<Question>
What is the largest integer less than $2010$ that has a remainder of $5$ when divided by $7,$ a remainder of $10$ when divided by $11,$ and a remainder of $10$ when divided by $13$?
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the largest integer less than 2010 that satisfies three conditions: when divided by 7, the remainder is 5; when divided by 11, the remainder is 10; and when divided by 13, the remainder is 10. Hmm, this sounds like a problem that can be approached using the Chinese Remainder Theorem (CRT). Let me recall how CRT works. It's a theorem that gives a solution for a system of simultaneous congruences with coprime moduli. The moduli here are 7, 11, and 13. Are these numbers coprime? Yes, because 7 is a prime number, and 11 and 13 are also primes, so none of them share any common factors other than 1. Therefore, CRT applies here, which means there is a unique solution modulo 7*11*13. Let me compute that product first.

7*11 is 77, and 77*13 is 1001. So the modulus is 1001. That means the solutions repeat every 1001 numbers. Since we need the largest such number less than 2010, maybe the solution is 2010 - something? Wait, but first, let's find the general solution.

The problem can be written as a system of congruences:
1. N ≡ 5 mod 7
2. N ≡ 10 mod 11
3. N ≡ 10 mod 13

So we need to solve these three congruences step by step. Let me start with the first two congruences and then incorporate the third one.

<prompt_1.1> Solve the system of congruences step by step to find the general solution. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's solve N ≡ 5 mod 7 and N ≡ 10 mod 11. Let me set N = 7k + 5 for some integer k. Then substitute this into the second congruence: 7k + 5 ≡ 10 mod 11. Subtract 5 from both sides: 7k ≡ 5 mod 11. Now, we need to find k such that 7k ≡ 5 mod 11. To solve for k, we can multiply both sides by the modular inverse of 7 mod 11. What's the inverse of 7 modulo 11? Let's find a number x where 7x ≡ 1 mod 11. Testing x=8: 7*8=56, which is 1 mod 11 (since 55 is divisible by 11). So the inverse is 8. Therefore, multiplying both sides by 8: k ≡ 5*8 mod 11 => k ≡ 40 mod 11. 40 divided by 11 is 3 with a remainder of 7, so k ≡ 7 mod 11. Therefore, k can be written as 11m + 7 for some integer m. Therefore, substituting back into N: N = 7*(11m +7) +5 = 77m +49 +5 = 77m +54. So the solutions satisfying the first two congruences are N ≡ 54 mod 77.

Now, we need to incorporate the third congruence: N ≡ 10 mod 13. So we have N = 77m +54 ≡ 10 mod 13. Let's compute 77 mod 13 and 54 mod 13 to simplify this. 13*5=65, so 77 -65=12, so 77 ≡12 mod13. Similarly, 54 divided by 13: 13*4=52, so 54-52=2, so 54≡2 mod13. Therefore, substituting into N: 12m +2 ≡10 mod13. Subtract 2 from both sides: 12m ≡8 mod13. Now, solve for m: 12m ≡8 mod13. Again, find the inverse of 12 mod13. Since 12 ≡-1 mod13, so inverse of -1 mod13 is -1, which is 12. So multiply both sides by 12: m ≡8*12 mod13. 8*12=96. 96 divided by13: 13*7=91, so 96-91=5. Therefore, m≡5 mod13. So m=13n +5 for some integer n. Therefore, substituting back into N: N=77*(13n +5)+54=77*13n +77*5 +54=1001n +385 +54=1001n +439. Therefore, the general solution is N=1001n +439.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, we need the largest N less than 2010. So let's find the maximum n such that 1001n +439 <2010. Let's solve for n: 1001n <2010 -439=1571. Therefore, n <1571/1001≈1.569. Since n must be an integer, the maximum possible n is 1. Therefore, N=1001*1 +439=1440. Wait, but let me check if 1001*1 +439 is 1440. 1001 +439: 1000+400=1400, 1+39=40, so 1400+40=1440.

