Number of polynomials which are divisible by $x+1$ Let $a,b,c,d$ be four integers (not necessarily distinct) in the set ${1,2,3,4,5}$ . The number of polynomials $f(x)=x^4+ax^3+bx^2+cx+d$ which are divisible by $x+1$ are:
$(A)$ Between 55 and 65
$(B)$ Between 65 and 85
$(C)$ Between 86 and 105
$(D)$ More than 105
I see that it will be divisible when $f(-1)=0$
i.e. $1-a+b-c+d=0$ but how should I proceed further ?
 A: On approach would be to brute-force the result by counting. Notice that you've reached the necessary and sufficient condition $a+c = 1+b+d$.
Since $1\le a,b,c,d \le 5$, then $a+c \in \{2, \dots, 10\}$, so $b+d \in \{1, \dots, 9\}$. Notice that $b+d=1$ is impossible, because it would force either $b$ or $d$ to be $0$ which is not allowed. Thus, $a+c \ne 2$.
For $n \ge 1$ natural, let $S_n$ be the number of ways of writing it as $a+c$, with $1 \le a,c \le 5$ and counting both $a+c$ and $c+a$ (i.e. order does matter).


*

*Every natural number $1 \le n \le 6$ can be written as $1 + (n-1) = 2 + (n-2) = \dots = (n-1) + 1$, so $S_n = n-1$ for $1 \le n \le 6$.

*For $7 \le n \le 10$ we can perform a case-by-case analysis: $7 = 2 + 5 = 3 + 4 = 4 + 3 = 5 + 2$, so $S_7 = 4$. $8 = 3 + 5 = 4 + 4 = 5 + 3$, so $S_8 = 3$. $9 = 4 + 5 = 5 + 4$, so $S_9 = 2$. $10 = 5 + 5$, so $S_{10} = 1$.

*No natural number $n \ge 11$ can be written in the desired way because at least one summand would be $\ge 6$, so $S_n = 0$ for $n \ge 6$.
The number of such polynomials, then, is $\sum \limits _{n = 3} ^{10} S_n S_{n-1}$ because $n$ plays the role of $a+c$ and $n-1$ plays the role of $b+d$. This produces
$$\sum \limits _{n = 3} ^6 S_n S_{n-1} + S_7 S_6 + S_8 S_7 + S_9 S_8 + S_{10} S_9 = \\
\sum \limits _{n = 3} ^6 (n-1) (n-2) + 4 \cdot 5 + 3 \cdot 4 + 2 \cdot 3 + 1 \cdot 2 = 40 + 20 + 12 + 6 + 2 = 80 .$$
The correct answer, then, is B.
A: Your approach of looking for the number of solutions to $1-a+b-c+d=0$ is excellent.
Let $f(n)$ be the number of ways of choosing two (possibly identical) numbers $h,k$ from $\{1,2,3,4,5\}$ so that $h-k=n$. It is easy to check that $f(n)=1,2,3,4,5,4,3,2,1$ for $n=-4,-3,-2,\dots,3,4$ and 0 for all other $n$.
So $1-a+b$ must be one of $-3,-2,\dots,4,5$. The number of $(a,b)$ giving each of these is $1,2,3,4,5,4,3,2,1$. The number of $(c,d)$ giving minus those values is $2,3,4,5,4,3,2,1,0$. So the total number of solutions is $1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot4+4\cdot3+3\cdot2+2\cdot1=80$.
So the answer to the multiple choice question is (B).
