Explanation:
Okay, so I need to solve the equation √(2x)/√(3x - 1) = 3/2. Hmm, let me think about how to approach this. First, I remember that when dealing with equations involving square roots, it's often helpful to eliminate the radicals by squaring both sides. But I need to be careful because squaring can sometimes introduce extraneous solutions. Let me write down the equation again to make sure I have it right:

√(2x) / √(3x - 1) = 3/2

Alright, so maybe I can start by cross-multiplying to get rid of the fractions. Wait, actually, since it's a ratio of square roots, maybe squaring both sides would be better. Let me try that. If I square both sides, the left side becomes [√(2x)/√(3x - 1)]², which simplifies to (2x)/(3x - 1). The right side becomes (3/2)², which is 9/4. So now the equation is:

(2x)/(3x - 1) = 9/4

Okay, that looks more manageable. Now I can solve for x by cross-multiplying. Let me do that. Multiply both sides by (3x - 1) and by 4 to eliminate denominators. So:

4 * 2x = 9 * (3x - 1)

Calculating the left side: 4*2x is 8x. The right side: 9*(3x - 1) is 27x - 9. So now the equation is:

8x = 27x - 9

Hmm, let me subtract 8x from both sides to get all the x terms on one side. That would be:

0 = 27x - 9 - 8x

Simplify that: 0 = 19x - 9. Then, adding 9 to both sides:

9 = 19x

So, x = 9/19. Wait, is that right? Let me check my steps again to make sure I didn't make a mistake.

Starting from the original equation:

√(2x)/√(3x - 1) = 3/2

Square both sides:

(2x)/(3x - 1) = 9/4

Cross-multiplying:

4*2x = 9*(3x - 1)

Which is 8x = 27x - 9. Then subtract 8x:

0 = 19x - 9 => 19x = 9 => x = 9/19. Hmm, seems correct. But I need to check if this solution is valid in the original equation because sometimes squaring can introduce extraneous solutions. Let me plug x = 9/19 back into the original equation.

First, compute the left side: √(2x)/√(3x - 1). Let's compute 2x and 3x - 1.

2x = 2*(9/19) = 18/19

3x - 1 = 3*(9/19) - 1 = 27/19 - 19/19 = 8/19

So the left side becomes √(18/19) / √(8/19). Let me simplify that. Since both denominators are √(1/19), they will cancel out. Wait, actually:

√(18/19) divided by √(8/19) is equal to √(18/19) * √(19/8) = √[(18/19)*(19/8)] = √(18/8) = √(9/4) = 3/2. Which is exactly the right side of the equation. So that checks out. Therefore, x = 9/19 is indeed a valid solution.

Wait, but let me make sure that the expressions under the square roots are positive because square roots of negative numbers aren't real. Let me check the original equation's domain. The expressions inside the square roots must be positive.

For √(2x), 2x > 0 => x > 0.

For √(3x - 1), 3x - 1 > 0 => 3x > 1 => x > 1/3.

So the domain is x > 1/3. Now, x = 9/19. Let me check if 9/19 is greater than 1/3. Let me convert them to decimals. 1/3 is approximately 0.333..., and 9 divided by 19 is approximately 0.473... So yes, 0.473 is greater than 0.333, so it's in the domain. Therefore, x = 9/19 is valid.

Wait, but just to be thorough, let me check if there are any other solutions. When I squared both sides, sometimes equations can have more solutions, but in this case, since we had a linear equation after squaring, there's only one solution. So I think that's the only one. Therefore, the answer is 9/19.

But let me just recap all steps again to make sure I didn't skip anything. Original equation: square roots. Squared both sides, cross-multiplied, solved linear equation, checked solution in original equation, checked domain. All steps seem okay. So I think that's correct.

Alternatively, maybe I can try another approach. Let me try cross-multiplying first before squaring. Let me see:

Original equation: √(2x)/√(3x - 1) = 3/2. Cross-multiplying gives 2√(2x) = 3√(3x - 1). Then, maybe square both sides here. Let me try that.

Left side squared: [2√(2x)]² = 4*(2x) = 8x

Right side squared: [3√(3x - 1)]² = 9*(3x - 1) = 27x - 9

So again, 8x = 27x - 9, leading to same equation. So same result. So same answer. So that's consistent. Therefore, x = 9/19 is correct.

