[Calvin] Which solution will you feature (10)?

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Below, we present a problem from the 2/18 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong.

Degree 99 Polynomial f(x)f(x) is a polynomial of degree 99. For exactly 100 (out of 101) integer values ranging from 0 0 to 100 100, we have f(x)=1x+1 f(x) = \frac {1}{x+1} . Also, f(101)=0f(101) = 0. For what value of aa, 0a100 0 \leq a \leq 100 is f(a)1a+1 f(a) \neq \frac {1}{a+1} ?

This problem is proposed by Aakash K, and Solution B is presented by Nathan.

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and has not been edited.

About 60% of those who answered this problem got it correct.

Solution A - This solution is completely wrong, despite the votes for it. The biggest tipoff is that it doesn't use the fact that f(x) f(x) is a polynomial of degree 99. (Yes, it defines Q(x)Q(x) as a polynomial of degree 98, but does nothing with it.) While Q(x)=P(x) Q(x) = P(x) for several values, that tells us nothing about the behavior of Q(x)Q(x), since P(x)P(x) is not a polynomial, but a rational function. Recall the question Polynomial powered by 2, where knowing that f(n)=2n f(n) = 2^n for some values didn't imply that f(9)=29 f(9) = 2^9, as pointed out in the discussion.

Note that I often do not penalize typos as I care more about your thought process. However, if your typos carry through or have massive repercussions, then you may be penalized accordingly.

Solution B - This solution is clear in it's presentation, explaining how the function g(x)g(x) is created, and how to calculate the value of cc. In this problem, having each of the main equations take up a line increases readability. This solution is presented by Nathan.

Solution C - I had no idea what was happening here. The degree of f(x) f(x) had nothing to do with the value of f(1) f(-1) . In fact, it is not clear how to calculate the value of f(1)f(-1) , without figuring out what g(x)g(x) (as defined in Solution B) is, which requires knowing the value of aa.

Note that f(x)=1/x+1 f(x) = 1/x + 1 is often interpreted as f(x)=1x+1 f(x) = \frac {1}{x} + 1 rather than f(x)=1x+1 f(x) = \frac {1}{x+1} . Be clear in your presentation, and say f(x)=1/(x+1) f(x) = 1/ (x+1) instead.

Pop quiz: What is the polynomial f(x) f(x) ?

Note: If you want to submit a problem, please ensure that you the problem is properly phrased, and that you include a proper complete solution. In this case, because the problem was interesting, I figured out my own proper solution.

Note by Calvin Lin
8 years, 6 months ago

No vote yet
5 votes

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Solution B - We know that f(x)=1x+1 f(x)=\frac{1}{x+1} for all integers ranging from 00 to 100100 except aa. Thus, (x+1)f(x)1=0(x+1)f(x)-1=0 for all integers ranging from 00 to 100100 except aa. This leads us to consider the 100 degree polynomial g(x)=(x+1)f(x)1g(x)=(x+1)f(x)-1, whose roots are the integers ranging from 00 to 100100 except aa. (There are no more roots because gg is only of degree 100.) Therefore, we can write g(x)=cx(x1)(x2)(x100)xa, g(x)=c\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a}, for some nonzero constant cc. But we are also given that f(101)=0f(101)=0, which implies g(101)=1g(101)=-1. Plugging in 101101 to the above equation yields c101!101a=1    c=101a101! c\cdot \frac{101!}{101-a}=-1\implies c=-\frac{101-a}{101!} Thus we can now write g(x)g(x) as g(x)=101a101!x(x1)(x2)(x100)xa g(x)=-\frac{101-a}{101!}\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a} Now notice that from our definition of g(x)g(x) we have g(x)+1=(x+1)f(x)g(x)+1=(x+1)f(x), which means that 1-1 is a root of g(x)+1g(x)+1. Thus plugging in x=1x=-1 into g(x)+1g(x)+1 should give us 00: 101a101!(1)(2)(3)(101)1a+1=0, -\frac{101-a}{101!}\cdot \frac{(-1)(-2)(-3)\cdots(-101)}{-1-a}+1=0, and after some cancellation and rearranging we get a=50a=\boxed{50}.

