I keep on disputing numerous problems of this type. I don't understand why the answer is so obvious to the author.

In my opinion, these questions are not really good.

I'll take Spiked Math's help to illustrate this out:

Moral:There are infinite formulae that fits a finite number of elements. There is no best fit formula. It is just your perception

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## Comments

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Fit this polynomial

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"much solution"

"very logic"

LMAO!!!

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Find the next number of the sequence

\(1, 1, 1, 120, ?\)

Ans: \(25852016738884976640000\), because

\(\Gamma (\Gamma (1))=1\)

\(\Gamma (\Gamma (2))=1\)

\(\Gamma (\Gamma (3))=1\)

\(\Gamma (\Gamma (4))=120\)

\(\Gamma (\Gamma (5))=25852016738884976640000\)

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Oh man, I just saw this post, I should have used this for TKC.

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Hi; whenever I see those type questions I always just fit a polynomial through them. Admittedly, it was tough with Q2.

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:D I agree with you. I fit the polynomial too. But when I try to enter the answer, Brilliant says that only integer values are allowed.

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bhaiya is it worth to join a dummy school in 11 th and 12 th

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Sorry, I do not know what that means. What is a

dummyschool?Log in to reply

agnishom bhaiya what moderator at brilliant.org means

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Moderators are responsible for helping improve the community experience, like curating nice problems, resolving reports, or seed community discussions

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how to be a moderator

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However, you can still help the community by actively participating in notes and solution discussions.

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Loved it .

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So, here's the ultimate way to cure this problem:

Let us come up with a way to put a polynomial through any set of inputs and outputs x and y.

Any clues?

For example, I defined one as follows:

Find the next number in the sequence: \(0,0,0,0,0,0,0,0,0,__\)

Ans: 10!

Now if there was a way to, say, put a function through something like 1, 4, 9, 16, 25, __ - that'd be great.

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This is the standard W|A/Mathematica command for fitting a polynomial

replace {1,4,9,16} with a set of your own data

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Well I am against those" odd number out of the set" type questions. I mean there we will always be able to choose a suitable prime p for which any 3 are quadratic residues while left one isn't

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1,3, and guess what's next? Something incomputable. Its the tree function. Next, 1, 1, 2, and guess what? Depends on the number of !'s. 0!!...=1, 1!!...=1, 2!!...=2, 3!!...=???

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