Alternatively, maybe I can rationalize or manipulate the original equation differently, but I think the steps are solid. So I think the answer is 9/19. Let me just check once more with substitution.

Original equation: √(2x)/√(3x - 1) = 3/2. Let me compute left side with x = 9/19.

Compute numerator: √(2*(9/19)) = √(18/19). Denominator: √(3*(9/19) - 1) = √(27/19 - 19/19) = √(8/19). Therefore, the ratio is √(18/19)/√(8/19) = √(18/8) = √(9/4) = 3/2. Perfect, matches the right side. So that's correct.

Therefore, I think I can confidently say that x = 9/19 is the solution.

**Final Answer**
The solution is \boxed{\dfrac{9}{19}}.

To solve the equation:

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{3}{2}
$$

we begin by eliminating the square roots. A common and effective strategy is to square both sides of the equation to remove the radicals. Let's proceed step by step.

---

### Step 1: Square Both Sides

$$
\left( \frac{\sqrt{2x}}{\sqrt{3x - 1}} \right)^2 = \left( \frac{3}{2} \right)^2
$$

$$
\frac{2x}{3x - 1} = \frac{9}{4}
$$

---

### Step 2: Cross-Multiply to Eliminate Denominators

$$
4 \cdot 2x = 9 \cdot (3x - 1)
$$

$$
8x = 27x - 9
$$

---

### Step 3: Solve the Linear Equation

$$
8x - 27x = -9
$$

$$
-19x = -9
$$

$$
x = \frac{9}{19}
$$

---

### Step 4: Verify the Solution

We must ensure that the solution satisfies the original equation and that the expressions under the square roots are positive (since square roots of negative numbers are not real).

- Check the domain:
  - $ 2x > 0 \Rightarrow x > 0 $
  - $ 3x - 1 > 0 \Rightarrow x > \frac{1}{3} $

Since $ \frac{9}{19} \approx 0.473 > \frac{1}{3} $, the solution is within the domain.

- Now, substitute $ x = \frac{9}{19} $ into the original equation:

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{\sqrt{2 \cdot \frac{9}{19}}}{\sqrt{3 \cdot \frac{9}{19} - 1}} = \frac{\sqrt{\frac{18}{19}}}{\sqrt{\frac{8}{19}}}
$$

$$
= \sqrt{\frac{18}{19}} \div \sqrt{\frac{8}{19}} = \sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2}
$$

This matches the right-hand side of the original equation, confirming that the solution is valid.

---

### Final Answer

$$
\boxed{\dfrac{9}{19}}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation √(2x)/√(3x - 1) = 3/2. Hmm, let me think about how to approach this. First, I remember that when dealing with equations involving square roots, it's often helpful to eliminate the radicals by squaring both sides. But I need to be careful because squaring can sometimes introduce extraneous solutions. Let me write down the equation again to make sure I have it right:

√(2x) / √(3x - 1) = 3/2

Alright, so maybe I can start by cross-multiplying to get rid of the fractions. Wait, actually, since it's a ratio of square roots, maybe squaring both sides would be better. Let me try that. If I square both sides, the left side becomes [√(2x)/√(3x - 1)]², which simplifies to (2x)/(3x - 1). The right side becomes (3/2)², which is 9/4. So now the equation is:

(2x)/(3x - 1) = 9/4

Okay, that looks more manageable. Now I can solve for x by cross-multiplying. Let me do that. Multiply both sides by (3x - 1) and by 4 to eliminate denominators. So:

4 * 2x = 9 * (3x - 1)

Calculating the left side: 4*2x is 8x. The right side: 9*(3x - 1) is 27x - 9. So now the equation is:

8x = 27x - 9

Hmm, let me subtract 8x from both sides to get all the x terms on one side. That would be:

0 = 27x - 9 - 8x

Simplify that: 0 = 19x - 9. Then, adding 9 to both sides:

9 = 19x

So, x = 9/19. Wait, is that right? Let me check my steps again to make sure I didn't make a mistake.

