Variety of interesting problems

[Deleted member 4993]

 

[JeffM]

View attachment 32548

Spoiler

How many ways can we draw two balls without regard to order?

[math]\dbinom{12}{2} = 66.[/math]
How many ways can we draw two balls of identical parity?

[math]\dbinom{6}{2} + \dbinom{6}{2} = 15 + 15 = 30.[/math]
How many ways can we draw two balls of differing parity?

[math]\dbinom{6}{1} * \dbinom{6}{1} = 6 * 6 = 36.[/math]
That looks right: 30 + 36 = 66.

So the probability of drawing two balls with an odd sum exceeds the probability of drawing two balls with an even sum, a counter-intuitive result.

 

[BigBeachBanana]

Spoiler: #2

[math] \sin^2(1\degree)+\sin^2(2\degree)+\dots+\sin^2(90\degree)=\\ [\sin^2(1\degree)+\sin^2(89\degree)]+[\sin^2(2\degree)+\sin^2(88\degree)]+\dots+ [\sin^2(44\degree)+\sin^2(46\degree)]+\sin^2(45\degree)+\sin^2(90\degree)=\\ [\sin^2(1\degree)+\cos^2(1\degree)]+[\sin^2(2\degree)+\cos^2(2\degree)]+\dots+[\sin^2(44\degree)+\cos^2(44\degree)]+ \sin^2(45\degree)+\sin^2(90\degree)=\\ \underbrace{1+1+\dots+1}_{\text{44 times}}+\sin^2(45\degree)+\sin^2(90\degree)=44+0.5+1=45.5 [/math]

 

[Deleted member 4993]

Spoiler: #2

[math] \sin^2(1\degree)+\sin^2(2\degree)+\dots+\sin^2(90\degree)=\\ [\sin^2(1\degree)+\sin^2(89\degree)]+[\sin^2(2\degree)+\sin^2(88\degree)]+\dots+ [\sin^2(44\degree)+\sin^2(46\degree)]+\sin^2(45\degree)+\sin^2(90\degree)=\\ [\sin^2(1\degree)+\cos^2(1\degree)]+[\sin^2(2\degree)+\cos^2(2\degree)]+\dots+[\sin^2(44\degree)+\cos^2(44\degree)]+ \sin^2(45\degree)+\sin^2(90\degree)=\\ \underbrace{1+1+\dots+1}_{\text{44 times}}+\sin^2(45\degree)+\sin^2(90\degree)=44+0.5+1=45.5 [/math]

Spoiler: #2 Alternate Hint

sin²(Θ) = (1/2)[1 - cos(2Θ)]

 

[lookagain]

For the first problem:

Spoiler

2,009 = 7*7*41, so \(\displaystyle \ \sqrt{2,009} = 7\sqrt{41}.\)

Therefore, 7 square roots of 41 can be split into the positive integer combos
of 1 & 6, 6 & 1, 2 & 5, 5 & 2, 3 & 4, and 4 & 3 square roots of 41, respectively.

For example with the first one, \(\displaystyle \ 1\sqrt{41} \ + \ 6\sqrt{41} \ \ \) gives \(\displaystyle \ \ \sqrt{41} \ + \ \sqrt{1476}.\)

Here are six solutions I found for (A, B): (41, 1476), (1476, 41), (164, 1025),
(1025, 164), (369, 656), (656, 369).

 

[AvgStudent]

For the first problem:

Spoiler

2,009 = 7*7*41, so \(\displaystyle \ \sqrt{2,009} = 7\sqrt{41}.\)

Therefore, 7 square roots of 41 can be split into the positive integer combos
of 1 & 6, 6 & 1, 2 & 5, 5 & 2, 3 & 4, and 4 & 3 square roots of 41, respectively.

For example with the first one, \(\displaystyle \ 1\sqrt{41} \ + \ 6\sqrt{41} \ \ \) gives \(\displaystyle \ \ \sqrt{41} \ + \ \sqrt{1476}.\)

Here are six solutions I found for (A, B): (41, 1476), (1476, 41), (164, 1025),
(1025, 164), (369, 656), (656, 369).

