We are to find a two-digit natural number $a$ such that $a + 9$ is a multiple of 7 and $a + 8$ is a multiple of 13. First of all, $a + 9$ and $a + 8$ can be represented as $$a + 9 = \mathbf { M } m , \quad a + 8 = \mathbf { N O } n ,$$ where $m$ and $n$ are natural numbers. From these two equalities, we have $$\mathbf { M } m - \mathbf { N O } n = \mathbf { P } .$$ Since the pair of $m = \mathbf { Q }$ and $n = \mathbf { R }$ is an integral solution of (1), we have $$\mathbf { M } ( m - \mathbf { Q } ) = \mathbf { NO } ( n - \mathbf { R } ) .$$ From (2), a natural number $n$ satisfying (1) can be represented as $$n = \mathbf { S }$$ where $k$ is an integer. Thus $$a = \mathbf { U V } k + \mathbf { W } ,$$ and since $a$ is a two-digit natural number, $a = \mathbf { X Y }$.
We are to find a two-digit natural number $a$ such that $a + 9$ is a multiple of 7 and $a + 8$ is a multiple of 13.
First of all, $a + 9$ and $a + 8$ can be represented as
$$a + 9 = \mathbf { M } m , \quad a + 8 = \mathbf { N O } n ,$$
where $m$ and $n$ are natural numbers. From these two equalities, we have
$$\mathbf { M } m - \mathbf { N O } n = \mathbf { P } .$$
Since the pair of $m = \mathbf { Q }$ and $n = \mathbf { R }$ is an integral solution of (1), we have
$$\mathbf { M } ( m - \mathbf { Q } ) = \mathbf { NO } ( n - \mathbf { R } ) .$$
From (2), a natural number $n$ satisfying (1) can be represented as
$$n = \mathbf { S }$$
where $k$ is an integer. Thus
$$a = \mathbf { U V } k + \mathbf { W } ,$$
and since $a$ is a two-digit natural number, $a = \mathbf { X Y }$.