Starting from the original equation:

√(2x)/√(3x - 1) = 3/2

Square both sides:

(2x)/(3x - 1) = 9/4

Cross-multiplying:

4*2x = 9*(3x - 1)

Which is 8x = 27x - 9. Then subtract 8x:

0 = 19x - 9 => 19x = 9 => x = 9/19. Hmm, seems correct. But I need to check if this solution is valid in the original equation because sometimes squaring can introduce extraneous solutions. Let me plug x = 9/19 back into the original equation.

First, compute the left side: √(2x)/√(3x - 1). Let's compute 2x and 3x - 1.

2x = 2*(9/19) = 18/19

3x - 1 = 3*(9/19) - 1 = 27/19 - 19/19 = 8/19

So the left side becomes √(18/19) / √(8/19). Let me simplify that. Since both denominators are √(1/19), they will cancel out. Wait, actually:

√(18/19) divided by √(8/19) is equal to √(18/19) * √(19/8) = √[(18/19)*(19/8)] = √(18/8) = √(9/4) = 3/2. Which is exactly the right side of the equation. So that checks out. Therefore, x = 9/19 is indeed a valid solution.

Wait, but let me make sure that the expressions under the square roots are positive because square roots of negative numbers aren't real. Let me check the original equation's domain. The expressions inside the square roots must be positive.

For √(2x), 2x > 0 => x > 0.

For √(3x - 1), 3x - 1 > 0 => 3x > 1 => x > 1/3.

So the domain is x > 1/3. Now, x = 9/19. Let me check if 9/19 is greater than 1/3. Let me convert them to decimals. 1/3 is approximately 0.333..., and 9 divided by 19 is approximately 0.473... So yes, 0.473 is greater than 0.333, so it's in the domain. Therefore, x = 9/19 is valid.

Wait, but just to be thorough, let me check if there are any other solutions. When I squared both sides, sometimes equations can have more solutions, but in this case, since we had a linear equation after squaring, there's only one solution. So I think that's the only one. Therefore, the answer is 9/19.

But let me just recap all steps again to make sure I didn't skip anything. Original equation: square roots. Squared both sides, cross-multiplied, solved linear equation, checked solution in original equation, checked domain. All steps seem okay. So I think that's correct.

Alternatively, maybe I can try another approach. Let me try cross-multiplying first before squaring. Let me see:

Original equation: √(2x)/√(3x - 1) = 3/2. Cross-multiplying gives 2√(2x) = 3√(3x - 1). Then, maybe square both sides here. Let me try that.

Left side squared: [2√(2x)]² = 4*(2x) = 8x

Right side squared: [3√(3x - 1)]² = 9*(3x - 1) = 27x - 9

So again, 8x = 27x - 9, leading to same equation. So same result. So same answer. So that's consistent. Therefore, x = 9/19 is correct.

Alternatively, maybe I can rationalize or manipulate the original equation differently, but I think the steps are solid. So I think the answer is 9/19. Let me just check once more with substitution.

Original equation: √(2x)/√(3x - 1) = 3/2. Let me compute left side with x = 9/19.

Compute numerator: √(2*(9/19)) = √(18/19). Denominator: √(3*(9/19) - 1) = √(27/19 - 19/19) = √(8/19). Therefore, the ratio is √(18/19)/√(8/19) = √(18/8) = √(9/4) = 3/2. Perfect, matches the right side. So that's correct.

Therefore, I think I can confidently say that x = 9/19 is the solution.

**Final Answer**
The solution is \boxed{\dfrac{9}{19}}.

To solve the equation:

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{3}{2}
$$

we begin by eliminating the square roots. A common and effective strategy is to square both sides of the equation to remove the radicals. Let's proceed step by step.

---

### Step 1: Square Both Sides

$$
\left( \frac{\sqrt{2x}}{\sqrt{3x - 1}} \right)^2 = \left( \frac{3}{2} \right)^2
$$

$$
\frac{2x}{3x - 1} = \frac{9}{4}
$$

---

### Step 2: Cross-Multiply to Eliminate Denominators

$$
4 \cdot 2x = 9 \cdot (3x - 1)
$$

$$
8x = 27x - 9
$$

---

### Step 3: Solve the Linear Equation

$$
8x - 27x = -9
$$

$$
-19x = -9
$$

$$
x = \frac{9}{19}
$$

---

### Step 4: Verify the Solution

We must ensure that the solution satisfies the original equation and that the expressions under the square roots are positive (since square roots of negative numbers are not real).

