Explanation:
Okay, so I need to find the largest integer less than 2010 that satisfies three conditions: when divided by 7, it leaves a remainder of 5; when divided by 11, it leaves a remainder of 10; and when divided by 13, it also leaves a remainder of 10. Hmm, let me think about how to approach this step by step.

First, let me recall that these types of problems are usually solved using the Chinese Remainder Theorem (CRT). The CRT allows us to solve systems of congruences with pairwise coprime moduli. However, I need to check if the moduli here are coprime. The moduli given are 7, 11, and 13. Since 7, 11, and 13 are all prime numbers, they are pairwise coprime. That means the system of congruences should have a unique solution modulo 7×11×13. Let me compute that modulus first.

Calculating 7×11×13. Let me do 7×11 first, which is 77. Then 77×13. Let me compute 70×13 and 7×13. 70×13 is 910, and 7×13 is 91. Adding those together, 910 + 91 = 1001. So the modulus is 1001. That means the solutions will be congruent modulo 1001. Therefore, if I find the smallest positive solution, all solutions can be expressed as that number plus multiples of 1001. Since we need the largest number less than 2010, perhaps I can find the smallest solution, then add 1001 as many times as needed without exceeding 2010.

But first, let me write down the system of congruences:

1. x ≡ 5 mod 7
2. x ≡ 10 mod 11
3. x ≡ 10 mod 13

So, I need to find x such that all three congruences are satisfied. Let me try solving these step by step.

Starting with the first two congruences:

x ≡ 5 mod 7

and

x ≡ 10 mod 11.

Let me solve these two first. Let me denote x = 7k + 5 for some integer k. Then substitute this into the second equation:

7k + 5 ≡ 10 mod 11

Subtract 5 from both sides:

7k ≡ 5 mod 11

Now, I need to solve for k. To do that, I can find the multiplicative inverse of 7 modulo 11. Let me recall that the inverse of 7 mod 11 is a number m such that 7m ≡ 1 mod 11.

Let me try m=8: 7*8=56. 56 divided by 11 is 5*11=55, so 56-55=1. Yes, so 7*8 ≡1 mod 11. Therefore, the inverse is 8.

Therefore, multiplying both sides of 7k ≡5 mod 11 by 8:

k ≡ 5*8 mod 11 => k ≡40 mod 11. Now, 40 divided by 11 is 3*11=33, remainder 7. Therefore, k ≡7 mod 11. So k can be written as k = 11m +7, where m is some integer.

Therefore, going back to x:

x = 7k +5 = 7*(11m +7) +5 = 77m +49 +5 = 77m +54.

Therefore, the solutions to the first two congruences are x ≡54 mod 77. So now, we have x ≡54 mod 77. Now, we need to incorporate the third congruence: x ≡10 mod 13.

So now, let me set x = 77m +54. Then substitute into the third congruence:

77m +54 ≡10 mod 13.

First, compute 77 mod 13 and 54 mod 13.

Let me compute 13*5=65, so 77 -65=12. Therefore, 77 ≡12 mod13. Similarly, 54 divided by 13: 13*4=52, so 54-52=2. Therefore, 54≡2 mod13. Therefore, substituting:

12m + 2 ≡10 mod13.

Subtract 2 from both sides: 12m ≡8 mod13.

Now, need to solve for m. Let me find the inverse of 12 mod13. Since 12 ≡ -1 mod13, so inverse of -1 mod13 is -1, because (-1)*(-1)=1. Therefore, inverse of 12 mod13 is 12. Therefore, multiply both sides by 12:

m ≡8*12 mod13.

Calculate 8*12=96. 96 divided by 13: 13*7=91, so 96-91=5. Therefore, m≡5 mod13. Therefore, m can be written as m=13n +5, where n is some integer.

Therefore, going back to x:

x=77m +54=77*(13n +5) +54= 77*13n + 385 +54= 1001n + 439.

Therefore, the solutions to all three congruences are x ≡439 mod1001. Therefore, the general solution is x=1001n +439, where n is an integer. Therefore, the solutions are ... 439, 1440, 2441, ... etc.

But the problem asks for the largest integer less than 2010. So let's find the largest n such that 1001n +439 <2010.

Let me compute 1001n +439 <2010 => 1001n <2010 -439=1571 => n <1571 /1001. Let me compute 1571 divided by 1001. 1001*1=1001, 1571 -1001=570. So 1571 /1001=1 + 570/1001 ≈1.569. Therefore, the maximum integer n is 1. Therefore, the largest x less than 2010 is when n=1: x=1001*1 +439=1440. Let me check if 1440 is less than 2010: yes, 1440 <2010. What about n=2? 1001*2 +439=2002+439=2441. Which is greater than 2010. Therefore, n=1 gives 1440, which is the largest one less than 2010.

Wait, but let me check if there is another solution between 1440 and 2010. Wait, the next solution after 1440 would be 1440 +1001=2441, which is beyond 2010. Therefore, 1440 is the largest one less than 2010. However, let me verify that 1440 actually satisfies all three conditions. Let me check.

First, check remainder when divided by 7: 1440 divided by 7. Let me compute 7*200=1400. 1440 -1400=40. 40 divided by 7 is 5*7=35, remainder 5. Therefore, 1440 divided by 7 leaves remainder 5. Good.

Next, check remainder when divided by 11: 1440 divided by 11. Let me compute 11*130=1430. 1440 -1430=10. Therefore, the remainder is 10. Perfect.

Third, check remainder when divided by 13: 1440 divided by 13. Let me compute 13*110=1430. 1440 -1430=10. Therefore, remainder is 10. Perfect. So 1440 satisfies all three conditions. Therefore, that's the answer?

Wait, but before I conclude, let me check if there is a larger number less than 2010. For example, maybe 1440 + something? But according to our earlier calculation, the next solution is 2441, which is beyond 2010. However, maybe there is another solution between 1440 and 2010? Wait, the modulus is 1001, so the solutions are spaced 1001 apart. Therefore, the solutions are 439, 1440, 2441,... So indeed, between 1440 and 2010, there is no other solution. Because 1440 + 1001=2441>2010. Therefore, 1440 is the answer. Wait, but just to be thorough, let me check 1440 + something smaller? For example, maybe there is another number less than 2010 which satisfies the congruence? But according to the Chinese Remainder Theorem, since 7, 11, 13 are coprime, there's a unique solution mod 1001. Therefore, all solutions are 439 + 1001k, so between 0 and 1001, there's only 439, then 1440, 2441, etc. Therefore, 1440 is the one less than 2010. Therefore, the answer is 1440. But let me check if there is a mistake here. Wait, perhaps I miscalculated the value of 1001n +439 when n=1?

Yes, 1001*1 +439. Let me compute 1001 + 439. 1000 + 400 = 1400, 1 + 39=40, so total 1400 +40=1440. Correct. So that is correct. Let me check again if 1440 is indeed less than 2010. 1440 is 1440, and 2010 is 2010, so yes. So 1440 is less. Is there a larger number? For example, is there a number between 1440 and 2010 which satisfies the conditions?

