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new math puzzle

milagros317

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It has been a long while since I posted a mathematics puzzle, so here goes:

A jar has some red balls and some green balls in it, all the same size and randomly mixed up.
There are at least two balls in the jar, but no more than 100.

If two balls are randomly chosen from the jar, then the probability that they are one red and one green is exactly 1/2.


How many balls of each color are in the jar?

The answer is not unique; give all possible solutions.


Note: That probability, p, will never be 1/2 if there are equally many of each color. For example, with 5 of each color, then p = 25/45 = 5/9.

Answer will be given in one week, on April 21[sup]st[/sup].
 
1. 3 and 1 (4 in total)
2. 6 and 3 (9)
3. 10 and 6 (16)
4. 15 and 10 (25)
5. 21 and 15 (36)
6. 28 and 21 (49)
7. 36 and 28 (64)
8. 45 and 36 (81)
9. 55 and 45 (100)
 
Could just be one of each (I know, the easy way out....).
 
1. 3 and 1 (4 in total)
2. 6 and 3 (9)
3. 10 and 6 (16)
4. 15 and 10 (25)
5. 21 and 15 (36)
6. 28 and 21 (49)
7. 36 and 28 (64)
8. 45 and 36 (81)
9. 55 and 45 (100)

Yes! This is a complete solution! Very well done.
 
Could just be one of each (I know, the easy way out....).

With only two balls in the jar, one of each color, then p=1. (It is certain,not p=1/2, that you will get one of each color when you randomly select two balls from the jar.)
 
Yes! This is a complete solution! Very well done.

Actually, I used Excel scripts to screen all the possible combinations and find suitable ones. My math is rusty, so I had no recourse but the brutal force method.

But the resulting figures form such a harmonic pattern, that there must be some elegant solution in terms of logic, still elusive for me.
 
Actually, I used Excel scripts to screen all the possible combinations and find suitable ones. My math is rusty, so I had no recourse but the brutal force method.

But the resulting figures form such a harmonic pattern, that there must be some elegant solution in terms of logic, still elusive for me.
High school algebra will reduce the problem to a quadratic Diophantine equation in two variables. To sole it analytically requires a course in number theory, which I last had in 1973 and have forgotten. So I just played with the small numbers (total number of balls at most ten), found that there were solutions only for 4 and 9, guessed that the higher solutions would also be perfect squares, and found them.
Well, it's a sequence of squares... that's all I noticed.
Yes. The infinite family of solutions, once we remove the upper limit of 100 balls, is as follows:

s[sup]2[/sup] total number of balls.

(s[sup]2[/sup] + s)/2 of one color

(s[sup]2[/sup] -s)/2 of the other color

High school algebra will verify that this is a valid solution for all integers s grater than or equal to 2.
 
As the original puzzle has been solved, here is a more difficult one of the same type:

A jar has some red balls and some green balls in it, all the same size and randomly mixed up.
There are at least two balls in the jar and there is no upper limit.

If two balls are randomly chosen from the jar, then the probability that they are one red and one green is exactly 9/19.


How many balls of each color are in the jar?

The answer is unique as far as I know; I have only one solution but I would be glad to see more.
 
‘...A weight of mass M hangs on a string of length L. A bullet of mass m moves at velocity V and hits the weight. Find the angle it swings...’
‘But teacher, is there any simpler task – something with a squirrel and nuts?’
‘Well – a squirrel is hanging on a string, while hit by a nut, so find the angle.’

The first five results are 21 and 36; 36 and 60; 365 and 585; 585 and 936; 936 and 1496. Piece a cake.
 
‘...A weight of mass M hangs on a string of length L. A bullet of mass m moves at velocity V and hits the weight. Find the angle it swings...’
‘But teacher, is there any simpler task – something with a squirrel and nuts?’
‘Well – a squirrel is hanging on a string, while hit by a nut, so find the angle.’

The first five results are 21 and 36; 36 and 60; 365 and 585; 585 and 936; 936 and 1496. Piece a cake.
Wonderful! :D
I knew 21 and 36 when I posted this originally and found 36 and 60 later. I had not found the three other solutions, above 100.
Your Excel scripts really did the job! :D
 
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