I think you missed the trivial one (0,2009) or by your method 0+7. But clever solution.

 

[lookagain]

I think you missed the trivial one . . .

"... for positive integer values ..."

 

[Dr.Peterson]

For the first problem:

Spoiler

2,009 = 7*7*41, so \(\displaystyle \ \sqrt{2,009} = 7\sqrt{41}.\)

Therefore, 7 square roots of 41 can be split into the positive integer combos
of 1 & 6, 6 & 1, 2 & 5, 5 & 2, 3 & 4, and 4 & 3 square roots of 41, respectively.

For example with the first one, \(\displaystyle \ 1\sqrt{41} \ + \ 6\sqrt{41} \ \ \) gives \(\displaystyle \ \ \sqrt{41} \ + \ \sqrt{1476}.\)

Here are six solutions I found for (A, B): (41, 1476), (1476, 41), (164, 1025),
(1025, 164), (369, 656), (656, 369).

This doesn't necessarily show that these are the only solutions ...

But I found the same ones by solving for y in terms of x, which showed directly that 41x has to be a perfect square.

 

[lookagain]

This doesn't necessarily show that these are the only solutions ...

But they do necessarily show that they are the only solutions, because there
must be integer multiples of the same radical added together. That forced
working with \(\displaystyle \ \sqrt{2,009} \ \ as \ \ 7\sqrt{41} \ \) and all of its 2-number sum combinations.

There cannot be any other radicands, because then there would not be any perfect square (greater than 1) to take outside in front of the radical.

 

[Cubist]

But they do necessarily show that they are the only solutions, because there
must be integer multiples of the same radical added together. That forced
working with \(\displaystyle \ \sqrt{2,009} \ \ as \ \ 7\sqrt{41} \ \) and all of its 2-number sum combinations.

There cannot be any other radicands, because then there would not be any perfect square (greater than 1) to take outside in front of the radical.

Very interesting. I don't remember considering this before. Can there be any other solutions? Given

[math]\sqrt{a} + \sqrt{b} = \sqrt{c}[/math]
With a, b positive integers with no repeating prime factors (therefore [imath]\sqrt{a}[/imath] can't be written as [imath]x\sqrt{y}[/imath] with integer x and y). Squaring gives...
[math]c=( \sqrt{a} + \sqrt{b} )^2\\ = a + b + 2\sqrt{ab}[/math]
therefore c can only be integer if [imath]2\sqrt{ab}[/imath] is integer which implies a=b because there isn't any other way to make every prime factor of [imath]ab[/imath] repeat (so that it's a perfect square). Therefore @lookagain is correct

And if a=b then [imath]\sqrt{c}=\sqrt{a}+\sqrt{a} = 2\sqrt{a} [/imath] which is the same as [imath]\sqrt{4a}[/imath]

 

[Dr.Peterson]

Very interesting. I don't remember considering this before. Can there be any other solutions?

This is exactly my thinking. When lookagain said "Here are six solutions I found", that in itself doesn't tell us that there are no other solutions; it takes additional thought (unless one happens to know such a theorem, and I don't trust my memory of such things enough to be sure that what seems right is right). So I chose not to make the assumption.

My work in solving the long way, however, closely paralleled your proof of the general case.

 

[Steven G]

p(1st ball is odd) =1/2
p(even sum) = p(2nd ball is odd) = 5/11
P(odd sum) = p(2nd ball is even) = 6/11

p(1st ball is even)= 1/2
p(even sum) = p(2nd ball is even) = 5/11
P(odd sum) = p(2nd ball is odd) = 6/11

P(same parity) = (1/2)(5/11) + (1/2)(5/11) = 5/11 = 30/66

P(different parity) = (1/2)(6/11) + (1/2)(6/11) = 6/11 = 36/66

Very strange results (not even close).

Note that my results matches JeffM's results.

 

[Steven G]

I think in the way I solved the problem above shows why we get different results for even and odd (same or different parities).
No matter which ball you pick 1st, there are more balls of the other parity left for the 2nd ball....