Wait, suppose there is another solution. For example, if there is a number x where x ≡439 mod1001, but between 1440 and 2010. Since 1001*1=1001, 1001*2=2002. So 439 + 1001*1=1440, and 439 +1001*2=2440. 2440 is way beyond 2010. Therefore, there is no solution between 1440 and 2010 except 1440 itself. Therefore, the answer is 1440. But hold on, let me check with another approach. Maybe there is another way to approach this problem, just to confirm.

Alternatively, since x ≡10 mod11 and x≡10 mod13, so x ≡10 mod lcm(11,13). Since 11 and 13 are coprime, lcm(11,13)=143. Therefore, x ≡10 mod143. Therefore, x can be written as x=143k +10, where k is an integer. Now, this x also needs to satisfy x ≡5 mod7. Therefore, substituting:

143k +10 ≡5 mod7.

Compute 143 mod7 and 10 mod7. Let me calculate 143 divided by 7. 7*20=140, so 143-140=3. Therefore, 143≡3 mod7. 10 mod7 is 3. Therefore:

3k +3 ≡5 mod7 => 3k ≡5 -3=2 mod7 => 3k≡2 mod7. Now solve for k.

Multiply both sides by inverse of 3 mod7. The inverse of 3 mod7 is 5 because 3*5=15≡1 mod7. Therefore:

k≡2*5=10≡3 mod7. Therefore, k=7m +3 for some integer m. Therefore, x=143k +10=143*(7m +3)+10=143*7m + 429 +10=1001m +439. Which is the same as before. Therefore, x=1001m +439. Therefore, same result. Therefore, the solutions are 439, 1440, 2441,... So again, 1440 is the one less than 2010.

So that seems to confirm the previous result. Therefore, 1440 is the answer. But let me check if maybe there is another solution that I haven't considered? For example, could there be a number larger than 1440 but less than 2010 which is congruent to 439 mod1001? Let's see. 1001*1 +439=1440. Then adding 1001 again would give 2441, which is over. So between 1440 and 2010, the next solution would be 1440 + 1001=2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010.

Alternatively, maybe I should check numbers near 2010. For example, let me check if 2010 - something is a solution. But maybe that's tedious. Alternatively, since we have the general solution x=1001m +439, and we need x <2010. So solving for m:

1001m +439 <2010 => 1001m <1571 => m <1571/1001≈1.569. So m can be 0,1. Therefore, maximum m is 1, giving x=1440. Hence, confirmed again.

But just to be absolutely thorough, let me check if there's any mistake in solving the congruences. Let me go back again.

First, solving x ≡5 mod7 and x≡10 mod11. I set x=7k +5, substituted into the second equation: 7k +5 ≡10 mod11 => 7k≡5 mod11. Then inverse of 7 mod11 is 8, so k≡5*8=40≡7 mod11. Therefore, k=11m +7, leading to x=77m +54. Then solving x≡10 mod13: 77m +54 ≡10 mod13. Then 77 mod13= 77-6*13=77-78=-1≡12 mod13. 54 mod13= 54-4*13=54-52=2 mod13. Therefore, 12m +2 ≡10 mod13 =>12m≡8 mod13. Then inverse of 12 mod13 is 12, since 12*12=144≡144-11*13=144-143=1 mod13. Therefore, m≡8*12=96≡96-7*13=96-91=5 mod13. Therefore, m=13n +5. Therefore x=77*(13n +5) +54= 1001n + 385 +54=1001n +439. All steps check out.

Alternatively, let me try plugging n=1 into x=1001n +439 gives 1440. Let me check if 1440 is indeed less than 2010. Yes. What if I tried n=2? 1001*2 +439=2002 +439=2441, which is above 2010. Therefore, n=1 is the maximum. Therefore, the answer is 1440.

But just to check if maybe there is a larger number less than 2010, perhaps not of the form 1001n +439? But according to the Chinese Remainder Theorem, since the moduli 7, 11, 13 are coprime, the solution is unique modulo 1001. Therefore, all solutions are of the form 1001n +439. Therefore, there can't be another solution between 1440 and 2010. Therefore, 1440 is the correct answer.

Wait, but let me check if maybe the initial solution was wrong. For example, when solving for the first two congruences, maybe I made a mistake?

Let me check x ≡5 mod7 and x≡10 mod11. Let me take x=54. Let me check:

54 divided by 7: 7*7=49, 54-49=5, so remainder 5. Good. 54 divided by 11: 11*4=44, 54-44=10. So remainder 10. Perfect. Therefore, x=54 is a solution for the first two congruences. Then, adding 77 (which is 7*11) gives the next solution. So 54 +77=131. Check 131: 131/7=18*7=126, remainder 5. 131/11=11*11=121, remainder 10. Correct. Similarly, adding 77 each time. Then when solving for the third congruence, we found that x=54 +77m, and then found m≡5 mod13, leading to x=1001n +439. Therefore, x=439 is the solution when n=0. Let me check x=439.

Check 439 divided by 7: 7*62=434, remainder 5. Correct. 439 divided by 11: 11*39=429, remainder 10. Correct. 439 divided by 13: 13*33=429, remainder 10. Correct. So x=439 satisfies all three. Then adding 1001, which is 7*11*13, gives the next solution. Therefore, 439 +1001=1440. Which we checked earlier. Then 1440 +1001=2441.

Therefore, all steps are correct. Therefore, 1440 is indeed the answer.

But just to be 100% sure, let me check another number close to 2010. For example, suppose I try to find x such that x ≡5 mod7, x≡10 mod11, x≡10 mod13, and x is as large as possible less than 2010.

Alternatively, maybe there is another way to compute it. Let me think. Since x ≡10 mod11 and x≡10 mod13, then x ≡10 mod143 as before. Therefore, x=143k +10. Then, x ≡5 mod7. So let me compute 143k +10 ≡5 mod7. As before, 143 mod7 is 3, 10 mod7 is 3, so 3k +3 ≡5 mod7 => 3k≡2 mod7 => k≡(2/3) mod7. Since 3*5=15≡1 mod7, so inverse is 5. Therefore, k≡2*5=10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m +439. Therefore same result. Therefore, same conclusion. Therefore, x=1001m +439. Therefore, same answer. Therefore, 1440 is correct.

Alternatively, maybe I can think of numbers less than 2010 which satisfy x ≡10 mod143. Let me try to find such numbers. Since x=143k +10. Let me find maximum k such that 143k +10 <2010. So 143k <2000 =>k <2000/143≈13.986. Therefore, maximum k=13. So x=143*13 +10. Let me compute that. 143*10=1430, 143*3=429, so 1430+429=1859. Then 1859 +10=1869. Wait, 143*13= 143*(10+3)=1430 + 429=1859. Then x=1859 +10=1869? Wait, no, x=143k +10. So when k=13, x=143*13 +10= 1859 +10=1869. Wait, but then we need to check if this x=1869 satisfies the first congruence x ≡5 mod7. Let me check. So 1869 divided by 7. Let me compute 7*267=1869? Let me check 7*200=1400, 7*60=420, 7*7=49. 1400+420=1820, 1820+49=1869. Yes! Therefore, 7*267=1869. Therefore, 1869 divided by 7 leaves remainder 0. But we needed remainder 5. Therefore, this contradicts. Wait, so what is happening here?

