8,9
Let $N$ be an even number that is not divisible by 10. Find the tens digit of the number $N^{20}$.
From the condition, it follows that $N$ is not divisible by 5. Therefore, according to Fermat's Little Theorem, $N^{4} \equiv 1(\bmod 5)$. Therefore, $N^{20}-1=\left(N^{4}-1\right)\left(N^{16}+N^{12}+N^{8}+N^{4}+1\right)$ is divisible by $5^{2}$. Since the number $N^{20}$ is even, it ends in 26 or 76. But a number ending in 26 is not divisible by 4.

## Answer

7

![](https://cdn.mathpix.com/cropped/2024_05_06_bf0205b4cbbe4b58a1adg-29.jpg?height=151&width=1252&top_left_y=13&top_left_x=0)

It is known that in the decimal representation of the number $2^{29}$, all digits are different. Is there a digit 0 among them?

## Solution

Notice that $2 \cdot 2^{29}=2^{30}=(1024)^{3}5 \cdot 10^{8}$. Therefore, the number $2^{29}$ has exactly nine digits.

If there is no zero among them, then the sum of the digits in the decimal representation of this number is $1+2+\ldots+9=45$. From this, it follows that $2^{29}$ is divisible by 3, which is not true. Contradiction.

## Answer

There is.