<prompt_2.1> Verify the solution by checking if N=1440 satisfies all the original conditions. </prompt_2.1>

But wait, let me check if there is a larger number than 1440 but still less than 2010. Since the modulus is 1001, the next solution would be 1440 +1001=2441, which is more than 2010, so indeed 1440 is the largest solution less than 2010.

<prompt_2.2> Verify if the next solution after 1440 would exceed 2010 and confirm 1440 as the largest solution. </prompt_2.2>

But wait, is that correct? Let me verify again.

Wait, but let me check if 1440 is indeed less than 2010 and satisfies all three conditions.

<prompt_2.3> Re-verify the solution by checking if 1440 satisfies all conditions and if there could be a larger solution. </prompt_2.3>

But wait, let me check another way. Maybe I made a mistake in the calculations. Let's retrace the steps.

<prompt_2.4> Re-evaluate the solution by retracing the steps and confirming the calculations. </prompt_2.4>

But let me check again with another approach. Let's try to solve the system step by step again.

<prompt_2.5> Re-check the solution using a different approach to ensure consistency. </prompt_2.5>

But just to make sure, let's check if 2010 - 564 = 1446. Wait, 1440 +6=1446. Let me check 1446.

<prompt_2.6> Perform a final verification and check if 1440 is the correct solution. </prompt_2.6>

But wait, let me check once more. Maybe I miscalculated when solving for m. Let's go back to when we had 12m ≡8 mod13. So 12m ≡8 mod13. Since 12≡-1 mod13, so -m ≡8 mod13 => m≡-8 mod13 => m≡5 mod13. That's correct. Then m=13n +5. Then N=77*(13n +5)+54=1001n +385 +54=1001n +439. So for n=1, N=1440. For n=2, 2441. So yes, 1440 is correct.

Alternatively, maybe there's a mistake in combining the congruences. Let me check another way.

<prompt_2.7> Explore alternative methods to confirm the solution and ensure its correctness. </prompt_2.7>

**Final Answer**
The largest integer less than $2010$ that satisfies the given conditions is \boxed{1440}.
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, 1001 + 439: 1000 + 400 = 1400, 1 + 39 = 40, so total 1440. Then 1440 is less than 2010. But wait, is there a larger number less than 2010? Let's check n=1 gives 1440, n=2 would be 1001*2 +439=2002 +439=2441, which is over 2010. So yes, 1440 is the answer?
</execution_2.1>

<execution_2.2>
Wait, but 1001*1 +439=1440. Then 1001*2 +439=2441, which is over 2010. So 1440 is the answer? Wait, but let me check if maybe 1440 + some multiple of 1001 but less than 2010. But 1440 +1001=2441>2010, so no. So 1440 is the largest one below 2010.
</execution_2.2>

<execution_2.3>
First, check if 1440 is less than 2010: yes. Then check the three congruences:

1. 1440 ÷7: 7*205=1435, so 1440-1435=5. Correct, remainder 5.
2. 1440 ÷11: 11*130=1430, 1440-1430=10. Correct, remainder 10.
3. 1440 ÷13: 13*110=1430, 1440-1430=10. Correct, remainder 10.
</execution_2.3>

<execution_2.4>
First, solving N ≡5 mod7 and N≡10 mod11. Let N=7k+5. Then 7k+5≡10 mod11 =>7k≡5 mod11. Inverse of 7 mod11 is 8, so k≡5*8=40≡7 mod11. So k=11m+7. Then N=7*(11m+7)+5=77m+49+5=77m+54. So N≡54 mod77. Then combining with N≡10 mod13. So 54 mod13 is 2, and 77 mod13 is 12. So 12m +2 ≡10 mod13 =>12m≡8 mod13. 12m≡8 mod13. Since 12≡-1 mod13, so -m≡8 mod13 =>m≡-8≡5 mod13. So m=13n+5. Then N=77*(13n+5)+54=77*13n +77*5 +54=1001n +385 +54=1001n +439. So general solution is N=1001n +439. Then the largest N less than 2010: 1001n +439 <2010 =>1001n <1571 =>n<1.57, so n=1. Then N=1001+439=1440. Then n=1 gives 1440, which is less than 2010. If n=2, 2441, which is over. So 1440 is the answer.
</execution_2.4>