- Check the domain:
  - $ 2x > 0 \Rightarrow x > 0 $
  - $ 3x - 1 > 0 \Rightarrow x > \frac{1}{3} $

Since $ \frac{9}{19} \approx 0.473 > \frac{1}{3} $, the solution is within the domain.

- Now, substitute $ x = \frac{9}{19} $ into the original equation:

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{\sqrt{2 \cdot \frac{9}{19}}}{\sqrt{3 \cdot \frac{9}{19} - 1}} = \frac{\sqrt{\frac{18}{19}}}{\sqrt{\frac{8}{19}}}
$$

$$
= \sqrt{\frac{18}{19}} \div \sqrt{\frac{8}{19}} = \sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2}
$$

This matches the right-hand side of the original equation, confirming that the solution is valid.

---

### Final Answer

$$
\boxed{\dfrac{9}{19}}
$$

Solution:
We can start by cross-multiplying: \begin{align*} 3\sqrt{3x-1}&=2\sqrt{2x}
\\\Rightarrow \qquad (3\sqrt{3x-1})^2 &=(2\sqrt{2x})^2
\\\Rightarrow \qquad 9(3x-1)& =4(2x)
\\\Rightarrow \qquad 27x-9& =8x
\\ \Rightarrow \qquad19x&=9
\\ \Rightarrow \qquad x&=\boxed{\frac9{19}}.
\end{align*}Checking, we see that this value of $x$ does indeed work, so it is not an extraneous solution.


Explanation:
Okay, so I need to find the value of n where the sum 1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 is congruent to n modulo 9. And n has to be between 0 and 8. Alright, let me think about how to approach this.

First, I remember that when dealing with modulo 9, there's a trick where you can sum the digits of a number and then take that sum modulo 9. This is because 10 ≡ 1 mod 9, so any power of 10 is also congruent to 1 mod 9. Therefore, a number like 123... is congruent to the sum of its digits mod 9. So maybe I can use this property to simplify each term in the sum before adding them up?

Let me verify that. For example, take the number 12. Its digits are 1 and 2, so 1 + 2 = 3. And 12 mod 9 is 3, which matches. Similarly, 123: 1 + 2 + 3 = 6, and 123 divided by 9 is 13*9=117, 123-117=6. Yep, that works. So this seems like a valid approach.

Therefore, instead of calculating each term's actual value and then adding them up, which might be tedious and prone to errors, I can just compute the sum of the digits of each term, then add those sums together, and then take that total modulo 9. That should give me the same result as the original sum modulo 9. That sounds manageable.

So let me list out each term and compute the sum of their digits:

1. The first term is 1. Its digit sum is 1.
2. The second term is 12. Digits are 1 and 2, sum is 1 + 2 = 3.
3. The third term is 123. Digits are 1, 2, 3. Sum is 1 + 2 + 3 = 6.
4. The fourth term is 1234. Digits are 1, 2, 3, 4. Sum is 1 + 2 + 3 + 4 = 10.
5. The fifth term is 12345. Digits are 1, 2, 3, 4, 5. Sum is 1 + 2 + 3 + 4 + 5 = 15.
6. The sixth term is 123456. Digits are 1, 2, 3, 4, 5, 6. Sum is 1 + 2 + 3 + 4 + 5 + 6 = 21.
7. The seventh term is 1234567. Digits are 1, 2, 3, 4, 5, 6, 7. Sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
8. The eighth term is 12345678. Digits are 1, 2, 3, 4, 5, 6, 7, 8. Sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.

Okay, so now I have the digit sums for each term. Let me write them down again:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 10

Term 5: 15

Term 6: 21

Term 7: 28

Term 8: 36

Now, I need to add all these numbers together and then take the result modulo 9. Let me compute the total sum first.

Let me add them step by step:

Start with 1 (from term 1).

Add term 2: 1 + 3 = 4.

Add term 3: 4 + 6 = 10.

Add term 4: 10 + 10 = 20.

Add term 5: 20 + 15 = 35.

Add term 6: 35 + 21 = 56.

Add term 7: 56 + 28 = 84.

Add term 8: 84 + 36 = 120.

