Question

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

• A. 110
• B. 143
• C. 165
• D. 220
• E. 286

Answer 1

Answer: (D)

220

Answer 2

The polynomial P(x) is of degree at most 3, so it can be written as:

$P(x) = ax^3 + bx^2 + cx + d$

The coefficients a, b, c, d are elements of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

We are given the condition P(-1) = -9. Substitute x = -1 into the polynomial:

$P(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d$

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

So, we have the equation:

$-a + b - c + d = -9$

Rearranging the terms, we get:

$b + d = a + c - 9$

Let $S_1 = b+d$ and $S_2 = a+c$. The equation becomes $S_1 = S_2 - 9$.

Since a, b, c, d are integers from 0 to 9, the possible range for $S_1$ and $S_2$ is:

Minimum sum: $0+0 = 0$ Maximum sum: $9+9 = 18$ So, $0 \le S_1 \le 18$ and $0 \le S_2 \le 18$.

From $S_1 = S_2 - 9$: Since $S_1 \ge 0$, we must have $S_2 - 9 \ge 0 \implies S_2 \ge 9$. Since $S_1 \le 18$, we must have $S_2 - 9 \le 18 \implies S_2 \le 27$. Combining with $S_2 \le 18$, the valid range for $S_2$ is $9 \le S_2 \le 18$.

Now, we need to find the number of ways to obtain a sum $k$ from two coefficients chosen from {0, 1, ..., 9}. Let this be $N(k)$. For two non-negative integers $x, y$ such that $x,y \in \{0, 1, ..., n\}$, the number of solutions to $x+y=k$ is:

• If $0 \le k \le n$: $k+1$ ways
• If $n < k \le 2n$: $2n - k + 1$ ways

In our case, $n=9$. So, $N(k)$ is:

• For $0 \le k \le 9$: $N(k) = k+1$
• For $10 \le k \le 18$: $N(k) = 2 \times 9 - k + 1 = 19 - k$

We need to sum the product of the number of ways to get $S_2$ and the number of ways to get $S_1 = S_2 - 9$, for all valid values of $S_2$. The total number of polynomials is $\sum_{S_2=9}^{18} N(S_2) \times N(S_2 - 9)$.

Let's calculate the terms:

1. If $S_2 = 9$: $S_1 = 0$. $N(9) = 9+1 = 10$. $N(0) = 0+1 = 1$. Product = $10 \times 1 = 10$.
2. If $S_2 = 10$: $S_1 = 1$. $N(10) = 19-10 = 9$. $N(1) = 1+1 = 2$. Product = $9 \times 2 = 18$.
3. If $S_2 = 11$: $S_1 = 2$. $N(11) = 19-11 = 8$. $N(2) = 2+1 = 3$. Product = $8 \times 3 = 24$.
4. If $S_2 = 12$: $S_1 = 3$. $N(12) = 19-12 = 7$. $N(3) = 3+1 = 4$. Product = $7 \times 4 = 28$.
5. If $S_2 = 13$: $S_1 = 4$. $N(13) = 19-13 = 6$. $N(4) = 4+1 = 5$. Product = $6 \times 5 = 30$.
6. If $S_2 = 14$: $S_1 = 5$. $N(14) = 19-14 = 5$. $N(5) = 5+1 = 6$. Product = $5 \times 6 = 30$.
7. If $S_2 = 15$: $S_1 = 6$. $N(15) = 19-15 = 4$. $N(6) = 6+1 = 7$. Product = $4 \times 7 = 28$.
8. If $S_2 = 16$: $S_1 = 7$. $N(16) = 19-16 = 3$. $N(7) = 7+1 = 8$. Product = $3 \times 8 = 24$.
9. If $S_2 = 17$: $S_1 = 8$. $N(17) = 19-17 = 2$. $N(8) = 8+1 = 9$. Product = $2 \times 9 = 18$.
10. If $S_2 = 18$: $S_1 = 9$. $N(18) = 19-18 = 1$. $N(9) = 9+1 = 10$. Product = $1 \times 10 = 10$.

Summing these products: Total = $10 + 18 + 24 + 28 + 30 + 30 + 28 + 24 + 18 + 10$ Total = $(10+10) + (18+18) + (24+24) + (28+28) + (30+30)$ Total = $20 + 36 + 48 + 56 + 60$ Total = $56 + 104 + 60$ Total = $160 + 60 = 220$.