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Math 194, problem set #1, SOLUTIONS
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1. Show that there is no polynomial P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 of degree
at least one, with integer coefficients ai , such that P (k) is prime for every nonnegative
integer k.
(Andreescu & Gelca)
Solution: Suppose P (x) is such a polynomial. Thus a0 = P (0) is prime, and by
plugging into our expression for P (x), we see that P (ma0 ) is a multiple of a0 for every
nonnegative integer m. By our hypothesis, P (ma0 ) is also prime for every nonnegative
integer m, so we have that P (ma0 ) = ±a0 for all m. Since either P (x)−a0 or P (x)+a0
has infinitely many zeros, P (x) = a0 , a contradiction.
2. Suppose n objects are arranged in a row. A subset of these objects is called unfriendly
if no two of its elements are consecutive.
Show that the number of unfriendly subsets
n−k+1
each having k elements is
.
(Larson, 1.3.12)
k
Solution: We proceed by induction on n, at each step proving the result for all k (notice
that for k < 0 and k > n, the number of unfriendly subsets and n−k+1
are both zero).
k
For n = k = 0, we see that ∅ has only itself as an unfriendly subset, and so 1 = 0−0+1
0
holds. For n = 1, we see that
a set with one element has only itself and ∅ as unfriendly
subsets, and so 1 = 1−1+1
and 1 = 0−0+1
hold. For n ≥ 2, assume that the number
1
0
of unfriendly subsets having k elements is N −k+1
for all N < n. Without loss of
k
generality, let our n objects arranged in a row be Sn = {1, 2, . . . , n}. An unfriendly
subset U ⊂ Sn either contains n or it does not. If n ∈ U , then U = {n} ∪ U 0 for
0
some unfriendly
of size k − 1; by the induction hypothesis, there are
subset
U ⊂ Sn−2
(n−2)−(k−1)+1
n−k
0
=
such
U
.
If
n 6∈ U , then U = U 00 for some unfriendly subset
k−1
k−1
n−k
U 00 ⊂ Sn−1 of size k; by the induction hypothesis, there are (n−1)−k+1
=
such
k
k
n−k
n−k
n−k+1
00
U . Thus, the number of unfriendly subsets of Sn is k−1 + k =
.
k
3. On an arbitrarily large chessboard, a generalized knight moves by jumping p squares
in one direction and q squares in a perpendicular direction, p, q > 0. Show that such
a knight can return to its original position only after an even number of moves.
(German Mathematical Olympiad)
Solution: Case 1: p + q is odd: Let f be the function that associates to each square
the sum of its x and y coordinates. Then f changes parity with each step, so it takes
an even number of steps to return.
Case 2: p, q both odd: Let f be the function that associates to each square its xcoordinate. Then f changes parity with each step, so it takes an even number of steps
to return.
Case 3: p, q both even: Scale by power of 2, i.e., replace p by p/2k and q by q/2k for
largest possible k. This reduces to previous cases.
4. Use the graph of y = sin(x) to show the following. Given a triangle with angles A, B, C,
(a) sin(B)+sin(C)
,
≤ sin B+C
2
2
(b) z sin(B) + (1 − z) sin(C) ≤ sin(zB + (1 − z)C), if 0 ≤ z ≤ 1.
(Larson, 1.3.12)
Solution: Without loss of generality, assume 0 ≤ B ≤ C ≤ π since A, B, and C are
the angles of a triangle. A proper solution should include a graph of y = sin(x) on
[0, π], the two points (B, sin(B)) and (C, sin(C)), and the secant line between those
two points.
(a): At x = (B + C)/2, sin(B)+sin(C)
is the height of the secant line, while sin B+C
2
2
is the height of y = sin(x). Since y = sin(x) is concave down here, we must have
sin(B)+sin(C)
B+C
≤
sin
.
2
2
(b): At x = zB + (1 − z)C, z sin(B) + (1 − z) sin(C) is the height of the secant line,
while sin(zB + (1 − z)C) is the height of y = sin(x). Since y = sin(x) is concave down
here, we must have z sin(B) + (1 − z) sin(C) ≤ sin(zB + (1 − z)C).
5. Every point of 3-dimensional space is colored red, green, or blue. Prove that one of the
colors attains all distances, meaning that for every positive real number d, two points
of that color are exactly distance d apart.
