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).

Problem 12: Handshakes

Bill and Hillary go to dinner with two other couples. Some handshakes are exchanged, but no one shakes hands with their spouse, with themselves, or with anyone else more than once. When Bill asked each person how many handshakes they had exchanged, they reported 0, 1, 2, 3, and 4. How many handshakes had Hillary exchanged?
(Due April 18, 2008)

Problem 11 winners are ...

Christopher P. and Ho Yi L.

Congratulations!!

Problem 11: Easter Egg Hunt

Ms. Jabbour's rectangular yard is to be used for the Easter egg hunt this year. The yard is 30 feet wide, 40 feet long and has two 3-foot wide cement paths as shown in the diagram. Her dog Franky is tied to one corner with a 6-foot long leash (assume that Franky cannot reach the cement paths). What is the maximum possible area in which the eggs can be placed if they cannot be placed on the cement walkways or where Franky can reach? (Due March 28, 2008)

Problem 10 winner and solution

Congratulations to Manjit from the main office for submitting a correct solution!

Solution:
Let's first list all possible sets of three cards with different numbers in increasing order having a sum of 13:

(1, 2, 10)(1, 3, 9)(1, 4, 8)(1, 5, 7)(2, 3, 8)(2, 4, 7)(2, 5, 6)(3, 4, 6)

When Alyssa looked at the card on the left but could not tell what the other two cards were, we knew she was not looking at “3” since 3-4-6 is a unique solution and she would know the other two cards in that case. So we can rule out the last case.

(1, 2, 10)(1, 3, 9)(1, 4, 8)(1, 5, 7)(2, 3, 8)(2, 4, 7)(2, 5, 6)

Brandon then looked at the rightmost card and said he could not tell what the other two cards were. By the same reasoning, we can rule out 1-2-10, 1-3-9, and 2-5-6 because each has a unique rightmost card.

(1, 4, 8)(1, 5, 7)(2, 3, 8)(2, 4, 7)

When Christina looked at the middle card and said she could not tell what the other two cards were, by the same reasoning, we can rule out 1-5-7 and 2-3-8 since 5 and 3 are unique numbers in the middle. That leaves us 4 as the only possible middle number. So the answer is 4.

(1, 4, 8)(2, 4, 7)

Problem 10: Three Cards

Three cards are drawn from a regular deck of 52 cards and are placed face-down on a table. Alyssa, Brandon, and Christina are told that

a. the numbers are all different (Ace = 1, Jack = 11, Queen = 12, King = 13)
b. the sum of the three numbers is 13, and
c. they are in increasing order, left to right

First Alyssa looks at the number on the leftmost card and says, “I don’t have enough information to determine the other two numbers.” Then Brandon looks at the number on the rightmost card and says, “I don’t have enough information to determine the other two numbers.” Finally, Christina looks at the number on the middle card and says, “I don’t have enough information to determine the other two numbers.” Assume that each person knows that the other two reason perfectly and hears their comments, what is the number on the middle card? (Due March 7, 2008)

Problem 9 Solution:

This is a variation of the famous Monty Hall problem or Monty Hall paradox. The original problem was based on the game show Let's Make a Deal where you are given a choice of three doors where behind one door is a new car, and behind the other two, goats. No matter which door you choose, the host, knowing what is behind each door, always opens one of the remaining doors that has the goat, and asks if you want to switch. Is it to your advantage to switch?

The answer to this problem is counterintuitive and has generated many heated debates. You can find more information about this interesting problem by searching for “Monty Hall” on the internet. Here is an explanation of why switching is better:

Below are the three possible scenarios:

Door1 Door2 Door3
Goat Goat Car
Goat Car Goat
Car Goat Goat

If the contestant picks door 3, and the host opens a door with the goat (either door 1 or 2 in case 1, door 1 in case 2, and door 2 in case 3), then in two of the three cases shown above, he will win the car by switching. So the probability of winning by switching is 2/3. Thus it is advantageous to switch.

Problem 9: Valentine's Day Gift

Suppose for a Valentine's Day gift you are asked to choose only one of three identical looking boxes of chocolates, but you are told that inside one of the boxes are chocolates AND a diamond ring! You pick a box, say the first on the left, and your Valentine, who knows what's inside the boxes, opens another box, say the middle one, which has only chocolates. He then says to you, "Do you want to pick the box on the right?" Is it to your advantage to switch your choice? Please explain.
(Due February 15, 2008)

Problem 8 winner is ...

Vivian! Good job.

This is an altered version of the famous "Census Taker Problem". There are exactly two triplets with a product of 72 and equal sums. This is necessary in order to confuse the census taker -- even though he sees the house number, he still cannot tell which triplet is the answer, until he finds out that there is an "oldest child". Can you find another number that has this property and can be used in a similar problem?

Problem 8: The Guitar Problem

Ms. Zambrano told a census-taker during the recent census that she had three boys. When asked their ages, she replied, “The product of their ages is 72. The sum of their ages is the same as my house number. 1�7

The census-taker looked at the house number and complained, “I still can’t tell! 1�7

Ms. Zambrano replied, “Oh, that’s right. I forgot to tell you that the oldest one likes to play guitar. 1�7

The census-taker promptly wrote down the ages
of the three children.

How old are they? (Due February 1st, 2008)

Problem 7 winners are ...

Kelvin Rodriguez, Diego Portillo, and Kathrine Fernandez are the first three students to submit a correct solution for version 1, 2, and 3, respectively. Congratulations! There are also many other correct solutions submitted by different students. Good job everyone!