4. Four managers of responsibility shifting reported: "If they are arranged in pairs, there will be 1 left. If they are arranged in threes, there will also be 1 left. If they are arranged in fours, there will be 2 left, and if they are arranged in fives, there will also be 2 left." Should the head of the report reception department believe such a report? Determine the maximum number of true statements among these four statements (possibly all) and for each maximum set of non-contradictory statements, find the smallest number of objects being arranged, considering that there are at least a thousand.
# Solution

Let $a$ be the desired quantity. If when dividing by 4, 2 remains, then $a$ is even. But then when dividing by two, 1 cannot remain. Therefore, these two statements are contradictory and the message as a whole is false.

Remove one of the contradictory statements.

1st var.: leave the division by 3, by 4, and by 5. From the condition, it follows that the number $a-2$ is divisible by 4 and by 5. Therefore, $a=20n+2$. Also, by the condition, $a=3k+1$. Form the equation $3k+1=20n+2$. Or, equivalently,

$$
3k=20n+1 \quad \Rightarrow \quad k=\frac{20n+1}{3}
$$

The smallest $n$ for which $a \geq 1000$ is $n=50$. By simple enumeration, we find that at $n=52$ the fraction gives an integer $k$. Thus, in this case, $a=1042$.

2nd var.: leave the division by 2, by 3, and by 5. From the condition, it follows that the number $a-1$ is divisible by 2 and by 3. Therefore, $a=6n+1$. Also, by the condition, $a=5k+2$. Form the equation $5k+2=6n+1$. Or, equivalently,

$$
6n=5k+1 \Rightarrow n=\frac{5k+1}{6}
$$

The smallest $k$ for which $a \geq 1000$ is $k=200$. By simple enumeration (over odd $k$), we find that at $k=205$ the fraction gives an integer $n$. Thus, in this case, $a=1027$.

Answer: Should not believe.

If leaving the division by 3, by 4, and by 5, then $a=1042$.

If leaving the division by 2, by 3, and by 5, then $a=1027$.