Wait, if x=143k +10, then according to previous calculations, x must also satisfy x≡5 mod7. But here, when k=13, x=1869≡0 mod7. Which is not 5. Therefore, why is this happening? Because not all numbers of the form x=143k +10 satisfy the first congruence. Therefore, only certain k's will satisfy x≡5 mod7. Therefore, my previous approach was correct, that we have to find k such that 143k +10 ≡5 mod7, leading to k≡3 mod7. Therefore, k=7m +3. Therefore, for k=7m +3, x=143*(7m +3) +10=1001m +439. Therefore, for example, when m=0, k=3: x=143*3 +10=429 +10=439. Which works. For m=1, k=10: x=143*10 +10=1430 +10=1440. Which works. For m=2, k=17: 143*17 +10. Let me compute that: 143*10=1430, 143*7=1001, so 143*17=1430 +1001=2431. Then 2431 +10=2441. Which is over 2010.

But when I took k=13, which is not of the form 7m +3. Since k=13, and 13≡6 mod7. Therefore, not 3 mod7. Therefore, x=143k +10 for k=13 doesn't satisfy the congruence mod7. Therefore, that's why it failed. Therefore, to find the maximum x less than 2010 of the form x=143k +10 which also satisfies x≡5 mod7, we need k=7m +3. Therefore, let me find the maximum m such that x=143*(7m +3) +10 <2010.

So x=143*(7m +3) +10= 1001m + 429 +10=1001m +439 <2010.

Which is same as before. Therefore, m=1 gives x=1440, as before. Therefore, even when considering k=7m +3, then the maximum x is 1440. Therefore, even if I take k=13, which is not in the required form, it doesn't work. Therefore, 1440 is indeed the correct answer. So maybe my previous calculation was correct, and the confusion was due to taking k=13 which doesn't satisfy the congruence.

Therefore, I think I can be confident that 1440 is the correct answer.

But just to make sure, let me check if there are any other numbers between 1440 and 2010 that satisfy all three conditions. Let me check x=1440 + 1001=2441 which is over. What about numbers between 1440 and 2010? Let me pick a number, say, 1869 which we saw earlier. As we saw, 1869≡0 mod7, which is not 5. What about 1869 - something? For example, 1869 - 1001=868. Which is less than 1440. Not helpful. Alternatively, maybe adding 7 to 1440? Let's check 1440 +7=1447. Check if 1447 satisfies the conditions.

First, check remainder when divided by 7: 1440 is 5 mod7, adding 7 keeps the remainder same. Therefore, 1447≡5 mod7. Good. Check remainder mod11: 1440≡10 mod11. Adding 7: 1447≡10 +7=17≡6 mod11. Not 10. Not good. Therefore, not satisfying. Similarly, adding 11 to 1440: 1440 +11=1451. Check mod7: 1451 mod7. Since 1440≡5 mod7, 1440 +11≡5 +11 mod7. 11 mod7=4, so 5 +4=9≡2 mod7≠5. Not good. Similarly, adding 13: 1440 +13=1453. Check mod7: 5 +13 mod7. 13 mod7=6. So 5 +6=11≡4 mod7≠5. Not good. Therefore, adding multiples of 7, 11, or 13 doesn't help. Therefore, only numbers congruent to 1440 +1001k will satisfy all three congruences.

Therefore, 1440 is indeed the only solution less than 2010.

Wait, another idea: perhaps there is a mistake in the modulus? Let me check if 7×11×13=1001. Yes, because 7×11=77, 77×13. 70×13=910, 7×13=91, so 910+91=1001. Correct. Therefore, the modulus is correct.

Alternatively, let me check x=1440. Let me divide by 7: 1440 ÷7. 7*200=1400, 1440-1400=40. 40 ÷7=5*7=35, remainder 5. Correct. Divided by 11: 1440 ÷11. 11*130=1430. 1440 -1430=10. Correct. Divided by 13: 1440 ÷13. 13*110=1430. 1440-1430=10. Correct. Therefore, all conditions are satisfied.

Therefore, I think the answer is 1440. Therefore, the largest integer less than 2010 is 1440. Hence, I can conclude.

**Final Answer**
\boxed{1440}

To solve the problem, we are asked to find the **largest integer less than 2010** that satisfies the following **system of congruences**:

- $ x \equiv 5 \pmod{7} $
- $ x \equiv 10 \pmod{11} $
- $ x \equiv 10 \pmod{13} $

---

### Step 1: Recognize the Structure of the Problem

We are given three congruences, each modulo a distinct prime number: 7, 11, and 13. These moduli are **pairwise coprime**, which means that the **Chinese Remainder Theorem (CRT)** can be applied.

According to the CRT, since 7, 11, and 13 are each prime and distinct, a unique solution exists modulo:
$$
M = 7 \times 11 \times 13 = 1001
$$

---

### Step 2: Solve the System with the Chinese Remainder Theorem

Let’s first solve the congruences pairwise to find the general solution.

#### Combine the first two congruences:

- $ x \equiv 5 \pmod{7} $
- $ x \equiv 10 \pmod{11} $

Let $ x = 7k + 5 $. Substitute into the second congruence:
$$
7k + 5 \equiv 10 \pmod{11} \Rightarrow 7k \equiv 5 \pmod{11}
$$

We find the modular inverse of 7 modulo 11. Since $7 \cdot 8 = 56 \equiv 1 \pmod{11}$, the inverse is 8. Multiply both sides:
$$
k \equiv 5 \cdot 8 = 40 \equiv 7 \pmod{11}
$$

So $ k = 11m + 7 $, and then:
$$
x = 7k + 5 = 7(11m + 7) + 5 = 77m + 54
$$

This gives:
$$
x \equiv 54 \pmod{77}
$$

#### Now combine with the third congruence:

- $ x \equiv 54 \pmod{77} $
- $ x \equiv 10 \pmod{13} $

Let $ x = 77m + 54 $. Substitute into the third congruence:
$$
77m + 54 \equiv 10 \pmod{13}
$$

We compute:
- $ 77 \mod 13 = 12 $
- $ 54 \mod 13 = 2 $

So:
$$
12m + 2 \equiv 10 \pmod{13} \Rightarrow 12m \equiv 8 \pmod{13}
$$

Now solve for $ m $. The inverse of 12 modulo 13 is 12, since $12 \cdot 12 = 144 \equiv 1 \mod 13$. Multiply both sides:
$$
m \equiv 8 \cdot 12 = 96 \equiv 5 \pmod{13}
$$

So $ m = 13n + 5 $, and:
$$
x = 77(13n + 5) + 54 = 1001n + 385 + 54 = 1001n + 439
$$

---

### Step 3: Find the Largest Solution Less than 2010

Our general solution is:
$$
x = 1001n + 439
$$

We want the largest such $ x $ **less than 2010**:
$$
1001n + 439 < 2010 \Rightarrow 1001n < 1571
$$

Divide both sides:
$$
n < \frac{1571}{1001} \approx 1.568
$$

So the largest integer $ n $ is 1.

