Do you know the reverse process? 2

Algebra Level 5

$f(x)=ax^3+bx^2+cx+d$ such that $a,b,c,d \epsilon N$

Gives,

$f(1)+f(2)=51$

$f(2)+f(3)=111$

$f(3)+f(4)=213$ and,

$f(5)=231$

Then evaluate ,

$(\sum a,b,c,d)(\prod a,b,c,d)$

This problem is original.

The answer is 1350.

1 solution

Chew-Seong Cheong
Oct 23, 2015

It is given that $f(x) = ax^3 + bx^2 + cx + d$ , where $a,b,c,d \in N$ . It is also given that:

$\begin{cases} f(1)+f(2) & = 9a+5b+3c+2d & = 51 &...(1) \\ f(2)+f(3) & = 35a + 13b + 5c + 2d & = 111 &...(2) \\ f(3) + f(4) & = 91a+25b+7c+2d & = 213 &...(3) \\ f(5) & = 125a+25b+5c+d & = 231 &...(4) \end{cases}$

From (4), for $a,b,c,d \in N$ , $a$ can only be $1$ because if $a=2$ and the minimum value of $f(5) = 250 + 25+5+1 = 281 > 231$ , when $b=c=d=1$ . Substituting $\color{#3D99F6}{a = 1}$ in the rest of the equations, we have:

$\begin{cases} (1): & 5b+3c+2d & = 42 &...(1a) \\ (2): & 13b + 5c + 2d & = 76 &...(2a) \\ (3): & 25b+7c+2d & = 122 &...(3a) \end{cases}$

$\begin{array} {rlll} (2a)-(3a): & 8b+2c = 34 & \Rightarrow 4b+c = 17 &...(5) \\ (3a)-(2a): & 12b+2c = 46 & \Rightarrow 6b+c = 23 &...(6) \\ (6)-(5): & 2b = 6 & \Rightarrow \color{#3D99F6}{b = 3} \\ (5): & 4(3)+c = 17 & \Rightarrow \color{#3D99F6}{c = 5} \\ (1a): & 5(3)+3(5)+2d = 42 & \Rightarrow \color{#3D99F6}{d = 6} \end{array}$

$\Rightarrow (a+b+c+d)abcd = (1+3+5+6)(1\times 3\times 5\times 6) = 15 \times 90 = \boxed{1350}$

$\color{#3D99F6}{\text{Previous solution using matrix calculations.}}$

$\begin{cases} f(1) = a+b+c+d \\ f(2) = 8a + 4b + 2c + d \\ f(3) = 27a + 9b + 3c + d \\ f(4) = 64a + 16b + 4c + d \\ f(5) = 125a + 25b + 5c + d \end{cases}$

$\Rightarrow \begin{cases} f(1)+f(2) & = 9a + 5b + 3c + 2d & =51 \\ f(2)+f(3) & = 35a + 13b + 5c + 2d & =111 \\ f(3)+f(4) & = 91a + 25b + 7c + 2d & =213 \\ f(5) & = 125a + 25b + 5c + 1d & =231 \end{cases}$

$\begin{aligned} \Rightarrow \begin{pmatrix} 9 & 5 & 3 & 2 \\ 35 & 13 & 5 & 2 \\ 91 & 25 & 7 & 2 \\ 125 & 25 & 5 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} & = \begin{pmatrix} 51 \\ 111 \\ 213 \\ 231 \end{pmatrix} \\ \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} & = \begin{pmatrix} 9 & 5 & 3 & 2 \\ 35 & 13 & 5 & 2 \\ 91 & 25 & 7 & 2 \\ 125 & 25 & 5 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 51 \\ 111 \\ 213 \\ 231 \end{pmatrix} \\ & = \frac{1}{90} \begin{pmatrix}-7 & 20 & -17 & 7 \\ 74 & -195 & 150 & -60 \\ -254 & 564 & -379 & 136 \\ 270 & -450 & 269 & -90 \\ \end{pmatrix} \begin{pmatrix} 51 \\ 111 \\ 213 \\ 231 \end{pmatrix} \\ & = \begin{pmatrix} 1 \\ 3 \\ 5 \\ 6 \end{pmatrix} \end{aligned}$

$\Rightarrow (a+b+c+d)abcd = 15 \times 90 = \boxed{1350}$

Any other method? This is too tedius . That's why i chose to click discuss solution

Aakash Khandelwal - 5 years, 7 months ago

The other method I know is Gaussian elimination, which is as tedious. I actually used Excel spreadsheet to do the matrix calculations.

Chew-Seong Cheong - 5 years, 7 months ago

Aakash, I have found a simple solution. See the revised solution.

Chew-Seong Cheong - 5 years, 6 months ago

Do you know the reverse process? 2

The answer is 1350.

1 solution

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