Runner $\underline{\text { A.V. }}$.

Find the smallest natural number divisible by 80, in which two of its different digits can be swapped so that the resulting number is also divisible by 80.

#
Note that both before and after the rearrangement of digits, the number is divisible by 10 and therefore must end in 0. We will show that there are no three-digit numbers with the property described in the condition. Indeed, if \(100a + 10b = 80k\) and \(100b + 10a = 80l\), where \(a \geq 1\). In this case, \(900(a-1) = 80m\), or \(45(a-1) = 4m\). Therefore, \(a-1\) is divisible by 4, which means \(a = 5\) or 9. Already with \(a = 5\) and \(b = 2\), we get the number \(1520 = 19 \cdot 80\), which satisfies the condition: \(5120 = 64 \cdot 80\).

## Answer

Author: Beguni A.V.

Nезнайка (Nезнайка) is only familiar with decimal logarithms and believes that the logarithm of the sum of two numbers is equal to the product of their logarithms, and the logarithm of the difference of two numbers is equal to the quotient of their logarithms. Can Nезнайка find at least one pair of numbers for which both of these equalities are indeed true simultaneously?

## Solution

Suppose such numbers \(x\) and \(y\) exist, where \(y > 1\). In this case, all logarithms are positive. Adding the equations of the system and applying the Cauchy inequality:

\[
\lg \left(x^2 - y^2\right) = \lg (x + y) + \lg (x - y) = \lg x \lg y + \lg x / \lg y \geq 2 \lg x = \lg x^2
\]
From this, \(x^2 - y^2 \geq x^2\), which is impossible.

## Answer

No, it cannot.