Problem 15 Solution

The answer is 22.

Howie needs 2 white socks and 2 black socks. So let's say he was very unlucky. He takes out all his Halloween socks(8), then he takes out all his white socks(12), and then he needs 2 black socks.

so the answer is 8+12+2=22
If you did it the other way around it would be 8+9+2=17

--Harry

Problem 15: Socks

(Final Problem of the school year 2007-2008)
Howie is leaving for a trip at 3:00 A.M. At the last minute he remembers that he forgot to pack one pair of white socks and one pair of black socks for his trip. In order not to wake his brother, who shares the same room with him, he decides not to turn on the light in the room (and therefore, he cannot see anything) and simply reach into the sock drawer and take as many socks as he needs with him. The drawer contains 12 identical individual white socks, 9 identical individual black socks, and 8 individual Halloween socks, all mixed together. What is the least number of socks he must take with him to guarantee that he has a pair of white socks and a pair of black socks? An answer must be accompanied by an explanation.
(Due June 13, 2008)

Problem 14 Solution:

Let’s list a few possible paths, using N to indicate walking one block North and E to indicate walking one block East:

NNNEEEEEEE
NNEEEEEEEN
NENENEEEEE
EEEEEEENNN
… and so on.

This problem then becomes the number of ways of arranging this 10-letter word, or the number of ways of positioning the letter N out of 10 possible spots in a row, which is 10C3= 120.

Problem 14: Shortest Paths

David’s house is at the corner of 9th Avenue and 14th Street. Every morning he walks to the school which is located at the corner of 2nd Avenue and 17th Street. He wants to take the shortest path to school, but a different path each day. For example, one day he walked 3 blocks north and 7 blocks east, and on another day, he walked 2 blocks east, 2 blocks north, 5 blocks east, and 1 block north. How many different shortest paths are possible?
(Due May 30, 2008)

Problem 13 Solution

Here’s an algebraic solution:

We can write five equations, one for each row, as follows:

M+1+4+G=28
M+3+5+I=28
A+1+3+C=28
A+4+2+I=28
C+5+2+G=28

Now if we add these five equations together, we get:

2M + 2A + 2G + 2I + 2C + 30 = 140
2(M + A + G + I + C) = 110
M + A + G + I + C = 55

Therefore the answer is 55. (Note that I didn’t need to find out what each variable represents)

Problem 13: Five Point Star

In the accompanying diagram, each row adds up to 28. Find the values for the letters in the word MAGIC. (Due May 9, 2008)

Problem 12 solution:

Thalia R. Hoyi L., Eric P. and Christopher P. all submitted the same answer, 2. Although it is the correct answer, no one has an acceptable solution. Please see the display case outside the main office for their solutions.

Here is one way to reason this out:

Let’s focus on Hillary first. Hillary either shook four hands, or she did not. If she shook four hands, then it would not be possible for anyone to report shaking 0 hands. Therefore Hillary did not shake four hands. If Hillary did not shake four hands, then one of the remaining four people did. Let’s call this person A, and let S(A) represent the “spouse of A”. Let’s also define the second couple as B and S(B). If A shook four hands, then they must be with Bill, Hillary, B, and S(B) since A could not shake hands with his spouse. This also means S(A) shook no hands. We now need to figure out who shook 1, 2, and 3 hands. If Hillary shook only one hand (that would be A), then neither B nor S(B) could shake 3 hands since the only person remaining to shake hands with them is Bill. So Hillary did not shake hands just once. Without lost of generality, let’s assume B shook hands once, then the only possible placement for 2 and 3 hand shakes would be for S(B) to shake hands with Hillary and Bill and for Hillary to have 2 hand shakes (that would be with S(B) and A).