## Task 3 - 251213

Determine all natural numbers $n$ that have the following properties:

(1) $n$ leaves a remainder of 1 when divided by 3,

(2) $n^{2}$ leaves a remainder of 1 when divided by 11,

(3) it holds that: $100<n<200$.
When $n=11 k+r$ with a natural number $k$ and $r=0,1,2, \ldots, 10$, then

$$
n^{2}=11\left(11 k^{2}+2 k r\right)+r^{2}=11 m+s
$$

with a natural number $m$ and $s=0,1,4,9,5,3,3,5,9,4,1$. Therefore, (2) is exactly satisfied by the numbers of the form

$$
n=11 k+1 \quad \text { or } \quad n=11 k+10
$$

The conditions (3) and (2) are thus exactly satisfied by the 18 numbers 111, 122, ..., 199 and 109, 120, ..., 197. For these numbers, (1) needs to be verified.

A number of the form (4) satisfies (1) exactly if, with a natural number $p$, also $n=3 p+1$, that is,

$$
11 k+1=3 p+1 \quad \text { or } \quad 11 k+10=3 p+1
$$

or, equivalently,

$$
11 k=3 p \quad \text { or } \quad 11 k=3(p-3)
$$

For this, $3 \mid k$, so $k=3 q$ with a natural number $q$, is necessary and sufficient; hence and according to (4), (1) and (2) are exactly satisfied by the numbers of the form

$$
n=33 q+1 \quad \text { or } \quad n=33 q+10
$$

(1), (2), (3) are thus exactly satisfied by the numbers 133, 166, 199 and 109, 142, 175.

Adapted from [5]