A number theory problem by Andy Hayes

Relevant wiki: Harmonic Number

$\begin{aligned} P & = \prod_{k=1}^{1000} \sqrt[k]{10} \\ & = \exp \left(\ln \prod_{k=1}^{1000} 10^\frac 1k \right) \\ & = \exp \left( \ln 10 \sum_{k=1} ^{1000} \frac 1k \right) \\ & = \exp \left( \ln 10 \cdot {\color{#3D99F6}H_{1000}} \right) & \small \color{#3D99F6} \text{where }H_n \text{ is the }n \text{th harmonic number.} \\ & \approx \exp \left( \ln 10 \cdot {\color{#3D99F6}(\gamma + \ln 1000} \right)) & \small \color{#3D99F6} \text{By: Euler-Macheroni constant }\gamma = \lim_{n \to \infty} \left(H_n - \ln n\right) \\ & \approx \exp \left( \ln 10 \cdot ({\color{#3D99F6}0.5772 + 6.9078} \right)) \\ & \approx \exp \left( \ln 10 \cdot 7.4850 \right) \\ & \approx 10^{7.4850} \end{aligned}$

$\implies \lfloor P \rfloor = \lfloor 10^{7.4850} \rfloor$ has $\boxed{8}$ digits.

Note that the product notation can be rewritten as floor( 10 * 10^ $\frac{1}{2}$ * 10^ $\frac{1}{3}$ * ... * 10^ $\frac{1}{1000}$ ).

Since the number of digits in N is given by floor( log(N) ) + 1, so:

log( 10 * 10^ $\frac{1}{2}$ * 10^ $\frac{1}{3}$ * ... * 10^ $\frac{1}{1000}$ )

= log(10) + log(10^ $\frac{1}{2}$ ) + log(10^ $\frac{1}{3}$ ) + ... + log(10^ $\frac{1}{1000}$ )

= 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{1000}$ .

... which is approximately equal to 7.485. (I used Desmos. There's probably a more math-y way to do it that I don't know about =/. ) Taking the floor function and adding 1 gives 8 .