7. Determine all natural numbers of the form $\overline{a b c d e f}$ such that in their representation they have the digits 1, 2, 3, 4, 5, and 6, none of which repeat, and the following conditions are met:

$\overline{a b}$ is divisible by 2,

$\overline{a b c}$ is divisible by 3,

$\overline{a b c d}$ is divisible by 4,

$\overline{a b c d e}$ is divisible by 5,

$\overline{a b c d e f}$ is divisible by 6.

## Ministry of Science and Education of the Republic of Croatia  Agency for Education and Upbringing  Croatian Mathematical Society

## SCHOOL/CITY COMPETITION IN MATHEMATICS  January 26, 2023.  8th grade - elementary school

In addition to the final result, the procedure is also graded. To receive full points, it is necessary to find all solutions and confirm that there are no others, write down the procedure, and justify your conclusions. Problems 1. - 5. are worth six points each, and 6. and 7. are worth ten points each.

## Problems worth 6 points:
First solution.

Since we have digits from 1 to 6 and the number $\overline{a b c d e}$ is divisible by 5, then the digit $e$ must be 5.

$$
e=5
$$

1 POINT

The numbers $\overline{a b}, \overline{a b c d}$, and $\overline{a b c d e f}$ are divisible by 2, so the digits $b, d$, and $f$ can be 2, 4, or 6. The digits $a$ and $c$ can be 1 or 3.

$$
\begin{gathered}
b, d, f \in\{2,4,6\} \\
a, c \in\{1,3\}
\end{gathered}
$$

2 POINTS

Let's consider the cases we get by choosing the digit $b \in\{2,4,6\}$.

First case.

Let $b=2$.

Then the number is of the form $\overline{a 2 c d 5 f}$.

Since the number $\overline{a 2 c}$ is divisible by 3, the sum of its digits $a+2+c$ must be divisible by 3.

The only possibility is that the digits $a$ and $c$ are 1 or 3.

If $a=1$ and $c=3$, then the number is of the form $\overline{123 d 5 f}$.

If $a=3$ and $c=1$, then the number is of the form $\overline{321 d 5 f}$.

1 POINT

The numbers $\overline{123 d}$ and $\overline{321 d}$ must be divisible by 4. This is only possible if $d=6$.

1 POINT

In that case, $f=4$.

1 POINT

The numbers 123654 and 321654 are divisible by 6 because they are even and the sum of their digits is 21, which is divisible by 3.

2 POINTS

The desired numbers are 123654 and 321654.

Second case.

Let $b=4$.

Then the number is of the form $\overline{a 4 c d 5 f}$.

Since the number $\overline{a 4 c}$ is divisible by 3, the sum of its digits $a+4+c$ must be divisible by 3.

There are no digits $a$ and $c$ that satisfy this condition.

1 POINT

Note 1: The student is awarded 1 POINT if they argue in any way that $b \neq 4$.

Third case.

Let $b=6$.

Then the number is of the form $\overline{a 6 c d 5 f}$.

Since the number $\overline{a 6 c}$ is divisible by 3, the sum of its digits $a+6+c$ must be divisible by 3.

There are no digits $a$ and $c$ that satisfy this condition.

1 POINT

Note 2: The student is awarded 1 POINT if they argue in any way that $b \neq 6$.

..TOTAL 10 POINTS