2021/073)Consider polynomials P(x) of degree at most 3 each of whose coefficients is an element of { 0,1,2,3,4,5,6,7,8,9}. How many such polynomials satisfy P(-1) = 9.

Because the polynomial is of degrees at most 3 with above coefficients we have 

$P(x) = ax^3+bx^2 + cx +d$ where a, b ,c, d are  element of { 0,1,2,3,4,5,6,7,8,9}.

now $P(-1) = -a + b -c + d = (b+d) - (a+c)$

now maximum sum of b+d can be 18 and as $b+d - (a+c) = 9$ so we have

$a+c =k $ and $b+d = 9+k$ and hence $0 <=<=9$

as the digits are from 0 to 9 for the sum to be k <= 9 one of the numbers can be x = 0 to k and another number can be k -x so there are k+1 possibilities

for the sum to be  $ k >=9$ one of the numbers a should be $k-9$ to $9$ and another number c shall be $k-9$ and there are $(9+1) - (k-9)$ or $19-k$ choices.

for the sum to be  $ k >=9= 9 + m$ there are  $19-(9+m)$ or $10-m$ choices.

For the value  $b+d - (a+c) = 9$ a + c can be k and b+d can be k+9 so there are (k+1)(10-k) ways 

now the value k can be from 0 to 9 so we have sum

$\sum_{k=0}^9(k+1)(10-k) =\sum_{n=1}^{10}n(11-n) = 11\sum_{n=1}^{10}n - \sum_{n=1}^{10}n^2 = 11 * \frac{10*11}{2} - \frac{10 * 11 * 21}{6} = 605- 385 = 220 $

So number of polynomials = 220