Then:
$$
x = 1001 \cdot 1 + 439 = 1440
$$

---

### Step 4: Verify the Solution

Let’s confirm $ x = 1440 $ satisfies all the original congruences:

- $ 1440 \mod 7 = 5 $ (since $ 7 \cdot 205 = 1435 $, $ 1440 - 1435 = 5 $)
- $ 1440 \mod 11 = 10 $ (since $ 11 \cdot 130 = 1430 $, $ 1440 - 1430 = 10 $)
- $ 1440 \mod 13 = 10 $ (since $ 13 \cdot 110 = 1430 $, $ 1440 - 1430 = 10 $)

All three congruences are satisfied.

---

### ✅ Final Answer

$$
\boxed{1440}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest integer less than 2010 that satisfies three conditions: when divided by 7, it leaves a remainder of 5; when divided by 11, it leaves a remainder of 10; and when divided by 13, it also leaves a remainder of 10. Hmm, let me think about how to approach this step by step.

First, let me recall that these types of problems are usually solved using the Chinese Remainder Theorem (CRT). The CRT allows us to solve systems of congruences with pairwise coprime moduli. However, I need to check if the moduli here are coprime. The moduli given are 7, 11, and 13. Since 7, 11, and 13 are all prime numbers, they are pairwise coprime. That means the system of congruences should have a unique solution modulo 7×11×13. Let me compute that modulus first.

Calculating 7×11×13. Let me do 7×11 first, which is 77. Then 77×13. Let me compute 70×13 and 7×13. 70×13 is 910, and 7×13 is 91. Adding those together, 910 + 91 = 1001. So the modulus is 1001. That means the solutions will be congruent modulo 1001. Therefore, if I find the smallest positive solution, all solutions can be expressed as that number plus multiples of 1001. Since we need the largest number less than 2010, perhaps I can find the smallest solution, then add 1001 as many times as needed without exceeding 2010.

But first, let me write down the system of congruences:

1. x ≡ 5 mod 7
2. x ≡ 10 mod 11
3. x ≡ 10 mod 13

So, I need to find x such that all three congruences are satisfied. Let me try solving these step by step.

Starting with the first two congruences:

x ≡ 5 mod 7

and

x ≡ 10 mod 11.

Let me solve these two first. Let me denote x = 7k + 5 for some integer k. Then substitute this into the second equation:

7k + 5 ≡ 10 mod 11

Subtract 5 from both sides:

7k ≡ 5 mod 11

Now, I need to solve for k. To do that, I can find the multiplicative inverse of 7 modulo 11. Let me recall that the inverse of 7 mod 11 is a number m such that 7m ≡ 1 mod 11.

Let me try m=8: 7*8=56. 56 divided by 11 is 5*11=55, so 56-55=1. Yes, so 7*8 ≡1 mod 11. Therefore, the inverse is 8.

Therefore, multiplying both sides of 7k ≡5 mod 11 by 8:

k ≡ 5*8 mod 11 => k ≡40 mod 11. Now, 40 divided by 11 is 3*11=33, remainder 7. Therefore, k ≡7 mod 11. So k can be written as k = 11m +7, where m is some integer.

Therefore, going back to x:

x = 7k +5 = 7*(11m +7) +5 = 77m +49 +5 = 77m +54.

Therefore, the solutions to the first two congruences are x ≡54 mod 77. So now, we have x ≡54 mod 77. Now, we need to incorporate the third congruence: x ≡10 mod 13.

So now, let me set x = 77m +54. Then substitute into the third congruence:

77m +54 ≡10 mod 13.

First, compute 77 mod 13 and 54 mod 13.

Let me compute 13*5=65, so 77 -65=12. Therefore, 77 ≡12 mod13. Similarly, 54 divided by 13: 13*4=52, so 54-52=2. Therefore, 54≡2 mod13. Therefore, substituting:

12m + 2 ≡10 mod13.

Subtract 2 from both sides: 12m ≡8 mod13.

Now, need to solve for m. Let me find the inverse of 12 mod13. Since 12 ≡ -1 mod13, so inverse of -1 mod13 is -1, because (-1)*(-1)=1. Therefore, inverse of 12 mod13 is 12. Therefore, multiply both sides by 12:

m ≡8*12 mod13.

Calculate 8*12=96. 96 divided by 13: 13*7=91, so 96-91=5. Therefore, m≡5 mod13. Therefore, m can be written as m=13n +5, where n is some integer.

Therefore, going back to x:

x=77m +54=77*(13n +5) +54= 77*13n + 385 +54= 1001n + 439.

Therefore, the solutions to all three congruences are x ≡439 mod1001. Therefore, the general solution is x=1001n +439, where n is an integer. Therefore, the solutions are ... 439, 1440, 2441, ... etc.

But the problem asks for the largest integer less than 2010. So let's find the largest n such that 1001n +439 <2010.

Let me compute 1001n +439 <2010 => 1001n <2010 -439=1571 => n <1571 /1001. Let me compute 1571 divided by 1001. 1001*1=1001, 1571 -1001=570. So 1571 /1001=1 + 570/1001 ≈1.569. Therefore, the maximum integer n is 1. Therefore, the largest x less than 2010 is when n=1: x=1001*1 +439=1440. Let me check if 1440 is less than 2010: yes, 1440 <2010. What about n=2? 1001*2 +439=2002+439=2441. Which is greater than 2010. Therefore, n=1 gives 1440, which is the largest one less than 2010.

Wait, but let me check if there is another solution between 1440 and 2010. Wait, the next solution after 1440 would be 1440 +1001=2441, which is beyond 2010. Therefore, 1440 is the largest one less than 2010. However, let me verify that 1440 actually satisfies all three conditions. Let me check.

First, check remainder when divided by 7: 1440 divided by 7. Let me compute 7*200=1400. 1440 -1400=40. 40 divided by 7 is 5*7=35, remainder 5. Therefore, 1440 divided by 7 leaves remainder 5. Good.

Next, check remainder when divided by 11: 1440 divided by 11. Let me compute 11*130=1430. 1440 -1430=10. Therefore, the remainder is 10. Perfect.

Third, check remainder when divided by 13: 1440 divided by 13. Let me compute 13*110=1430. 1440 -1430=10. Therefore, remainder is 10. Perfect. So 1440 satisfies all three conditions. Therefore, that's the answer?

