## Task 4 - 160924

In the following arrangement of characters

$$
\begin{array}{ccccc}
\mathrm{ab} & \mathrm{X} & \mathrm{ab} & = & \mathrm{cad} \\
\mathrm{Y} & & \mathrm{Y} & & \mathrm{Z} \\
\mathrm{ae} & \mathrm{X} & \mathrm{ae} & = & \mathrm{ffe} \\
\hline \mathrm{ff} & \mathrm{Y} & \mathrm{ff} & = & \mathrm{gg}
\end{array}
$$

the individual symbols should be replaced by elements of the respective base set so that true statements are formed in each case.

The base set for the lowercase letters $b, d, e$ is the set of digits from 0 to 9, for $a, c, f, g$ the set of digits from 1 to 9, and the base set for the uppercase letters $X, Y, Z$ is the set of operation symbols "+", "-", "." and ":". Identical symbols mean identical, different symbols mean different elements of the respective base set.

Determine whether such a substitution is possible, and if so, find all substitutions with the required properties!
Assuming a substitution has the required properties. Then $X$ cannot stand for the plus sign; because if the sum of two two-digit numbers is a three-digit number, then its first digit must be a 1; the results of

$$
\mathrm{abXab}=\mathrm{cad} \quad \text { and } \quad \text { aeXae }=\mathrm{ffe}
$$

begin with different digits.

Furthermore, neither the difference nor the quotient of two two-digit numbers can result in a three-digit number, so $X$ must stand for the sign $\cdot$.

From ffYff $=$ gg, it follows that because $ff - ff = 0$ and $ff : ff = 1$, $Y$ cannot represent the sign - or the sign :. Therefore, $Y$ stands for the sign +.

From cadZffe $=$ gg, it follows that $Z$ cannot stand for the sign :, because the quotient of two three-digit numbers cannot be a two-digit number. Therefore, $Z$ stands for the sign -. Thus, we have so far

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0991.jpg?height=184&width=404&top_left_y=1524&top_left_x=835)

Since $32^{2} > 1000$, it follows from ae $\cdot$ ae $=$ ffe that the number represented by ae can be at most 31. Because the last digits of the three numbers are the same, $e$ can only be one of the numbers 0, 1, 5, or 6.

From $\mathrm{ab} + \mathrm{ae} = \mathrm{ff}$, it follows that $e$ cannot be 0, because then $b$ and $f$ would represent the same number. Of the numbers 15, 16, 21, 25, 26, and 31 that are thus possible for ae, only the numbers 15 and 21 satisfy the condition that the hundreds and tens digits of their squares are the same.

If $a=1$ and $e=5$, then because ab + ae = ff, $f$ must be 3 and $b$ must be 8. Since $18^{2} = 324$, it follows from $\mathrm{ab} \cdot \mathrm{ab} = \mathrm{cad}$ that $a=2$, which is a contradiction.

For $a=2$ and $e=1$, it follows that $f=4$ and $b=3$. Therefore, the only substitution that can satisfy all the conditions of the problem is

$$
\begin{array}{ccccc}
23 & \cdot & 23 & = & 529 \\
+ & & + & & - \\
21 & \cdot & 21 & = & 441 \\
\hline 44 & + & 44 & = & 88
\end{array}
$$

Solutions of the 2nd round 1976 taken from [5]

### 6.18.3 III. Level 1976, Class 9