So the total sum of the digit sums is 120. Now, I need to compute 120 mod 9.

But wait, maybe I can simplify this even further. Since each digit sum was already congruent to the original number mod 9, then adding them up is equivalent to adding the original numbers mod 9. Therefore, the total sum mod 9 is equal to 120 mod 9. Let me compute that.

Alternatively, maybe I can compute each digit sum mod 9 first, then add those up mod 9. Let me try that approach as well to cross-verify.

Let me compute each digit sum mod 9:

Term 1: 1 mod 9 = 1

Term 2: 3 mod 9 = 3

Term 3: 6 mod 9 = 6

Term 4: 10 mod 9 = 1 (since 9*1=9, 10-9=1)

Term 5: 15 mod 9. 9*1=9, 15-9=6. So 6.

Term 6: 21 mod 9. 9*2=18, 21-18=3. So 3.

Term 7: 28 mod 9. Let's see, 9*3=27, 28-27=1. So 1.

Term 8: 36 mod 9. Well, 36 is 4*9, so 0.

So now, adding these up:

1 + 3 + 6 + 1 + 6 + 3 + 1 + 0.

Let me compute that step by step:

Start with 1.

Add 3: 1 + 3 = 4.

Add 6: 4 + 6 = 10.

Add 1: 10 + 1 = 11.

Add 6: 11 + 6 = 17.

Add 3: 17 + 3 = 20.

Add 1: 20 + 1 = 21.

Add 0: 21 + 0 = 21.

So the total is 21. Now, 21 mod 9 is... 9*2=18, 21-18=3. Therefore, 21 mod 9 is 3. Therefore, the total sum mod 9 is 3. Therefore, n is 3?

Wait, but earlier when I added all the digit sums to get 120, then 120 mod 9. Let me check that as well. 120 divided by 9: 9*13=117, 120-117=3. So 120 mod 9 is also 3. So both methods give me 3. Therefore, n is 3. So the answer is 3?

But let me check again, just to make sure I didn't make any mistakes in my calculations. It's easy to make an arithmetic error here.

First, let me recheck the digit sums:

Term 1: 1. Correct.

Term 2: 1+2=3. Correct.

Term 3: 1+2+3=6. Correct.

Term 4: 1+2+3+4=10. Correct.

Term 5: 1+2+3+4+5=15. Correct.

Term 6: 1+2+3+4+5+6=21. Correct.

Term 7: 1+2+3+4+5+6+7=28. Correct.

Term 8: 1+2+3+4+5+6+7+8=36. Correct.

So the digit sums are correct. Then adding them up:

1 + 3 = 4; 4 + 6 = 10; 10 + 10 = 20; 20 + 15 = 35; 35 + 21 = 56; 56 + 28 = 84; 84 + 36 = 120. Correct. 120 mod 9 is 3. Correct.

Alternatively, when I took each digit sum mod 9:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 10 mod 9 is 1

Term 5: 15 mod 9 is 6

Term 6: 21 mod 9 is 3

Term 7: 28 mod 9 is 1

Term 8: 36 mod 9 is 0

Adding those: 1 + 3 + 6 + 1 + 6 + 3 + 1 + 0. Let me add again:

1 + 3 = 4; 4 + 6 = 10; 10 + 1 = 11; 11 + 6 = 17; 17 + 3 = 20; 20 + 1 = 21; 21 + 0 = 21. 21 mod 9 is 3. Correct.

So both methods give me 3. Therefore, n is 3. Therefore, the answer is 3. But let me check once more by another approach, just to be thorough.

Another way: instead of calculating digit sums, maybe compute each term mod 9 directly, then add them up. Let me try that.

First, let me recall that for any number, its value mod 9 is equal to the sum of its digits mod 9. Therefore, each term mod 9 is equal to the sum of its digits mod 9. Therefore, this is the same as the previous approach. So, if I compute each term mod 9, then add them up, I should get the same result.

Alternatively, maybe I can compute each term mod 9 directly. Let me try that.

First term: 1 mod 9 = 1.

Second term: 12 mod 9. 12 divided by 9 is 1 with remainder 3. So 3.

Third term: 123 mod 9. Let me compute 123 divided by 9. 9*13=117, 123-117=6. So 6.