Wait, but before I conclude, let me check if there is a larger number less than 2010. For example, maybe 1440 + something? But according to our earlier calculation, the next solution is 2441, which is beyond 2010. However, maybe there is another solution between 1440 and 2010? Wait, the modulus is 1001, so the solutions are spaced 1001 apart. Therefore, the solutions are 439, 1440, 2441,... So indeed, between 1440 and 2010, there is no other solution. Because 1440 + 1001=2441>2010. Therefore, 1440 is the answer. Wait, but just to be thorough, let me check 1440 + something smaller? For example, maybe there is another number less than 2010 which satisfies the congruence? But according to the Chinese Remainder Theorem, since 7, 11, 13 are coprime, there's a unique solution mod 1001. Therefore, all solutions are 439 + 1001k, so between 0 and 1001, there's only 439, then 1440, 2441, etc. Therefore, 1440 is the one less than 2010. Therefore, the answer is 1440. But let me check if there is a mistake here. Wait, perhaps I miscalculated the value of 1001n +439 when n=1?

Yes, 1001*1 +439. Let me compute 1001 + 439. 1000 + 400 = 1400, 1 + 39=40, so total 1400 +40=1440. Correct. So that is correct. Let me check again if 1440 is indeed less than 2010. 1440 is 1440, and 2010 is 2010, so yes. So 1440 is less. Is there a larger number? For example, is there a number between 1440 and 2010 which satisfies the conditions?

Wait, suppose there is another solution. For example, if there is a number x where x ≡439 mod1001, but between 1440 and 2010. Since 1001*1=1001, 1001*2=2002. So 439 + 1001*1=1440, and 439 +1001*2=2440. 2440 is way beyond 2010. Therefore, there is no solution between 1440 and 2010 except 1440 itself. Therefore, the answer is 1440. But hold on, let me check with another approach. Maybe there is another way to approach this problem, just to confirm.

Alternatively, since x ≡10 mod11 and x≡10 mod13, so x ≡10 mod lcm(11,13). Since 11 and 13 are coprime, lcm(11,13)=143. Therefore, x ≡10 mod143. Therefore, x can be written as x=143k +10, where k is an integer. Now, this x also needs to satisfy x ≡5 mod7. Therefore, substituting:

143k +10 ≡5 mod7.

Compute 143 mod7 and 10 mod7. Let me calculate 143 divided by 7. 7*20=140, so 143-140=3. Therefore, 143≡3 mod7. 10 mod7 is 3. Therefore:

3k +3 ≡5 mod7 => 3k ≡5 -3=2 mod7 => 3k≡2 mod7. Now solve for k.

Multiply both sides by inverse of 3 mod7. The inverse of 3 mod7 is 5 because 3*5=15≡1 mod7. Therefore:

k≡2*5=10≡3 mod7. Therefore, k=7m +3 for some integer m. Therefore, x=143k +10=143*(7m +3)+10=143*7m + 429 +10=1001m +439. Which is the same as before. Therefore, x=1001m +439. Therefore, same result. Therefore, the solutions are 439, 1440, 2441,... So again, 1440 is the one less than 2010.

So that seems to confirm the previous result. Therefore, 1440 is the answer. But let me check if maybe there is another solution that I haven't considered? For example, could there be a number larger than 1440 but less than 2010 which is congruent to 439 mod1001? Let's see. 1001*1 +439=1440. Then adding 1001 again would give 2441, which is over. So between 1440 and 2010, the next solution would be 1440 + 1001=2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010.

Alternatively, maybe I should check numbers near 2010. For example, let me check if 2010 - something is a solution. But maybe that's tedious. Alternatively, since we have the general solution x=1001m +439, and we need x <2010. So solving for m:

1001m +439 <2010 => 1001m <1571 => m <1571/1001≈1.569. So m can be 0,1. Therefore, maximum m is 1, giving x=1440. Hence, confirmed again.

But just to be absolutely thorough, let me check if there's any mistake in solving the congruences. Let me go back again.

First, solving x ≡5 mod7 and x≡10 mod11. I set x=7k +5, substituted into the second equation: 7k +5 ≡10 mod11 => 7k≡5 mod11. Then inverse of 7 mod11 is 8, so k≡5*8=40≡7 mod11. Therefore, k=11m +7, leading to x=77m +54. Then solving x≡10 mod13: 77m +54 ≡10 mod13. Then 77 mod13= 77-6*13=77-78=-1≡12 mod13. 54 mod13= 54-4*13=54-52=2 mod13. Therefore, 12m +2 ≡10 mod13 =>12m≡8 mod13. Then inverse of 12 mod13 is 12, since 12*12=144≡144-11*13=144-143=1 mod13. Therefore, m≡8*12=96≡96-7*13=96-91=5 mod13. Therefore, m=13n +5. Therefore x=77*(13n +5) +54= 1001n + 385 +54=1001n +439. All steps check out.

Alternatively, let me try plugging n=1 into x=1001n +439 gives 1440. Let me check if 1440 is indeed less than 2010. Yes. What if I tried n=2? 1001*2 +439=2002 +439=2441, which is above 2010. Therefore, n=1 is the maximum. Therefore, the answer is 1440.

But just to check if maybe there is a larger number less than 2010, perhaps not of the form 1001n +439? But according to the Chinese Remainder Theorem, since the moduli 7, 11, 13 are coprime, the solution is unique modulo 1001. Therefore, all solutions are of the form 1001n +439. Therefore, there can't be another solution between 1440 and 2010. Therefore, 1440 is the correct answer.

Wait, but let me check if maybe the initial solution was wrong. For example, when solving for the first two congruences, maybe I made a mistake?

Let me check x ≡5 mod7 and x≡10 mod11. Let me take x=54. Let me check:

54 divided by 7: 7*7=49, 54-49=5, so remainder 5. Good. 54 divided by 11: 11*4=44, 54-44=10. So remainder 10. Perfect. Therefore, x=54 is a solution for the first two congruences. Then, adding 77 (which is 7*11) gives the next solution. So 54 +77=131. Check 131: 131/7=18*7=126, remainder 5. 131/11=11*11=121, remainder 10. Correct. Similarly, adding 77 each time. Then when solving for the third congruence, we found that x=54 +77m, and then found m≡5 mod13, leading to x=1001n +439. Therefore, x=439 is the solution when n=0. Let me check x=439.

Check 439 divided by 7: 7*62=434, remainder 5. Correct. 439 divided by 11: 11*39=429, remainder 10. Correct. 439 divided by 13: 13*33=429, remainder 10. Correct. So x=439 satisfies all three. Then adding 1001, which is 7*11*13, gives the next solution. Therefore, 439 +1001=1440. Which we checked earlier. Then 1440 +1001=2441.

Therefore, all steps are correct. Therefore, 1440 is indeed the answer.

But just to be 100% sure, let me check another number close to 2010. For example, suppose I try to find x such that x ≡5 mod7, x≡10 mod11, x≡10 mod13, and x is as large as possible less than 2010.

Alternatively, maybe there is another way to compute it. Let me think. Since x ≡10 mod11 and x≡10 mod13, then x ≡10 mod143 as before. Therefore, x=143k +10. Then, x ≡5 mod7. So let me compute 143k +10 ≡5 mod7. As before, 143 mod7 is 3, 10 mod7 is 3, so 3k +3 ≡5 mod7 => 3k≡2 mod7 => k≡(2/3) mod7. Since 3*5=15≡1 mod7, so inverse is 5. Therefore, k≡2*5=10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m +439. Therefore same result. Therefore, same conclusion. Therefore, x=1001m +439. Therefore, same answer. Therefore, 1440 is correct.

