Saturday, March 29, 2008

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!!

Saturday, March 8, 2008

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)