\section*{Problem 1 - 271031}

Determine the number of all pairs \((x ; y)\) that satisfy the following conditions (1) to (4)!

(1) \(x\) and \(y\) are three-digit natural numbers.

(2) The three digits of \(x\) are all different from each other.

(3) The three digits of \(x\) are also the three digits of \(y\), but in a different order.

(4) It holds that \(x-y=45\).
}

The conditions (1) to (4) are satisfied if and only if the digits of \(x\) are three natural numbers \(a, b, c\) for which the following holds: 

\[
1 \leq a \leq 9, \quad 0 \leq b \leq 9, \quad 0 \leq c \leq 9, \quad a \neq b, \quad a \neq c, \quad b \neq c
\]

and condition (4) is satisfied by the number \(x=100 a+10 b+c\) and the number \(y\) for which exactly one of the following cases I to V applies:

(I) It holds that \(y=100 a+10 c+b\)

(II) It holds that \(y=100 b+10 a+c\) and in addition to (5) also \(b \neq 0\).

(III) It holds that \(y=100 b+10 c+a\) and in addition to (5) also (6).

(IV) It holds that \(y=100 c+10 a+b\) and in addition to (5) also \(c \neq 0\).

(V) It holds that \(y=100 c+10 b+a\) and in addition to (5) also (7).

In case I, (4) is equivalent to \(x-y=9(b-c)\) and thus to \(b-c=5\), which under the conditions (5) is satisfied exactly by

\[
(b ; c)=(5 ; 0),(6 ; 1),(7,2),(8 ; 3),(9 ; 4)
\]

together with those numbers \(a=1, \ldots, 9\) that also satisfy \(a \neq b\) and \(a \neq c\). For each of the 9 numbers \(a=1, \ldots, 9\), exactly 4 pairs \((b ; c)\) remain.

In case II, (4) is not satisfiable because 45 is not a multiple of 90.

In case III, (4) is equivalent to \(x-y=9(11 a-10 b-c)\) and thus to \(11 a-10 b-c=5\) and this to \(11 a-5=10 b+c\).

Due to (5), (6), the values \(a \geq 5\) are excluded; for these values, calculating \(11 a-5\) would result in a two-digit number with the tens digit \(a\), leading to the contradiction \(a=b\). Furthermore, \(a=1\) is excluded, as it would result in a one-digit number.

The remaining values \(a=2,3,4\) satisfy \(11 a-5=17,28,39\) and each together with

\[
(b ; c)=(1 ; 7),(2 ; 8),(3 ; 9)
\]

all conditions (5), (6). Thus, (1) to (4) are satisfied in case III by exactly 3 pairs \((x ; y)\).

In case IV, (4) is equivalent to \(x-y=9(10 a+b-11 c)\) and thus to \(10 a+b-11 c=5\) and this to \(11 c+5=10 a+b\).

Due to (5), (7), the values \(c \leq 4\) and the value \(c=9\) are excluded; \(c=0\) does not satisfy (7), and for \(1 \leq c \leq 4\) or \(c=9\), \(11 c+5\) would result in a two-digit number with the tens digit \(a=c\) or a three-digit number.

The remaining values \(c=5,6,7,8\) satisfy \(11 c+5=60,71,82,93\) and each together with

\[
(a ; b)=(6 ; 0),(7 ; 1),(8 ; 2),(9 ; 3)
\]

all conditions (5), (7). Thus, (1) to (4) are satisfied in case IV by exactly 4 pairs \((x ; y)\).

In case V, (4) is not satisfiable because 45 is not a multiple of 99.

The total number of pairs \((x ; y)\) that satisfy (1) to (4) is thus \(36+3+4=43\).

Adapted from \([5]\)