An algebra problem by Justin Tuazon

Algebra Level 4

Let $P(x)$ be a polynomial with a degree of 6. $P(x)$ satisfies the following conditions: $\begin{cases} P(0)=5041 \\ P(1)=3 \\ P(2)=7 \\ P(3)=13 \\ P(4)=21 \\ P(5)=31 \\ P(6)=43 \\ \end{cases}$

Find the sum of the digits of the value of $P(9)$ .

The answer is 10.

1 solution

Ravi Dwivedi
Jul 13, 2015

Clearly $P(x)=x^2+x+1$ for $x=1,2,3,4,5,6$ We can write $P(x)=a(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)+x^2+x+1$ where $a$ is the constant to be determined.

$P(0)=5041$ $\implies a(6!)+1=5041\\$ $\implies a=7\\$

$P(x)=7(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)+x^2+x+1$

Putting $x=9$ yields $P(9)=141211$

Sum of digits= $1+4+1+2+1+1=\boxed{10}$

Moderator note:

Standard approach of using the Remainder-Factor Theorem to figure out the Polynomial Interpolation.

Correct me if I'm wrong, but shouldn't the final equation for $P(x)$ have $x^{2}+x+1$ added to it? Also, upon putting $x=9$ into the equation, I got $7 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3=141120$ ; then, once you add in $9^{2}+9+1=91$ , you get $141120+91=141211$ . $1+4+1+2+1+1=10.$ How did you get $20251$ ?

Yee-Lynn Lee - 5 years, 9 months ago

Sorry.......that was a typo........thanks for pointing it out

I have done correct in my copy and here i have mistaken.Now i have edited the solution.

Thanks once again

Ravi Dwivedi - 5 years, 9 months ago

Don't forget to change the final answer too!

Yee-Lynn Lee - 5 years, 9 months ago

An algebra problem by Justin Tuazon

The answer is 10.

1 solution

Moderator note:

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