Alternatively, maybe I can think of numbers less than 2010 which satisfy x ≡10 mod143. Let me try to find such numbers. Since x=143k +10. Let me find maximum k such that 143k +10 <2010. So 143k <2000 =>k <2000/143≈13.986. Therefore, maximum k=13. So x=143*13 +10. Let me compute that. 143*10=1430, 143*3=429, so 1430+429=1859. Then 1859 +10=1869. Wait, 143*13= 143*(10+3)=1430 + 429=1859. Then x=1859 +10=1869? Wait, no, x=143k +10. So when k=13, x=143*13 +10= 1859 +10=1869. Wait, but then we need to check if this x=1869 satisfies the first congruence x ≡5 mod7. Let me check. So 1869 divided by 7. Let me compute 7*267=1869? Let me check 7*200=1400, 7*60=420, 7*7=49. 1400+420=1820, 1820+49=1869. Yes! Therefore, 7*267=1869. Therefore, 1869 divided by 7 leaves remainder 0. But we needed remainder 5. Therefore, this contradicts. Wait, so what is happening here?

Wait, if x=143k +10, then according to previous calculations, x must also satisfy x≡5 mod7. But here, when k=13, x=1869≡0 mod7. Which is not 5. Therefore, why is this happening? Because not all numbers of the form x=143k +10 satisfy the first congruence. Therefore, only certain k's will satisfy x≡5 mod7. Therefore, my previous approach was correct, that we have to find k such that 143k +10 ≡5 mod7, leading to k≡3 mod7. Therefore, k=7m +3. Therefore, for k=7m +3, x=143*(7m +3) +10=1001m +439. Therefore, for example, when m=0, k=3: x=143*3 +10=429 +10=439. Which works. For m=1, k=10: x=143*10 +10=1430 +10=1440. Which works. For m=2, k=17: 143*17 +10. Let me compute that: 143*10=1430, 143*7=1001, so 143*17=1430 +1001=2431. Then 2431 +10=2441. Which is over 2010.

But when I took k=13, which is not of the form 7m +3. Since k=13, and 13≡6 mod7. Therefore, not 3 mod7. Therefore, x=143k +10 for k=13 doesn't satisfy the congruence mod7. Therefore, that's why it failed. Therefore, to find the maximum x less than 2010 of the form x=143k +10 which also satisfies x≡5 mod7, we need k=7m +3. Therefore, let me find the maximum m such that x=143*(7m +3) +10 <2010.

So x=143*(7m +3) +10= 1001m + 429 +10=1001m +439 <2010.

Which is same as before. Therefore, m=1 gives x=1440, as before. Therefore, even when considering k=7m +3, then the maximum x is 1440. Therefore, even if I take k=13, which is not in the required form, it doesn't work. Therefore, 1440 is indeed the correct answer. So maybe my previous calculation was correct, and the confusion was due to taking k=13 which doesn't satisfy the congruence.

Therefore, I think I can be confident that 1440 is the correct answer.

But just to make sure, let me check if there are any other numbers between 1440 and 2010 that satisfy all three conditions. Let me check x=1440 + 1001=2441 which is over. What about numbers between 1440 and 2010? Let me pick a number, say, 1869 which we saw earlier. As we saw, 1869≡0 mod7, which is not 5. What about 1869 - something? For example, 1869 - 1001=868. Which is less than 1440. Not helpful. Alternatively, maybe adding 7 to 1440? Let's check 1440 +7=1447. Check if 1447 satisfies the conditions.

First, check remainder when divided by 7: 1440 is 5 mod7, adding 7 keeps the remainder same. Therefore, 1447≡5 mod7. Good. Check remainder mod11: 1440≡10 mod11. Adding 7: 1447≡10 +7=17≡6 mod11. Not 10. Not good. Therefore, not satisfying. Similarly, adding 11 to 1440: 1440 +11=1451. Check mod7: 1451 mod7. Since 1440≡5 mod7, 1440 +11≡5 +11 mod7. 11 mod7=4, so 5 +4=9≡2 mod7≠5. Not good. Similarly, adding 13: 1440 +13=1453. Check mod7: 5 +13 mod7. 13 mod7=6. So 5 +6=11≡4 mod7≠5. Not good. Therefore, adding multiples of 7, 11, or 13 doesn't help. Therefore, only numbers congruent to 1440 +1001k will satisfy all three congruences.

Therefore, 1440 is indeed the only solution less than 2010.

Wait, another idea: perhaps there is a mistake in the modulus? Let me check if 7×11×13=1001. Yes, because 7×11=77, 77×13. 70×13=910, 7×13=91, so 910+91=1001. Correct. Therefore, the modulus is correct.

Alternatively, let me check x=1440. Let me divide by 7: 1440 ÷7. 7*200=1400, 1440-1400=40. 40 ÷7=5*7=35, remainder 5. Correct. Divided by 11: 1440 ÷11. 11*130=1430. 1440 -1430=10. Correct. Divided by 13: 1440 ÷13. 13*110=1430. 1440-1430=10. Correct. Therefore, all conditions are satisfied.

Therefore, I think the answer is 1440. Therefore, the largest integer less than 2010 is 1440. Hence, I can conclude.

**Final Answer**
\boxed{1440}

To solve the problem, we are asked to find the **largest integer less than 2010** that satisfies the following **system of congruences**:

- $ x \equiv 5 \pmod{7} $
- $ x \equiv 10 \pmod{11} $
- $ x \equiv 10 \pmod{13} $

---

### Step 1: Recognize the Structure of the Problem

We are given three congruences, each modulo a distinct prime number: 7, 11, and 13. These moduli are **pairwise coprime**, which means that the **Chinese Remainder Theorem (CRT)** can be applied.

According to the CRT, since 7, 11, and 13 are each prime and distinct, a unique solution exists modulo:
$$
M = 7 \times 11 \times 13 = 1001
$$

---

### Step 2: Solve the System with the Chinese Remainder Theorem

Let’s first solve the congruences pairwise to find the general solution.

