Week 7
November 3, 2015
1. At a strange party, each person knew exactly 22 others. For any pair
of people X and Y who knew one another, there was no other person
at the party that they both knew. For any pair of people X and Y
who did not know one another, there were exactly 6 other people that
they both knew. How many people were at the party?
2. So a student takes cosine rule a2 = b2 + c2 − 2bc.cosA and misapplies
the log function to get log(a2 ) = log(b2 ) + log(c2 ) − log(2bc.cosA). He
then correctly applies the rules of logarithms and exponentiates. The
result holds, shockingly enough. What can we say about the original
triangle?
3. So, take an 8x8 chessboard and chop off the opposite corners. Can
you then tile it with dominoes?
4. What about tiling it with n-ominoes?
5. Take an 8x8 chessboard. Tile with dominoes and show that the number
of horizontal dominoes whose left tile is on a black square equals the
number of horizontal dominoes whose right tile is on a black square.
6. Does this hold for a 2nx2n chessboard?
7. Find all polynomials such that f (x2 ) = f (x)2
8. Find 11 consectutive squares whose sum is also a square
9. Find all f : N → N such that
(a) f (2n) = 3f (n)
(b) f (n + 1) ≤ 3f (n)
10. In a tournament with N players, N < 10, each player plays once
against each other player scoring 1 point for a win and 0 points for
a loss. Draws do not occur. In a particular tournament only one
player ended with an odd number of points and was ranked fourth.
Determine whether or not this is possible. If so, how many wins did
the player have?
1
11. Prove that a triangle ABC is right-angled if and only if sin2 A+sin2 B+
sin2 C = 2
12. Find all f : R → R with f (1) = 1 satisfying f (xy+f (x)) = xf (y)+f (x)
13. Solve 2n = a! + b! + c! where n, a, b, c are positive integers!
14. Show that there is no perfect cube in base 10 of the form xyxy, where
xyxy are it’s digits
15. Is there any base b for which such a cube exists?
16. Four prisoners are arrested for a crime, but the jail is full and the jailor
has nowhere to put them. He eventually comes up with the solution
of giving them a puzzle so if they succeed they can go free but if they
fail they are executed. The jailor puts three of the men sitting in a
line. The fourth man is put behind a screen (or in a separate room).
He gives all four men party hats. The jailor explains that there are
two black hats, and two white hats; that each prisoner is wearing one
of the hats; and that each of the prisoners only see the hats in front
of him but not on himself or behind him. The fourth man behind the
screen can’t see or be seen by any other prisoner. No communication
between the prisoners is allowed. Can they all go free?
17. So if α is a root of x3 − x + 1 and β is a root of x3 − x + α, show there
is a polynomial with rational coefficients with β as a root.
18. Show a2 + b2 + c2 ≥ ab + bc + ca for a, b, c positive reals.
19. So we take a circle of radius 3 at (0, 0) and a circle of radius 1 at (5, 0)
and stick a bug on each circle, one at (3, 0) and one at (6, 0). Each
bug runs around their respective circles at 1 rad/s. Find the locus of
the midpoint of their positions as a function of time.
R1
20. Compute −1 xcos(xcos(xcos(x))) dx
2