1. For what smallest natural $k$ does the number 2016 $\cdot$ 20162016 $\cdot$ 201620162016 $\cdot$... 20162016...2016( $k$ factors) divide without remainder by $3^{67}$?
Solution.

We will show that in the case of 27 factors, the product is divisible by $3^{67}$, but not by $3^{68}$.

$2016 \cdot 20162016 \cdot \ldots \cdot \underbrace{20162016 . .2016}_{27 \text { times }}=$

$=2016^{27} \cdot(1 \cdot 1001 \cdot 1001001 \cdot \ldots \cdot \underbrace{1001001 \ldots 1001}_{27 \text { times }})=$

$=3^{54} \cdot 244^{27} \cdot(1001 \cdot 1001001 \cdot \ldots \cdot \underbrace{1001001 \ldots 1001}_{27 \text { times }})$.

244 is not divisible by 3.

$\underbrace{1001001 . . .1001}$ is not divisible by 3 if $\overline{k \vdots} 3$. $k$ ones

$\underbrace{1001001 \ldots 1001}: 3$, but not divisible by $9=3^{2}$, if $k$ ones
$k=3,6,12,15,21,24$. Their product is divisible by $3^{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_887daaf3f79b01753c00g-3.jpg?height=90&width=1038&top_left_y=159&top_left_x=194)

![](https://cdn.mathpix.com/cropped/2024_05_06_887daaf3f79b01753c00g-3.jpg?height=92&width=1080&top_left_y=261&top_left_x=195)

![](https://cdn.mathpix.com/cropped/2024_05_06_887daaf3f79b01753c00g-3.jpg?height=88&width=1113&top_left_y=372&top_left_x=195)

$3^{4}$.

$54+6+2+2+3=67$

$2016 \cdot 20162016 \cdot \ldots \cdot \underbrace{20162016 \ldots 2016}_{27 \text { times }}: 3^{67}$, but not divisible by $3^{68}$.

Answer: 27

## Recommendations for checking.

Answer without proof -0 points.

Proven that, $2016 \cdot 20162016 \cdot \ldots \cdot \underbrace{20162016 \ldots 2016}_{27 \text { times }}: 3^{67}-5$ points.