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#1 2006-02-22 19:36:47

James Nannen
Guest

3 problems

I have three hard math problems that I can not figure out no matter how hard I try. I would be nice if some one would help me and get an answer for each (or maybe just give me one) and how to do the problem.


1. Three people play a game in which one person loses and two people win each game. The one who loses must double the amount of money that each of the other two players has the time. The three players agree to play three games. At the end of the three games, each player has lost one game each person has $8. What was the original stake of each player?

2. In a ball game where only 10 and 13 points can be scored at any one time, what is the largest final score that cannot be obtained?

3. Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 102, 104, 105, 106, 107, 108, 109, 110, 112, 113. How much does each bale weigh? (Bale weight must be whole numbers.)

#2 2006-02-23 07:59:42

mathsyperson
Moderator

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Re: 3 problems

1. Let's say that the 3 people are Paul, Graham and Elliot. Paul loses the furst game, then Graham, then Elliot.

At the end, they have 8, 8, 8.
Working backwards, for Paul and Graham to have 8 points after winning, they must have each had 4 before the third game. 4+4 = 8, so Elliot would have lost 8 points altogether.

Therefore, before the third game, they have 4, 4, 16.
Using similar reasoning for the second game shows that before it they each have 2, 14, 8.
Finally, more similar reasoning shows that at the start they have 13, 7 and 4 points respectively.

2. The first few multiples of 13 are 13, 26, 39, 52, 65, 78, 91, 104, 117.
To decide whether a certain number can be scored, look at its last digit. Find the corresponding number in that list and if your number is equal to or bigger than the list number, then it is possible to be scored.

Therefore, the biggest impossible number is the biggest number in that list, minus 10. Namely, 107.

3. I can't really explain how I got this, except to say that I simplified it down so that the total weights were 2, 4, 5, 6, 7, 8, 9, 10, 12 and 13, and then using trial and error.

I got these: 20, 22, 24, 25 and 28.


Why did the vector cross the road?
It wanted to be normal.

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