#### Combine the first two congruences:

- $ x \equiv 5 \pmod{7} $
- $ x \equiv 10 \pmod{11} $

Let $ x = 7k + 5 $. Substitute into the second congruence:
$$
7k + 5 \equiv 10 \pmod{11} \Rightarrow 7k \equiv 5 \pmod{11}
$$

We find the modular inverse of 7 modulo 11. Since $7 \cdot 8 = 56 \equiv 1 \pmod{11}$, the inverse is 8. Multiply both sides:
$$
k \equiv 5 \cdot 8 = 40 \equiv 7 \pmod{11}
$$

So $ k = 11m + 7 $, and then:
$$
x = 7k + 5 = 7(11m + 7) + 5 = 77m + 54
$$

This gives:
$$
x \equiv 54 \pmod{77}
$$

#### Now combine with the third congruence:

- $ x \equiv 54 \pmod{77} $
- $ x \equiv 10 \pmod{13} $

Let $ x = 77m + 54 $. Substitute into the third congruence:
$$
77m + 54 \equiv 10 \pmod{13}
$$

We compute:
- $ 77 \mod 13 = 12 $
- $ 54 \mod 13 = 2 $

So:
$$
12m + 2 \equiv 10 \pmod{13} \Rightarrow 12m \equiv 8 \pmod{13}
$$

Now solve for $ m $. The inverse of 12 modulo 13 is 12, since $12 \cdot 12 = 144 \equiv 1 \mod 13$. Multiply both sides:
$$
m \equiv 8 \cdot 12 = 96 \equiv 5 \pmod{13}
$$

So $ m = 13n + 5 $, and:
$$
x = 77(13n + 5) + 54 = 1001n + 385 + 54 = 1001n + 439
$$

---

### Step 3: Find the Largest Solution Less than 2010

Our general solution is:
$$
x = 1001n + 439
$$

We want the largest such $ x $ **less than 2010**:
$$
1001n + 439 < 2010 \Rightarrow 1001n < 1571
$$

Divide both sides:
$$
n < \frac{1571}{1001} \approx 1.568
$$

So the largest integer $ n $ is 1.

Then:
$$
x = 1001 \cdot 1 + 439 = 1440
$$

---

### Step 4: Verify the Solution

Let’s confirm $ x = 1440 $ satisfies all the original congruences:

- $ 1440 \mod 7 = 5 $ (since $ 7 \cdot 205 = 1435 $, $ 1440 - 1435 = 5 $)
- $ 1440 \mod 11 = 10 $ (since $ 11 \cdot 130 = 1430 $, $ 1440 - 1430 = 10 $)
- $ 1440 \mod 13 = 10 $ (since $ 13 \cdot 110 = 1430 $, $ 1440 - 1430 = 10 $)

All three congruences are satisfied.

---

### ✅ Final Answer

$$
\boxed{1440}
$$

Solution:
We want a remainder of $10$ when divided by both $11$ and $13$. The least common multiple of $11$ and $13$ is $143$. We add $10$ to the number such that the remainder would be $10$ when divided by $11$ and $13$ so we get $143+10=153$. However, that does not give a remainder of $5$ when divided by $7$, so we add more $143$s until we get a value that works. We get that $153+143+143=439$ gives a remainder of $5$ when divided by $7$.

Since we want the largest integer less than 2010, we keep adding the least common multiple of $7$, $11$, and $13$ until we go over. The least common multiple is $7 \cdot 11 \cdot 13 =1001$. We add it to $439$ to get $1440$, adding it again would give a value greater than $2010$, so our answer is $\boxed{1440}$.


Explanation:
Okay, let me try to figure out this problem. So, it says that y varies inversely as the square root of x. Hmm, inverse variation. I remember that when something varies inversely, it means that as one increases, the other decreases. The general formula for inverse variation is y = k/x, right? But wait, in this case, it's not just x, it's the square root of x. So maybe the formula is different?

Let me think. If y varies inversely as sqrt(x), then that should mean y is equal to some constant divided by sqrt(x). So, the formula should be y = k / sqrt(x), where k is the constant of variation. Yeah, that makes sense. So, when x increases, sqrt(x) increases, so y decreases, which is inverse variation. Okay, got that part.

Now, they give me a specific case where when x is 24, y is 15. So I can use that to find the constant k. Let me write that down:

When x = 24, y = 15. So plugging into the equation:

15 = k / sqrt(24)

I need to solve for k. So, multiply both sides by sqrt(24):

k = 15 * sqrt(24)

Hmm, sqrt(24)... I can simplify that. sqrt(24) is sqrt(4*6) which is 2*sqrt(6). So, sqrt(24) = 2*sqrt(6). Therefore, k = 15 * 2 * sqrt(6) = 30*sqrt(6). Let me check that again. Yes, 15 times 2 is 30, so k = 30√6. Okay, so now I have the constant.

Now, the question is asking: What is x when y = 3? So, we need to find x such that when y is 3, the equation holds. Let me write that down.

We have the equation y = k / sqrt(x). We know y is 3, and k is 30√6. So:

3 = (30√6) / sqrt(x)

Now, solve for x. Let me rearrange this equation. Multiply both sides by sqrt(x):

3 * sqrt(x) = 30√6

Then, divide both sides by 3:

sqrt(x) = (30√6) / 3 = 10√6

So sqrt(x) equals 10√6. To solve for x, square both sides:

(sqrt(x))² = (10√6)²

Which simplifies to:

x = 10² * (√6)²

Calculating that:

10 squared is 100, and (√6) squared is 6. So:

x = 100 * 6 = 600

Wait, so x is 600? Let me verify if that makes sense. Let me check my steps again.

Starting with y varies inversely as sqrt(x), so y = k / sqrt(x). Given x=24, y=15. So k = 15*sqrt(24). Then sqrt(24) is 2*sqrt(6), so k is 15*2*sqrt(6) = 30√6. Correct.

Then, when y=3, set up equation 3 = 30√6 / sqrt(x). Multiply both sides by sqrt(x): 3 sqrt(x) = 30√6. Divide by 3: sqrt(x) = 10√6. Squared: x = 100*6 = 600. Seems right.

Let me check with substitution. If x=600, then sqrt(x) is sqrt(600). Let me compute sqrt(600). Well, sqrt(600) is sqrt(100*6) which is 10*sqrt(6). So then, y = k / sqrt(x) = 30√6 / (10√6). Let me compute that. The √6 cancels out, so 30 / 10 = 3. Which is the y we wanted. Perfect, that checks out. So x=600 is correct.

So, the answer should be 600. Wait, but let me make sure I didn't make any calculation errors. Let me go through once more.

Original equation: y = k / sqrt(x). Given x=24, y=15. So k = y * sqrt(x) = 15 * sqrt(24). sqrt(24) is approx 4.899, but exact value is 2*sqrt(6). So k is 15*2*sqrt(6) = 30√6. Correct.

Then when y=3, 3 = 30√6 / sqrt(x). So sqrt(x) = (30√6)/3 = 10√6. Then x = (10√6)^2 = 100*6 = 600. Yep. All steps are consistent. Therefore, the answer is 600.

Wait, but maybe I should write sqrt(600) as 10*sqrt(6) again. If we plug back into the original equation, we get y = 30√6 / (10√6) = 3, which is correct. So, all good.

I think that's solid. So, I don't see any mistakes here. Therefore, x is 600 when y is 3.

**Final Answer**
The value of $ x $ when $ y = 3 $ is \boxed{600}.

To solve the problem, we begin by understanding the nature of the relationship described:

> "The value of $ y $ varies inversely as $ \sqrt{x} $."

This means that $ y $ is inversely proportional to the square root of $ x $. In mathematical terms, the relationship can be expressed as:

$$
y = \frac{k}{\sqrt{x}}
$$

where $ k $ is the constant of proportionality.

