(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)
Friday, May 30, 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.
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.
Saturday, May 10, 2008
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)
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)
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