# CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

## Round 3: Problem 5

with one comment

You are given 8 unit cubes such that 24 of the faces are painted blue and 24 of the faces are painted red. Prove that it is always possible to use these cubes to form a 2x2x2 cube that has the same number of blue and red unit squares on its surface.

Submissions for round 3 are due by Sunday, April 12 at 11:59 p.m. Solution will be posted (if we solve it) immediately after submissions are over.

March 30, 2009 at 00:55

Posted in Round 3

## Round 3: Problem 4

Each CUNY Math Challenge participant is allocated a 50-digit identification number. The I.D. number must start with a 1 and only use the digits 0 and 1. Moreover, it may not contain two consecutive 1s or three consecutive 0s. How many different identification numbers that meet these requirements are possible?

Submissions for round 3 are due by Sunday, April 12 at 11:59 p.m. Solution will be posted (if we solve it) immediately after submissions are over.

March 30, 2009 at 00:54

Posted in Round 3

## Round 3: Problem 3

Castle

The interior of a castle has the shape shown in the diagram to the right and consists of 45 square-shaped rooms.

There are doors between every two rooms that share a common wall. A tourist starts from one of the rooms and desires to visit as many of the castle’s rooms as possible so that he returns to his starting room but does not visit any (other) room more than once. What’s the largest number of rooms that he can visit?

Submissions for round 3 are due by Sunday, April 12 at 11:59 p.m. Solution will be posted (if we solve it) immediately after submissions are over.

March 30, 2009 at 00:54

Posted in Round 3

## Round 3: Problem 2

A set of distinct positive integers is called “sweet” if it has the property that, for every 3 distinct elements—a,b and c—of the set, a+b+c is always prime. What is the largest size a “sweet” set can have?

Submissions for round 3 are due by Sunday, April 12 at 11:59 p.m. Solution will be posted (if we solve it) immediately after submissions are over.

March 30, 2009 at 00:51

Posted in Round 3

## Round 3: Problem 1

Two CUNY students (over the age of 21) are having an all-day study party. One student brings 3 bottles of beer and another brings 5 bottles of beer. A third student joins them later and they all enjoy the beer. When the “studying” ends, the one who came late, wanting to contribute his fair share of the total cost of the party, leaves $8. How should the$8 be divided up fairly amongst the two students who brought the beer?

Submissions for round 3 are due by Sunday, April 12 at 11:59 p.m. Solution will be posted (if we solve it) immediately after submissions are over.

March 30, 2009 at 00:50

Posted in Round 3

## Round 3 is (apparently) open.

Well, hot off the heels of round 2 solutions (just posted), apparently Round 3 is already up. I’ll post the questions in just a tick. Stay tuned!

March 30, 2009 at 00:46

Posted in Round 3

## Round 2: Problem 5 Solution

I spent a tiny bit of time on this problem, because the solution did not come immediately to mind.

Ultimately, realizing that being that the angle of incidence equals the angle of reflection helped to come up with a bit of magic that helps visualize the path. Because the angles are equal, we can unwind the path that the puck takes into a straight line by “mirroring” the rink. This will allow us to solve the problem with just a tiny bit of high school trigonometry.

Unwound Path

Essentially what we have done is mapped the bouncing path into a straight one. For the sake of rigorousness we could show that this straight line is a bijective mapping of a bouncing path, but I will assume that it is trivial for our purposes and trust that if you don’t believe me that you trace it out on one of the unused boxes.

This path represents a trajectory that bounces off each of the four walls and arrives in the “goal.” To solve the problem, all we need is the angle $\alpha$. We know that $b=58m$ (because both the puck and goal are 1m away from the walls) and we also know that $h=5+5+10=20m$, therefore $\angle \alpha =\text{tan}^{-1}\left[\frac{20}{58}\right]\approx .3321\text{ rad}$, or since the problem asks for degrees, $19.0256{}^{\circ}$.

What is interesting is that we can modify this method to work with an arbitrary amount of bounces and end up at any point along the rink. Check out some further reading on the topic of Dynamical Billiards.