2 math problems

[JamesNanne]

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.)

[Guest]

#2. The answer is 107.

Solution:
*THe multiples of 10 don't affect the units digit. It can only be affected by multiples of 13, 0-9 times 13 is:
0, 13, 26, 39, 52, 65, 78, 91, 104, 117.

*That list includes all possible values of the units digit, 0-9. For each value of the units digit, the figure given is the lowest number ending in that units digit that is achievable, for example 39 is the lowest achievable number ending in 9. Any higher number just needs sufficient multiples of 10.

*117 is the lowest number ending in 7. 107 is the highest number not achievable. (108-116 can all be achieved by adding multiples of 10 to lower achievable numbers).

[ryan]

#1

8 8 8


game 3:

2 win and in the end each have 8 dollars

that means that the loser had to double x to get 16, or 16/2 = 8

So each winner has 4 dollars before game 3, the loser had 8.

game 2:

4 4 8

4/2 + 8/2 = 2 + 4

so we have 2, 4, 4

game 1:

2 4 4

4/2 + 2/2 = 2 + 1

so we have 2, 1, 4