The product to 2027

Algebra Level 3

Consider

$\large a= \prod _{n=1}^{2027} \left(\sqrt [2027]2\right)^n$

Let $a = 2^k$ . Then

If $k$ is an integer, enter $k$ .
If $k$ can be expressed as a fraction in the form of $\frac{b}{c}$ , where $b$ and $c$ are coprime integers and $c > 1$ , enter $b + c$ .
If $k$ is irrational, enter $-1$ .

The answer is 1014.

1 solution

Chew-Seong Cheong
Sep 24, 2017

$\large \begin{aligned} a & = \prod_{n=1}^{2027} \left(\sqrt[2027]2\right)^n = 2^{\sum_{n=1}^{2027} n/2027} = 2^{2027 \times 2028/(2 \times 2027)} = 2^{1014} \end{aligned}$

$\implies k = \boxed{1014}$

You boxed the wrong number.

Ron Lauterbach - 3 years, 8 months ago

Thanks, I was thinking about $2^{10}$ .

Chew-Seong Cheong - 3 years, 8 months ago

The product to 2027

The answer is 1014.

1 solution

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