summer problems
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http://www.pleacher.com/mp/probweek/p2002/a010702.html
http://www.pleacher.com/mp/probweek/p2002/a010702.html
Try counting all the rectangles.
When you are finished, click on this link and see a picture
of a type of rectangle that you may have overlooked the first time.
Now go back and recount.
The most difficult thing in a problem like this is keeping
track (or labeling) of the ones that you counted.
By labeling each region you can list each rectangle as
a group of letters.
Click on this link to see a possible labeling and the
beginning of keeping track.
Now how many do you find?
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The Number Sequence Problem
What are the next three numbers in this sequence?
1, 8, 11, 18, 80, 81, 82, 83, 84, 85, 86, 87, 88, ...
Solution:
89, 100, 101.
The sequence consists of the positive integers (also called natural numbers) beginning with vowels.
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The Coin Problem
You have 64 coins, all of which appear identical. However, one of the coins is heavier than the others.
Using a two-pan balance, what is the least number of weighings that you need (in the worst case) to determine the false coin? Explain your process.
Solution:
You need only four weighings.
This proved to be a challenging problem. Most people sent in six for the solution because they divided the coins into two groups of 32.
To solve this in only four weighings,
Divide the coins into groups of 21, 21, and 22 coins.
Weigh the two groups of 21.
If one pan dips, then the false coin must be in that pan.
If the two pans balance, then the false coin must be in the group of 22.
Take the group with the heavier coin and split into 3 groups: 7, 7, and 8 (or 7).
Weigh two groups of 7. If they balance, the false coin is in the other group. If the pan dips, the false coin is in that group of coins.
Take the group of 7 (or 8) coins and split into three groups of 2, 2, and 3 (or 2, 3, and 3). Weigh the two equal groups of coins. If they balance, the false coin is in the other group. If the pan dips, the false coin is in that group.
You are now left with only 2 or 3 coins. Weigh two of them to determine the false coin.
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The Bellman Problem
Three people went to a hotel and rented a room for $30, each paying $10 for his share. Later, the clerk discovered that the price of the room was only $25. She handed the bellman five $1 bills and asked him to return them to the three people. The bellman, not knowing how to divide $5 among three people, instead gave each person $1 and kept the other $2 for himself.
Here's the question:
The three people originally paid $10 each, but each received $1 back, so they've now paid a total of $27 for the room.
Add to that the $2 that the bellman kept, and you have a total expenditure of $29 instead of $30. What happened to the other dollar?
Solution:
There is no missing dollar.
This is a classic problem -- it has been around for at least 30 years.
The total expenditure is now only $27, accounted for by adding the $25 in the hands of the hotel clerk to the $2 in the hands of the bellman.
In other words, the original $30 now is divided like this:
the hotel clerk has $25, the guests have $3, and the bellman has $2.
The error arose when an asset ($2) was added to an expense ($27) instead of the other asset ($25), thereby "mixing apples with oranges."
Solution:
A. 20 B. 6 C. 16 D. 10 E. 14
F. 18 G. 12 H. 22 I. 8
The magic number was 42 -- Elvis' age at his death.
