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Hello:

That's right. How did you get it?

**ElainaVW**- Replies: 6

A person picks from the set of numbers {1,2,3,...100} eighty of them without replacement that sum to 3690. In how many ways can they do that?

Merry Christmas to everyone!

Did it change your mind about formalism?

The meeting with BillG?

Hello:

I finished the video too. Did you see SW close to the end? What did you make of that?

You do eat tuna don't you?

There has been much discussion about that particular point. I do not believe it is posted anywhere in the rules clearly. I would suggest not giving the full answer away but rather try to get the OP do it for himself. Help him along with hints and suggestions or show him a bit of math that can get the answer.

Is that an answer?

Hello:

Since everyone else just gives the answer. I'll try it.

[hidden text by admin.]

Hello thickhead:

Too much notation for such a simple problem don't ya think? I thought there was a rule against just giving away the answer like that?

Give a guy a couple of fish and he's bloated for a day. Teach him how to fish and he can open a good seafood restaurant that bobbym can eat at.

Hello Bobby:

Did you really really really need MMA to solve

?Hello:

How about too easy. That will take care of tary-try-tree-tri-trey.

How do you say it, yall gots to come up with something a bit harder aye bumpkin.

Hello:

Quite elementary.

Hello:

Where is the spiel?

The old man in the forum wrote:

I suppose that code up there yields {}?

You know it!

You should like it, it's yours!

I'm hungry.

Hello:

Have you tested that?

Does it get all the solutions?

You are doing what he did. You know the answer before you asked the questions.

The first thing I ever taught you was to never ask a question you didn't already have the answer to. - The Verdict

The verification you showed me 2 hours ago:

`b=0;FindInstance[{81a^4-420a^3*b+216a^3*c+882a^2*b^2-756a^2*b*c+216a^2*c^2-732a*b^3+1260a*b^2*c-252a*b*c^2+96a*c^3+271b^4-404b^3+546b^2*c^2+196b*c^3+16c^4-1==0,c!=(-1-3a)/2,c!=(1-3a)/2},{a,c},Integers]`

Mathegocart wrote:

You know who he is in real life??

Yes

That is not factual.

I didn't say that the right answer wasn't posted. I said the correct solution hadn't been posted yet.

The solution you propose is longer and requires more work.

This solution uses the defining properties of both ellipse and hyperbola. This is shorter.

It can be shortened more because you don't even need to know

. He could've got that with EM. It is obvious that solving 2 linear equations is terse and therefore best.Hello:

@bobbym, it is amazing that you didn't get the right answer for 8. Your diagram shows you were working on the wrong hyperbola!

The red one is the right hypie.

The correct solution is not yet given so I'll put it down.

This solution uses the defining properties of both ellipse and hyperbola.

Hello thickhead:

That's wrong too. All the terms are integers so you can't get an irrational answer.

Mistake is when you did

1000*1001/2-1023=49977

your calculation got all squished together

1000 * 100 (1/2) -1023 = 49977

the right way

(1000 * 1001) / 2 - 1023 = 499,477

Hello thickhead:

thickhead wrote:

therefore required sum is 1000*1001/2-1023=49977

That's how we solved that but you've got to calculate as accurate as you think. That answer (49977) just can't be right.