<execution_2.5>
First, solve N ≡5 mod7 and N≡10 mod11. Let me write N=7k+5. Then 7k+5≡10 mod11 =>7k≡5 mod11. The inverse of 7 mod11 is 8, so k≡5*8=40≡7 mod11. So k=11m+7. Then N=7*(11m+7)+5=77m+54. So N≡54 mod77. Then, N≡10 mod13. So 77m +54 ≡10 mod13. 77 mod13: 13*5=65, 77-65=12, so 12m +54 mod13. 54 mod13: 13*4=52, 54-52=2. So 12m +2 ≡10 mod13 =>12m≡8 mod13. 12≡-1 mod13, so -m≡8 mod13 =>m≡-8≡5 mod13. So m=13n +5. Then N=77*(13n +5)+54=77*13n +77*5 +54=1001n +385 +54=1001n +439. So general solution is N=1001n +439. Then, to find the largest N<2010, solve 1001n +439 <2010. So 1001n <1571. n<1571/1001≈1.569. So n=1. Then N=1001+439=1440.
</execution_2.5>

<execution_2.6>
Wait, perhaps I made a mistake in solving for m. Let's re-examine that step. We had 12m ≡8 mod13. Since 12 is congruent to -1 mod13, so -m ≡8 mod13 => m ≡-8 mod13 => m≡5 mod13. That's correct. Then m=13n +5. Then N=77*(13n +5)+54=77*13n +77*5 +54=1001n +385 +54=1001n +439. So when n=1, N=1440. Then n=2 would be 1440+1001=2441, which is over 2010. So 1440 is the answer. But let me check 1440 against the original conditions.

First condition: 1440 divided by 7. Let's compute 1440 ÷7. 7*200=1400, so 1440-1400=40. 40 ÷7 is 5*7=35, remainder 5. So 1440 ≡5 mod7. Good.

Second condition: 1440 ÷11. 11*130=1430, so 1440-1430=10. So remainder 10. Correct.

Third condition: 1440 ÷13. 13*110=1430, so 1440-1430=10. Remainder 10. Correct.

So all conditions are satisfied. Therefore, 1440 is the correct answer.
</execution_2.6>

<execution_2.7>
Let me check if 1440 satisfies all three conditions:

1. 1440 ÷7: 7*205=1435, so remainder 1440-1435=5. Correct.

2. 1440 ÷11: 11*130=1430, remainder 1440-1430=10. Correct.

3. 1440 ÷13: 13*110=1430, remainder 1440-1430=10. Correct.

So yes, 1440 works. But wait, 1440 +1001=2441, which is over 2010, so the next one is too big. So 1440 is indeed the largest one below 2010. Wait, but 2010 -1440=570. So 1440 is 570 less than 2010. But maybe there is a larger number than 1440 but still less than 2010? For example, 1440 +1001=2441, which is too big, so no. Therefore, 1440 is the answer. But wait, let me check if there's another solution between 1440 and 2010. Since the modulus is 1001, the next solution is 1440 +1001=2441, which is over 2010, so no. Therefore, 1440 is indeed the correct answer.
</execution_2.7>
</Execution_2>

<Plan_3>
Based on execution_2.7:
**Final Answer**
The largest integer less than $2010$ that has a remainder of $5$ when divided by $7,$ a remainder of $10$ when divided by $13$?
\boxed{1440}
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the largest integer less than $2010$ that satisfies the given conditions is $\boxed{1440}$.
</Final_answer>
</think>