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)