MHB Find the Value of a Polynomial with Given Constraints - POTW #250 Jan 17th, 2017

[anemone]

Here is this week's POTW:

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Let $P$ be a polynomial such that $P(x)=a_0+a_1x+\cdots+a_nx^n$ where $a_0,\,a_1,\cdots$ are non-negative integer. Given that $P(1)=4$ and $P(5)=152$, find $P(6)$.

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[anemone]

Congratulations to the following members for their correct solution::)
1. kaliprasad
2. lfdahl

Solution from lfdahl:

Spoiler

All coefficients ($a_i$) belong to the set: $\{0,1,2,3,...\}$

It must be a cubic polynomial, because a quartic and higher would exceed $P(5) = 152$.

So, we´re left with the function: $P(x) = a_0+a_1x+a_2x^2+a_3x^3$.

$P(1) = 4$, implies $a_0,a_1,a_2,a_3 \in \{0,1,2,3,4\}$ under the restriction: $a_0+a_1+a_2+a_3 = 4.$

Clearly, $a_3 = 1$, because any higher value would imply: $P(5) > 152$.

On the other hand: $a_3 = 0$, is not possible, because a quadratic polynomial cannot reach the value $152$.

Thus far, we have the polynomial value: $5^3 = 125$. We still need 27 to reach $P(5)$.

The only way to obtain $27$ with the quadratic part of the polynomial is the combination:

$2 + x^2$ ($a_0=2$, $a_1=0$ and $a_2 = 1$).

Hence, $P$ has the explicit form: $P(x) = 2 + x^2+x^3$, and

$P(6) = 2 + 36 + 216 = 254.$