(German Mathematical Olympiad)
Solution: Suppose that none of the colors attain all distances. Thus, there exist r, g, b ∈
R+ such that no two red points are at distance r, no two blue points are at distance b,
and no two green points are at distance g. Without loss of generality, assume r ≥ b ≥ g.
Let x ∈ R3 be a red point if there is one, and an arbitrary point otherwise. Then the
sphere of radius r around x has only blue and green points. Let y be a blue point on
this sphere if there is one, or an arbitrary point on the sphere otherwise. Then the
sphere of radius b around y has only red and green points. Since b ≤ r, the intersection
of our two spheres is a circle of green points. This circle must have diameter greater
than b, and so green attains all distances up through b, including g, a contradiction.
6. Let a1 , a2 , . . . , an denote an arbitrary ordering of the numbers 1, 2, . . . , n. Prove that
if n is odd, then the product (a1 − 1)(a2 − 2) · · · (an − n) is even.
(Larson, 1.10.8)
Solution: (n+1)/2 of the ai ’s are odd, and (n+1)/2 of {1, . . . , n} are odd. Distributing
them among n factors, an odd ai and an odd i must go into same factor by PHP. Then
(ai − i) is even.
7. Show that no set of 9 consecutive integers can be partitioned into two subsets such
that the product of the elements in the first set is equal to the product of the elements
in the second set.
(Andreescu & Gelca)
Solution: Suppose that C = {s, s + 1, . . . , s + 8} can beQ
partitionedQinto two subsets
A = {a1 , a2 , . . . , aj } and B = {b1 , b2 , . . . , bk } such that jr=1 ar = kt=1 bt . Without
loss of generality, assume j > k. Notice that C cannot contain all negative elements,
since we would then have a product of an odd number of negative numbers equal to
the product of an even number of negative numbers. Notice also that C cannot contain
0, since we would then have 0 equal to a product of nonzero numbers. Thus s > 0.
Now observe that for any prime p, there cannot be only one multiple of p in C; if there
were, we would have a multiple of p equal to a non-multiple of p. Thus, considering
the prime p = 7, s 6∈ [1, 5]. Considering the prime p = 11, s 6∈ [3, 11]. We now have
s ≥ 12. Consider
Qj
ar
s(s + 1)(s + 2)(s + 3)(s + 4)
s
≥
= (s+5)
· (s+1)
· (s+2) · (s+3) · (s + 4)
Qr=1
(s+6) (s+7) (s+8)
k
(s + 5)(s + 6)(s + 7)(s + 8)
t=1 bt
≥ 12
· 13 · 14 · 15 · 16 > 1,
17 18 19 20
so
Qj
r=1
ar 6=
Qk
t=1 bt ,
a contradiction.
8. Find, with explanation, the maximum value of f (x) = x3 − 3x on the set of all real
numbers x satisfying x4 + 36 ≤ 13x2 .
(Putnam, 1986)
Solution: x4 + 36 ≤ 13x2 ⇐⇒ (x2 − 9)(x2 − 4) ≤ 0 ⇐⇒ |x| ∈ [2, 3]. f (−3) =
−18, f (−2) = −2, f (2) = 2, f (3) = 18. The critical points of f are x = ±1. So the
maximum is 18.
9. Define a selfish set to a set which has its own cardinality (number of elements) as
an element. Find, with proof, the number of subsets of 1, 2, . . . , n which are minimal
selfish sets, that is, selfish sets none of whose proper subsets is selfish. (Putnam, 1996)
Solution: A set is minimal selfish if and only if its order is its
least element. Let ni
be the number of selfish sets of cardinality i. Then ni = n−i
. If f (n) = number of
P n−i i−1
minimal selfish subsets of {1, . . . , n} then f (n) = i i−1 . Check: f (1) = f (2) = 1
and f (n + 1) = f (n) + f (n − 1). Thus f (n) = Fn . (Or check the recursion directly,
since the m.s. subsets of {1, . . . , n + 1} not containing n + 1 are the m.s. subsets of
{1, . . . , n}, and the ones containing n + 1 are the m.s. subsets of {1, . . . , n − 1} shifted
up by 1, together with n + 1.)