## Task $8 / 65$

a) Determine the smallest two natural numbers for which the number $Z=248011 n-1$ is divisible by 36.

b) Prove that there is no natural number $n$ for which the number $Z^{\prime}=248001 n-1$ is divisible by 36.
A number $x$ is divisible by 36 if and only if it is divisible by both 4 and 9.

1. According to the known divisibility rule, a number $x$ is divisible by 4 if and only if the number formed by the last two digits of $x$ is divisible by 4.

The relationship $4 \mid(11 n-1)$ is satisfied for every integer $n$ that is an element of the set $M_{1}=$ $\{3 ; 7 ; 11 ; \ldots\}$, i.e., the set consisting of the terms $a_{k}$ of the arithmetic sequence $\left\{a_{k}\right\}=\{4 k-1\}$ with $k=$ $1 ; 2 ; 3 ; \ldots$.

2. The number $Z_{1}=248011$ leaves a remainder of 7 when divided by 9, so the $n$-fold consequently leaves a remainder of $7 n$ (which is not the smallest possible remainder). The number $Z=248011 n-1$ therefore leaves a remainder of $7 n-1$ when divided by 9, and it is divisible by 9 if and only if $7 n-1$ is divisible by 9.
3. For those numbers $n$ that belong to both the set $M_{1}$ and the set $M_{2}$, i.e., the elements of the intersection $M_{1} \cap M_{2}$, both divisibility by 4 and divisibility by 9, and thus also divisibility by 36, are established. It is

$$
M=M_{1} \cap M_{2}=\{31 ; 67 ; 103 ; 139 ; \ldots\}
$$

The smallest two of these numbers are $n_{1}=31$ and $n_{2}=67$. Consequently, the two sought numbers are

$$
248011 n_{1}-1=7688340 \quad ; \quad 248011 n_{2}-1=16616736
$$

b) The number $Z_{1}^{\prime}=248001$ is divisible by 3 because its digit sum is divisible by 3. Then $Z_{2}^{\prime}=248001 n$ is also divisible by 3. Therefore, $Z^{\prime}=248001 n-1$ cannot be divisible by 3, and thus also not by 9 and consequently not by 36.