7. How many four-digit numbers $\overline{a b c d}$ are there such that the three-digit number $\overline{a b c}$ is divisible by 4 and the three-digit number $\overline{b c d}$ is divisible by 3?

The use of a pocket calculator or any reference materials is not allowed.

## 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.  7th grade - elementary school

In addition to the final result, the process is also graded. To receive full credit, all solutions must be found and it must be confirmed that there are no others, the process must be written down, and the conclusions must be justified. Problems 1. - 5. are worth six points each, and 6. and 7. are worth ten points each.

## Problems worth 6 points:
7. How many four-digit numbers $\overline{a b c d}$ are there such that the three-digit number $\overline{a b c}$ is divisible by 4 and the three-digit number $\overline{b c d}$ is divisible by 3?

## Solution.

$\overline{a b c d}$ is a four-digit number, and $\overline{a b c}$ and $\overline{b c d}$ are three-digit numbers, so $a, b \neq 0$.

1 POINT

If $\overline{a b c}$ is divisible by 4, then the two-digit ending $\overline{b c}$ is divisible by 4, so the two-digit endings can be:

12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96.

2 POINTS

For eight two-digit endings, 12, 24, 36, 48, 60, 72, 84, and 96, the sum of the digits is divisible by 3, so the digit $d$ can be chosen in four ways, i.e., $d \in\{0,3,6,9\}$.

1 POINT

Since the digit $a$ can be chosen in 9 ways, there are $8 \cdot 9 \cdot 4=288$ such four-digit numbers.

For the remaining fourteen two-digit endings, the sum of the digits is not divisible by 3, so the digit $d$ can be chosen in 3 ways, i.e., $d \in\{1,4,7\}$ or $d \in\{2,5,8\}$.

1 POINT

Since the digit $a$ can be chosen in 9 ways, there are $14 \cdot 9 \cdot 3=378$ such four-digit numbers.

There are $288+378=666$ such four-digit numbers.

Note: If a student does not list the two-digit numbers $\overline{b c}$ divisible by 4 but instead argues that there are 22 such numbers, and among them, 8 are divisible by 3, they receive 3 POINTS.

For example: The first two-digit number divisible by four is 12, and the last is 96.

Since $12=3 \cdot 4$, and $96=24 \cdot 4$, there are $24-2=22$ such multiples of four.

Of these 22 multiples of four, every third one is also a multiple of three. The first such is 12, and the last is 96.

Since $12=3 \cdot 4=(1 \cdot 3) \cdot 4$, and $96=24 \cdot 4=(8 \cdot 3) \cdot 4$, there are 8 such multiples of three.

## SCHOOL/CITY COMPETITION IN MATHEMATICS