Let $A, B, C, D$ be positive integers, each with $k$ digits, where every digit of $A$ is 3, every digit of $B$ is 4, every digit of $C$ is 6, and every digit of $D$ is 7. Additionally, the number $E$ has $2k$ digits, all of which are 5. Calculate the value of the following expression:

$$
E-A \cdot D-B \cdot C+1
$$
I. solution. Let the number written with $k$ ones be denoted by $Q$, so $A=3 Q, B=4 Q, C=6 Q, D=7 Q, E=5\left(10^{k} \cdot Q+Q\right)$, and

\[
K=5\left(10^{k}+1\right) \cdot Q-21 Q^{2}-24 Q^{2}+1=5 Q\left(10^{k}+1-9 Q\right)+1
\]

On the other hand, the number $10^{k}$ is one greater than the number written with $k$ nines, which is $9 Q$, so the expression in parentheses has a value of 2, and $K=10 Q+1$. This is obtained from $Q$ by appending a 1, so it is the number written with $k+1$ ones.

Judit Pongó (Makó, József A. g. I. o. t.)

II. solution. With the above notation, $A D+B C=45 Q^{2}=5 A^{2}$, which is five times the square of the number written with $k$ threes. In exercise 1003, we saw that this number is equal to the difference between the number written with $2k$ ones, denoted by $R$, and the number written with $k$ twos. Therefore,

\[
K=5 R-5 A^{2}+1=5 R-5(R-2 Q)+1=10 Q+1
\]

which is the number written with $k+1$ ones.

László Mérő (Budapest, Berzsenyi D. g. II. o. t.)[^0]


[^0]:    ${ }^{1}$ K. M. L. 32 (1966) 123. o.