20. Given integers $n$ and $k$, with $1 \leq k \leq n$, we define a polynomial of degree $n-1$ as follows:

$$
p(x)=\frac{(x+1)(x+2) \ldots(x+n)}{(x+k)} .
$$

For example, if $n=5$ and $k=2$, we have $p(x)=(x+1)(x+3)(x+4)(x+5)$. Suppose that for a certain choice of $n$ and $k$, the coefficient of $x^{n-2}$ in the polynomial $p(x)$ is equal to 67. What is the value of $n$ in this case?
(A) 68
(B) 10
(C) 12
(D) 11
(E) 69

## Unione Matematica Italiana  PROGETTO OLIMPIADI DI MATEMATICA

Ministry of Education, University and Research

Scuola Normale Superiore

## Archimedes Games - Triennio Competition

## November 23, 2016

- The test consists of 20 problems. Each question is followed by five answers indicated by the letters $(\mathbf{A}),(\mathbf{B}),(\mathbf{C}),(\mathbf{D}),(\mathbf{E})$: only one of these answers is correct, the other 4 are incorrect.
- For each of the problems, you must write the letter corresponding to the answer you believe is correct in the grid provided below. No erasures or corrections are allowed on the grid. The use of any type of calculator or communication device is not permitted.

You have 110 minutes to complete the test. Good luck and have fun!

NAME

LAST NAME

class:
![](https://cdn.mathpix.com/cropped/2024_04_17_82169a1997fb4f3442bcg-3.jpg?height=104&width=1314&top_left_y=1154&top_left_x=80)
Problem 20. The answer is (C).

The polynomial $(x+1) \ldots(x+n)$ has degree $n$, and its coefficient of degree $n-1$ is

$$
1+2+\cdots+n=n(n+1) / 2 \text { . }
$$

Consequently, the polynomial $p(x)$ has degree $n-1$, and its coefficient of degree $n-2$ is $n(n+1) / 2-k$. We then need to solve $n(n+1) / 2-k=67$. Recalling that $1 \leq k \leq n$, we have

$$
68 \leq n(n+1) / 2 \leq 67+n \text { . }
$$

The inequality $n(n+1) / 2 \geq 68$ is satisfied only when $n \geq 12$, while $n(n+1) / 2 \leq 67+n$ can also be written as $n(n-1) / 2 \leq 67$, from which $n \leq 12$. In conclusion, $n=12$ and $k=11$.

(Problem proposed by V. Ricciuti)