Calvin Lin Staff - 8 years, 6 months ago

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Solution A - Let SS be the set of all integers between 00 and 100100 for which f(x)=1x+1f(x) = \frac{1}{x+1} for all xSx \in S. From the problem statement, we have S=100|S| = 100. Since 101101 is a root of f(x)f(x), we can rewrite ff as f(x)=(101x)Q(x)f(x) = (101-x)Q(x), where Q(x)Q(x) is a polynomial of degree 9898. Hence, Q(x)=f(x)101x=1(x+1)(101x)Q(x) = \frac{f(x)}{101-x} = \frac{1}{(x+1)(101-x)} for all xSx \in S. Define P(x)=1(x+1)(101x)P(x) = \frac{1}{(x+1)(101-x)}. Then, substituting 100x100-x into the equation, we get P(x)=P(101x)P(x) = P(101-x). So xSx \in S if and only if 100xS100-x \in S. Let aa be an integer from 00 to 100100 which does not belong to SS. We conclude that both aa and 100a100-a do not belong to SS. But, there can be at most one such aa. So, we must have a=100aa=50a = 100-a \Leftrightarrow a = 50.

Calvin Lin Staff - 8 years, 6 months ago

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I think this is better solution than Solution B, just one typo though, it should be P(x)=P(100x)P(x)=P(100-x) instead of P(x)=P(101x)P(x) = P(101-x)

Aldrian Obaja - 8 years, 6 months ago

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Sir did you modify my solution.? i wrote it in very bad manner , but i doubt if i have written this.. :D

Shivang Jindal - 8 years, 6 months ago

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I do not announce whose solution was posted.

There were no edits made (apart from LaTeX for necessity), so calculation and editorial errors will remain. If you do not recognize it, it is likely not yours.

Calvin Lin Staff - 8 years, 6 months ago

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Is this solution wrong? I know it is mine and I just received an email saying that my solution to this problem was incorrect then I got -115 points. As I know, the only mistake is the typo that Aldrian mentioned above.

Ahmad Zaky - 8 years, 6 months ago

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I don't understand what do you mean by 'completely wrong'. Can you point out which part of my solution is wrong? Indeed, I don't use the fact that ff has degree 9999. But I use the fact that ff is a polynomial to define QQ. If ff is not a polynomial, we cannot say that f(x)=(101x)Q(x)f(x) = (101-x)Q(x) only by knowing f(101)=0f(101)=0. My solution above works for all polynomials ff despite of its degree. It even works for all functions ff having form (101x)Q(x)(101-x)Q(x) for some function QQ.

Ahmad Zaky - 8 years, 6 months ago

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@Ahmad It is an important skill for you to be able to figure out your mistakes (or if they are mistakes), especially if someone points out various errors. There are a lot of statements which you're saying, which doesn't make sense in the context of the question. For example, there is a unique polynomial ff of degree 99 which can satisfy the conditions of the question, while you seem to be suggesting that it will work regardless of the degree. The polynomial assumption is extremely restrictive (as opposed to a random curve), which is what allows us to draw a conclusion.

As I said, "QQ is a polynomial of degree 98", but you never used that fact anywhere else. Comparing a polynomial to anything else but a polynomial doesn't tell you about it's behavior except at those points. So what if P(x)=P(101x) P(x) = P(101-x)? How does that explain why xS101xS x \in S \Leftrightarrow 101-x \in S? Your statements have no logical implications whatsoever, and you're doing a false proof by "I have this amazing coincidence, hence this other irrelevant fact is true simply because I say it is".

Furthermore, note that the statement is actually that IF x,101xS x, 101-x \in S, then Q(x)=Q(101x) Q(x) = Q(101-x) , with which you are using to conclude that xS101xS x \in S \Leftrightarrow 101-x \in S. Your conclusion is simply a (weakened) restatement of your assumptions.

Calvin Lin Staff - 8 years, 6 months ago

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Solution A is just like how i solved it and it is quite clear and easy to get

Aakash Kansal - 8 years, 6 months ago

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Is this a level 5 question?

David Altizio - 8 years, 6 months ago

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This problem was posed to both Level 4 and 5. Not every level 4 or 5 would have seen it though.

Calvin Lin Staff - 8 years, 6 months ago

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Yes

Zi Song Yeoh - 8 years, 6 months ago

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Remarks have been added.

Calvin Lin Staff - 8 years, 6 months ago

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Solution C - f(x)=1/x+1f(x)= 1/x+1 for 100 integers between 0 to 100 we can write (x+1)f(x)1=p(xa1)(xa2)..............(xa100)(x+1)f(x)-1 = p(x-a_{1})(x-a_{2})..............(x-a_{100}) .....eqn 1 where a1,a2......a100a_{1}, a_{2}...... a_{100} are integers belonging to [0,1,2,3.....100][0,1,2,3.....100] . Since the degree of f(x)f(x) is 99, therefore f(1)=0f(-1)=0. It is given that f(101)=0f(101)=0. On applying the above 2 conditions on equation 1, we get that aia_{i} does not contain 50.

Calvin Lin Staff - 8 years, 6 months ago

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