A four-digit integer $m$ and the four-digit integer obtained by reversing the order of the digits of $m$ are both divisible by 45. If $m$ is divisible by 7, what is the greatest possible value of $m$?
Let the integer obtained by reversing the digits of $m$ be $n$. $m$ and $n$ are both divisible by $45$, which means that they are both divisible by $5$.  Thus, they both have units digits of $5$ or $0$.  If one has a units digit of $0$, the other will have a leading digit of $0$, which cannot be.  So both end in $5$; reversing them shows that both start with $5$ as well.  Since $m$ is divisible by $45$ and by $7$, it is divisible by $7(45)=315$.  There are four multiples of $315$ between $5000$ and $6000$: $5040$, $5355$, $5670$, and $5985$. $5985$ is the largest, and it is easy to see that it and its reversal, $5895$, meet all requirements.  So $\boxed{5985}$ is the answer.
5985
Level 5
Number Theory
Problem: A four-digit integer $m$ and the four-digit integer obtained by reversing the order of the digits of $m$ are both divisible by 45. If $m$ is divisible by 7, what is the greatest possible value of $m$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to understand the conditions for divisibility by 45. A number is divisible by 45 if it is divisible by both 5 and 9. 

For divisibility by 5, the last digit of the four-digit integer $m$ must be either 0 or 5. Since $m$ is a four-digit number, it cannot end with 0, as that would make it a three-digit number. Therefore, the last digit of $m$ must be 5.

Next, for divisibility by 9, the sum of the digits of $m$ must be divisible by 9. Let $m$ be represented as $abcd$, where $d = 5$. The sum of the digits is $a + b + c + 5$. 

Now, we also need to consider the reversed integer, which is $dcba$. This number must also be divisible by 45. The last digit of $dcba$ is $a$, so for $dcba$ to be divisible by 5, $a$ must be 0 or 5. Since $m$ is a four-digit number, $a$ cannot be 0, so $a$ must also be 5. 

Thus, we can express $m$ as $5bc5$. The sum of the digits now becomes $5 + b + c + 5 = b + c + 10$. This sum must be divisible by 9.

Next, we need to find the values of $b$ and $c$ such that $b + c + 10$ is divisible by 9. This can be rewritten as $b + c \equiv -10 \equiv -1 \equiv 8 \mod 9$. Therefore, $b + c$ must be congruent to 8 modulo 9.

Now, we also have the condition that $m$ must be divisible by 7. We can express $m$ as $5000 + 100b + 10c + 5$. 

To find the greatest possible value of $m$, we will maximize $b$ and $c$. The maximum value for $b$ and $c$ is 9. However, we need to ensure that $b + c \equiv 8 \mod 9$. 

Testing combinations:
- If $b = 9$, then $c$ must be $-1$, which is not valid.
- If $b = 8$, then $c$ must be $0$ (valid).
- If $b = 7$, then $c$ must be $1$ (valid).
- If $b = 6$, then $c$ must be $2$ (valid).
- If $b = 5$, then $c$ must be $3$ (valid).
- If $b = 4$, then $c$ must be $4$ (valid).
- If $b = 3$, then $c$ must be $5$ (valid).
- If $b = 2$, then $c$ must be $6$ (valid).
- If $b = 1$, then $c$ must be $7$ (valid).
- If $b = 0$, then $c$ must be $8$ (valid).

Now we check the combinations for divisibility by 7:
- For $m = 5805$, $5805 \div 7 = 830.7142857$ (not divisible).
- For $m = 5795$, $5795 \div 7 = 827.8571429$ (not divisible).
- For $m = 5785$, $5785 \div 7 = 826.4285714$ (not divisible).
- For $m = 5775$, $5775 \div 7 = 825$ (divisible).
- For $m = 5965$, $5965 \div 7 = 852.1428571$ (not divisible).
- For $m = 5985$, $5985 \div 7 = 855$ (divisible).

After checking all combinations, the greatest valid $m$ that meets all conditions is $5985$.<|END_OF_LATENT|> The final answer is $5985$.