Trailer $\underline{\text { P. }}$.
Find the smallest number of the form
a) $\left|11^{k}-5^{n}\right|$;
b) $\left|36^{k}-5^{n}\right|$;
c) $\left|53^{k}-37^{n}\right|$, where $k$ and $n$ are natural numbers.
a) If $11^{k}>5^{n}$, then the number $\left|11^{k}-5^{n}\right|$ ends in 6; if $11^{k}<5^{n}$, then it ends in 4.

Since $\left|11^{2}-5^{3}\right|=4$, then 4 is the smallest number of the form $\left|11^{k}-5^{n}\right|$.

b) The number $\left|36^{k}-5^{n}\right|$ ends in 1 or 9.

Note that $\left|36^{1}-5^{2}\right|=11$, and we will prove that 11 is the smallest number of this form, that is, that $\left|36^{k}-5^{n}\right|$ cannot equal 1 or 9.

1) If $36^{k}-5^{n}=1$, then $\left(6^{k}-1\right)\left(6^{k}+1\right)=5^{n}$. But $6^{k}+1$ cannot be a power of 5, as it ends in 7.
2) If $5^{n}-36^{k}=9$, then $5^{n}=9+36^{k}$, which is also impossible, as the right side is divisible by 9, but the left side is not.

c) $\left|53^{k}-37^{n}\right|=\left|(4 \cdot 13+1)^{k}-(4 \cdot 9+1)^{n}\right|$, so it is divisible by 4.

On the other hand, $\left|53^{k}-37^{n}\right|=\left|(6 \cdot 9-1)^{k}-(4 \cdot 9+1)^{n}\right|$. Therefore, when such a number is divided by 9, the remainder can only be 0, 2, or -2.

The smallest number that satisfies these two conditions is $16=53-37$.

## Answer

a) 4; b) 11; c) 16.