---

### Step 1: Use the given values to find the constant $ k $

We are given that:

$$
\text{When } x = 24, \quad y = 15
$$

Substitute into the equation:

$$
15 = \frac{k}{\sqrt{24}}
$$

Multiply both sides by $ \sqrt{24} $:

$$
k = 15 \sqrt{24}
$$

Since $ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} $, we can simplify:

$$
k = 15 \cdot 2\sqrt{6} = 30\sqrt{6}
$$

---

### Step 2: Use the constant to find $ x $ when $ y = 3 $

We now know that:

$$
y = \frac{30\sqrt{6}}{\sqrt{x}}
$$

We are given $ y = 3 $, so:

$$
3 = \frac{30\sqrt{6}}{\sqrt{x}}
$$

Now, solve for $ \sqrt{x} $:

$$
\sqrt{x} = \frac{30\sqrt{6}}{3} = 10\sqrt{6}
$$

Square both sides to solve for $ x $:

$$
x = (10\sqrt{6})^2 = 10^2 \cdot (\sqrt{6})^2 = 100 \cdot 6 = 600
$$

---

### Final Answer

$$
\boxed{600}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, it says that y varies inversely as the square root of x. Hmm, inverse variation. I remember that when something varies inversely, it means that as one increases, the other decreases. The general formula for inverse variation is y = k/x, right? But wait, in this case, it's not just x, it's the square root of x. So maybe the formula is different?

Let me think. If y varies inversely as sqrt(x), then that should mean y is equal to some constant divided by sqrt(x). So, the formula should be y = k / sqrt(x), where k is the constant of variation. Yeah, that makes sense. So, when x increases, sqrt(x) increases, so y decreases, which is inverse variation. Okay, got that part.

Now, they give me a specific case where when x is 24, y is 15. So I can use that to find the constant k. Let me write that down:

When x = 24, y = 15. So plugging into the equation:

15 = k / sqrt(24)

I need to solve for k. So, multiply both sides by sqrt(24):

k = 15 * sqrt(24)

Hmm, sqrt(24)... I can simplify that. sqrt(24) is sqrt(4*6) which is 2*sqrt(6). So, sqrt(24) = 2*sqrt(6). Therefore, k = 15 * 2 * sqrt(6) = 30*sqrt(6). Let me check that again. Yes, 15 times 2 is 30, so k = 30√6. Okay, so now I have the constant.

Now, the question is asking: What is x when y = 3? So, we need to find x such that when y is 3, the equation holds. Let me write that down.

We have the equation y = k / sqrt(x). We know y is 3, and k is 30√6. So:

3 = (30√6) / sqrt(x)

Now, solve for x. Let me rearrange this equation. Multiply both sides by sqrt(x):

3 * sqrt(x) = 30√6

Then, divide both sides by 3:

sqrt(x) = (30√6) / 3 = 10√6

So sqrt(x) equals 10√6. To solve for x, square both sides:

(sqrt(x))² = (10√6)²

Which simplifies to:

x = 10² * (√6)²

Calculating that:

10 squared is 100, and (√6) squared is 6. So:

x = 100 * 6 = 600

Wait, so x is 600? Let me verify if that makes sense. Let me check my steps again.

Starting with y varies inversely as sqrt(x), so y = k / sqrt(x). Given x=24, y=15. So k = 15*sqrt(24). Then sqrt(24) is 2*sqrt(6), so k is 15*2*sqrt(6) = 30√6. Correct.

Then, when y=3, set up equation 3 = 30√6 / sqrt(x). Multiply both sides by sqrt(x): 3 sqrt(x) = 30√6. Divide by 3: sqrt(x) = 10√6. Squared: x = 100*6 = 600. Seems right.

Let me check with substitution. If x=600, then sqrt(x) is sqrt(600). Let me compute sqrt(600). Well, sqrt(600) is sqrt(100*6) which is 10*sqrt(6). So then, y = k / sqrt(x) = 30√6 / (10√6). Let me compute that. The √6 cancels out, so 30 / 10 = 3. Which is the y we wanted. Perfect, that checks out. So x=600 is correct.

So, the answer should be 600. Wait, but let me make sure I didn't make any calculation errors. Let me go through once more.

Original equation: y = k / sqrt(x). Given x=24, y=15. So k = y * sqrt(x) = 15 * sqrt(24). sqrt(24) is approx 4.899, but exact value is 2*sqrt(6). So k is 15*2*sqrt(6) = 30√6. Correct.

Then when y=3, 3 = 30√6 / sqrt(x). So sqrt(x) = (30√6)/3 = 10√6. Then x = (10√6)^2 = 100*6 = 600. Yep. All steps are consistent. Therefore, the answer is 600.

Wait, but maybe I should write sqrt(600) as 10*sqrt(6) again. If we plug back into the original equation, we get y = 30√6 / (10√6) = 3, which is correct. So, all good.

I think that's solid. So, I don't see any mistakes here. Therefore, x is 600 when y is 3.

**Final Answer**
The value of $ x $ when $ y = 3 $ is \boxed{600}.

To solve the problem, we begin by understanding the nature of the relationship described:

> "The value of $ y $ varies inversely as $ \sqrt{x} $."

This means that $ y $ is inversely proportional to the square root of $ x $. In mathematical terms, the relationship can be expressed as:

$$
y = \frac{k}{\sqrt{x}}
$$

where $ k $ is the constant of proportionality.

---

### Step 1: Use the given values to find the constant $ k $

We are given that:

$$
\text{When } x = 24, \quad y = 15
$$

Substitute into the equation:

$$
15 = \frac{k}{\sqrt{24}}
$$

Multiply both sides by $ \sqrt{24} $:

$$
k = 15 \sqrt{24}
$$

Since $ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} $, we can simplify:

$$
k = 15 \cdot 2\sqrt{6} = 30\sqrt{6}
$$

---

### Step 2: Use the constant to find $ x $ when $ y = 3 $

We now know that:

$$
y = \frac{30\sqrt{6}}{\sqrt{x}}
$$

We are given $ y = 3 $, so:

$$
3 = \frac{30\sqrt{6}}{\sqrt{x}}
$$

Now, solve for $ \sqrt{x} $:

$$
\sqrt{x} = \frac{30\sqrt{6}}{3} = 10\sqrt{6}
$$

Square both sides to solve for $ x $:

$$
x = (10\sqrt{6})^2 = 10^2 \cdot (\sqrt{6})^2 = 100 \cdot 6 = 600
$$

---

### Final Answer

$$
\boxed{600}
$$

Solution:
Since $y$ and $\sqrt{x}$ are inversely proportional, this means that $y\sqrt{x}=k$ for some constant $k$.  Substituting the given values, when $x=24$ and $y=15$, we find that $15\sqrt{24}=30\sqrt{6}=k$.  Therefore, when $y=3$, we can solve for $x$: \begin{align*}
3\cdot\sqrt{x}&=30\sqrt{6}\\
\Rightarrow\qquad (\sqrt{x})^2&=(10\sqrt{6})^2\\
\Rightarrow\qquad x&=100\cdot6\\
&=\boxed{600}
\end{align*}