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

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 \( f(x) \) is a polynomial of degree 99. (Yes, it defines \(Q(x)\) as a polynomial of degree 98, but does nothing with it.) While \( Q(x) = P(x) \) for several values, that tells us nothing about the behavior of \(Q(x)\), since \(P(x) \) is not a polynomial, but a rational function. Recall the question Polynomial powered by 2, where knowing that \( f(n) = 2^n\) for some values didn't imply that \( f(9) = 2^9\), 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 \(g(x)\) is created, and how to calculate the value of \(c\). 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 \( f(x) \) had nothing to do with the value of \( f(-1) \). In fact, it is not clear how to calculate the value of \(f(-1) \), without figuring out what \(g(x)\) (as defined in Solution B) is, which requires knowing the value of \(a\).

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

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

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.

## Comments

Sort by:

TopNewestSolution B - We know that \( f(x)=\frac{1}{x+1}\) for all integers ranging from \(0\) to \(100\) except \(a\). Thus, \((x+1)f(x)-1=0\) for all integers ranging from \(0\) to \(100\) except \(a\). This leads us to consider the 100 degree polynomial \(g(x)=(x+1)f(x)-1\), whose roots are the integers ranging from \(0\) to \(100\) except \(a\). (There are no more roots because \(g\) is only of degree 100.) Therefore, we can write \[ g(x)=c\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a}, \] for some nonzero constant \(c\). But we are also given that \(f(101)=0\), which implies \(g(101)=-1\). Plugging in \(101\) to the above equation yields \[ c\cdot \frac{101!}{101-a}=-1\implies c=-\frac{101-a}{101!} \] Thus we can now write \(g(x)\) as \[ g(x)=-\frac{101-a}{101!}\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a} \] Now notice that from our definition of \(g(x)\) we have \(g(x)+1=(x+1)f(x)\), which means that \(-1\) is a root of \(g(x)+1\). Thus plugging in \(x=-1\) into \(g(x)+1\) should give us \(0\): \[ -\frac{101-a}{101!}\cdot \frac{(-1)(-2)(-3)\cdots(-101)}{-1-a}+1=0, \] and after some cancellation and rearranging we get \(a=\boxed{50}\). – Calvin Lin Staff · 4 years, 2 months ago

Log in to reply

Solution A - Let \(S\) be the set of all integers between \(0\) and \(100\) for which \(f(x) = \frac{1}{x+1}\) for all \(x \in S\). From the problem statement, we have \(|S| = 100\). Since \(101\) is a root of \(f(x)\), we can rewrite \(f\) as \(f(x) = (101-x)Q(x)\), where \(Q(x)\) is a polynomial of degree \(98\). Hence, \(Q(x) = \frac{f(x)}{101-x} = \frac{1}{(x+1)(101-x)}\) for all \(x \in S\). Define \(P(x) = \frac{1}{(x+1)(101-x)}\). Then, substituting \(100-x\) into the equation, we get \(P(x) = P(101-x)\). So \(x \in S\) if and only if \(100-x \in S\). Let \(a\) be an integer from \(0\) to \(100\) which does not belong to \(S\). We conclude that both \(a\) and \(100-a\) do not belong to \(S\). But, there can be at most one such \(a\). So, we must have \(a = 100-a \Leftrightarrow a = 50\). – Calvin Lin Staff · 4 years, 2 months ago

Log in to reply

– Aldrian Obaja · 4 years, 2 months ago

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

– Shivang Jindal · 4 years, 2 months ago

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

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 · 4 years, 2 months ago

Log in to reply

– Ahmad Zaky · 4 years, 2 months ago

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.Log in to reply

is a polynomialto define \(Q\). If \(f\) is not a polynomial, we cannot say that \(f(x) = (101-x)Q(x)\) only by knowing \(f(101)=0\). My solution above works for all polynomials \(f\) despite of its degree. It even works for all functions \(f\) having form \((101-x)Q(x)\) for some function \(Q\). – Ahmad Zaky · 4 years, 2 months agoLog in to reply

As I said, "\(Q\) 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 \( P(x) = P(101-x)\)? How does that explain why \( x \in S \Leftrightarrow 101-x \in S\)? 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 \( x, 101-x \in S\), then \( Q(x) = Q(101-x) \), with which you are using to conclude that \( x \in S \Leftrightarrow 101-x \in S\). Your conclusion is simply a (weakened) restatement of your assumptions. – Calvin Lin Staff · 4 years, 2 months ago

Log in to reply

Solution A is just like how i solved it and it is quite clear and easy to get – Aakash Kansal · 4 years, 2 months ago

Log in to reply

Remarks have been added. – Calvin Lin Staff · 4 years, 2 months ago

Log in to reply

Is this a level 5 question? – David Altizio · 4 years, 2 months ago

Log in to reply

– Calvin Lin Staff · 4 years, 2 months ago

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

– Zi Song Yeoh · 4 years, 2 months ago

YesLog in to reply

Solution C - \(f(x)= 1/x+1\) for 100 integers between 0 to 100 we can write \((x+1)f(x)-1 = p(x-a_{1})(x-a_{2})..............(x-a_{100})\) .....eqn 1 where \(a_{1}, a_{2}...... a_{100}\) are integers belonging to \([0,1,2,3.....100] \). Since the degree of \(f(x)\) is 99, therefore \(f(-1)=0\). It is given that \(f(101)=0\). On applying the above 2 conditions on equation 1, we get that \(a_{i}\) does not contain 50. – Calvin Lin Staff · 4 years, 2 months ago

Log in to reply