Bogdanov I.I.

A number not less than 10 digits, in its notation only two different digits are used, and identical digits do not stand next to each other.

By what highest power of two can such a number be divisible?

#
Let's cut off the "tail" of the last 8 digits from the number in the condition. The remaining number ends with 8 zeros, so it is divisible by $2^{8}$. The "tail" has the form $\frac{a b a b a b a b}{}=\overline{a b} \cdot 1010101$, where $a$ and $b$ are digits. It is clear that the twos in the prime factorization of the "tail" come only from the number $\overline{a b}$. The highest power of two by which it can be divisible is no more than the sixth ( $2^{7}$ is already three-digit). Then the original number is divisible by no more than the sixth power of two. This is achieved for the number 6464646464.

## Answer

By $2^{6}$.

## [Divisibility of numbers. General properties]

Little Man gave 111 candies to Karlson. Some of them they ate together, $45 \%$ of the remaining candies went to Karlson for lunch, and a third of the candies left after lunch were found by Fröken Bok during cleaning. How many candies did she find?

## Solution

$45 \%$ is $9 / 20$ of the number of candies remaining before Karlson's lunch. Therefore, this number is a multiple of 20. Let's denote it as 20n. Note that

$n \leq 5$. After lunch, 11n candies remained, and Fröken Bok found a third of this amount. Therefore, $n$ is a multiple of 3.

Thus, $n=3$, and Fröken Bok found 11 candies.

## Answer

11 candies.