# Determine the Maths Function [closed]

You found yourself in a weird room, there was no doors, but a limited computer with no internet that has open an app that is displaying the following message:

f(0) = 0
f(1) = 549
f(2) = 646
f(10) = 1350
f(100) = 450
Hence, find f(x) and f(75)

For people who don't know what a maths function is:

I will give a clue everyday if no one has the right answer, if alternative answer was found and has the same results as my answer, it counted as the right answer.

CLUE:

• the function contains squared number
• the function does not has complex number
• the function uses BODMAS

## closed as off-topic by skv, Joe, Gilles, Psychemaster, Rob WattsNov 17 '14 at 21:01

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• This question appears to be off-topic because it belongs to Math SE – skv Nov 15 '14 at 16:44
• @skv, it's the same as the password puzzles (spaghetti party etc), just the presentation is different. – A E Nov 15 '14 at 17:16
• Yes, I agree, and I recognised that too, but that word play is required to make it a puzzle – skv Nov 15 '14 at 17:17
• @skv: it is? why? – A E Nov 15 '14 at 19:25
• Anyway, there's the "You found yourself in a weird room, ..." stuff already. – A E Nov 15 '14 at 19:32

$f(1) = 549 = 1 * 100 + 450 - 1^2$
$f(2) = 646 = 2 * 100 + 450 - 2^2$
$f(10) = 1350 = 10 * 100 + 450 - 10^2$
$f(100) = 450 = 100 * 100 + 450 - 100^2$

So, $f(x) = 100x + 450 - x^2$, and $f(75) = 2325$.

Then I realized... maybe it should be the following, but then I think I cheated, right?

$f(x) = \begin{cases} 100x + 450 - x^2 &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x = 0\end{cases}$

• wow, so fast, and formula was close enough to mine, my f(x) = ((x+5) * 100 ) - (x^2 + 50) – wuiyang Nov 16 '14 at 0:35
• It's actually the same formula, except that f(0) part. The way I solve it is to realize f(1) differ from 1 from multiple of 10; f(2) is 4 differ from multiple of 10, f(10) is already a multiple of 10, so I just extrapolate the f(1) and f(2) pattern, which is x^2 away from multiple of 10, and assumed f(10) is 100 away from multiples of 10, and I did the same for f(100). Then I spot what should be a nice constant to take out from each number when they are added to the multiple of 10, which turns out to be 450. The coefficient of x is trivial afterwards. – nexolute Nov 16 '14 at 0:52

The problem with this type of puzzle is that you can match any dataset using a polynomial of sufficiently high degree. For this one:

$$f(x)=-0.225x^4+25.425x^3-300.7x^2+824.5x$$

giving

$$f(75)=1977431.25$$

Of course you probably have a different formula, but from the point of view answering the question we have no way of knowing which of the infinite number of formulae matching this data is the one you had in mind

• And, by extension, you can pick any $y \in \mathbb{R}$ and add a data point $(75, y)$ to your Lagrange interpolation, creating a function of order $n + 1$ that satisfies the requirements. – wchargin Nov 15 '14 at 23:09
• nope, the answer was integer – wuiyang Nov 16 '14 at 0:33
• As the previous comment points out you can make f(75) be whatever you want, and still fit all question's criteria – Michal Nov 16 '14 at 0:37
• Can confirm. In fact, those solutions will be more correct due to the fact that your solution does not fit your f(0)=0 case. Maybe adding the criterion for degree 2 polynomial will make these strict, but that will just reduce to a problem of 3 equations 3 unknown linear equations. – nexolute Nov 16 '14 at 6:11