. Now,
assume that in this subsequence of m numbers, there does not exist a
strictly increasing subsequence of length s + 1. If we partition the m
1
2
numbers into strictly increasing subsequences formed in the above way,
then we have at least r + 1 of them, and the last entries of these strictly
increasing subsequences will form a strictly decreasing subsequence.
Problem 4*.
(Problem 7, Chapter 4) The Independence N umber of a graph G,
denoted α(G), is the maximum number of pairwise disjoint vertices in
G.
Prove the following dual version of Turán’s theorem: If G has n vertices
n
and nk/2 edges (k ≥ 1), then α(G) ≥ k+1
.
Solution. Let n = (k + 1)q
+ r. If G has
Ḡ has
n(n−k−1)
2
≥ 1−
1
q
n2
2
nk
2
edges, then the complement
edges, where equality holds iff r = 0. By
Turán’s theorem, Ḡ can avoid Kq+1 but not Kq if r = 0, and Ḡ cannot
avoid Kq+1 if r 6= 0. Hence, α(G) ≥ q if r = 0 and α(G) ≥ q + 1 if
n
r 6= 0. In other words, α(G) ≥ d k+1
e.
Problem 5.
Let X be a subset of the positive integers with the property that the
sum of any two not necessarily distinct elements in X is again in X.
Suppose that {a1 , a2 , . . . , an } is the set of all positive integers not in
X. Prove that a1 + a2 + · · · + an ≤ n2 .
Solution. We can prove it by induction. WLOG, assume that ai ’s
are arranged in increasing order. When n = 1, it is trivial. Now, if
a1 + a2 + · · · + an ≤ n2 while a1 + a2 + · · · + an+1 > (n + 1)2 for some
n, then an+1 > 2n + 1. Since an+1 is missed in X, we have to miss
one term from each of the following pairs: {1, an+1 − 1}, {2, an+1 −
2}, . . . , {b an+1
c, d an+1
e}. As an+1 > 2n + 1, there are at least n + 1
2
2
pairs, contradicting that we only have n terms smaller than an+1 .