Fourth term: 1234 mod 9. Let me compute this. Alternatively, since 1234 is 123*10 + 4. Since 123 mod 9 is 6, then 123*10 mod 9 is 6*10 mod 9. 10 mod 9 is 1, so 6*1 = 6. Then add 4: 6 + 4 = 10. 10 mod 9 is 1. So 1234 mod 9 is 1. Which matches the previous result where the digit sum was 10, which mod 9 is 1. So that's consistent.

Fifth term: 12345 mod 9. Let me compute this. Alternatively, since 1234 mod 9 is 1, then 12345 is 1234*10 + 5. So 1*10 + 5 = 10 + 5 = 15. 15 mod 9 is 6. Which matches the digit sum of 15 mod 9 is 6. Correct.

Sixth term: 123456 mod 9. Let me compute. 12345 mod 9 is 6. Then 123456 is 12345*10 + 6. So 6*10 + 6 = 60 + 6 = 66. 66 mod 9. 9*7=63, 66-63=3. So 3. Which matches the digit sum of 21 mod 9 is 3. Correct.

Seventh term: 1234567 mod 9. Let me compute. 123456 mod 9 is 3. Then 1234567 is 123456*10 + 7. So 3*10 + 7 = 30 + 7 = 37. 37 mod 9. 9*4=36, 37-36=1. So 1. Which matches the digit sum of 28 mod 9 is 1. Correct.

Eighth term: 12345678 mod 9. Let me compute. 1234567 mod 9 is 1. Then 12345678 is 1234567*10 + 8. So 1*10 + 8 = 10 + 8 = 18. 18 mod 9 is 0. Which matches the digit sum of 36 mod 9 is 0. Correct.

So now, adding up all these mod 9 results:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 1

Term 5: 6

Term 6: 3

Term 7: 1

Term 8: 0

Adding them up: 1 + 3 + 6 + 1 + 6 + 3 + 1 + 0. Which is the same as before, 21. 21 mod 9 is 3. So same result. Therefore, n is 3. So seems like 3 is the answer.

But just to be absolutely thorough, let me compute the original sum and then take mod 9. Although the numbers are large, maybe I can compute them step by step.

Original sum: 1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678.

Let me compute this step by step:

Start with 1.

Add 12: 1 + 12 = 13.

Add 123: 13 + 123 = 136.

Add 1234: 136 + 1234 = 1370.

Add 12345: 1370 + 12345 = let's compute 1370 + 12345. 12345 + 1000 = 13345, then + 370 = 13345 + 370 = 13715.

Wait, let me check again:

Wait 1370 + 12345. Let me do it properly:

12345 + 1370:

Breakdown:

12345 + 1000 = 13345

13345 + 300 = 13645

13645 + 70 = 13715. Yes, that's correct.

So now total is 13715.

Next term: add 123456. So 13715 + 123456.

Let me compute that:

123456 + 13715. Let me think:

123456 + 10000 = 133456

133456 + 3000 = 136456

136456 + 700 = 137156

137156 + 15 = 137171. So total is 137171.

Next term: add 1234567. So 137171 + 1234567.

Let me compute:

1234567 + 137171. Let me break it down:

1234567 + 100000 = 1334567

1334567 + 30000 = 1364567

1364567 + 7000 = 1371567

1371567 + 171 = 1371738. So total is 1,371,738.

Next term: add 12345678. So 1,371,738 + 12,345,678.

Let me compute that:

1,371,738 + 12,345,678. Let me add them:

First, 1,371,738 + 12,345,678.

Adding the numbers:

Let me write them aligned:

1,371,738

+12,345,678

= ?

Adding:

Starting from the right:

8 + 8 = 16

3 + 7 = 10, plus 1 = 11

7 + 6 = 13, plus 1 = 14

1 + 5 = 6, plus 1 = 7

7 + 4 = 11, plus 1 = 12

3 + 3 = 6, plus 1 = 7

1 + 2 = 3

Wait, maybe I should do it more carefully:

Let me write the numbers:

1,371,738

+12,345,678

Let me convert them to the same number of digits:

1,371,738 is 01,371,738

Adding to 12,345,678:

So:

01,371,738

+12,345,678

= ?

Adding digit by digit:

Starting from the rightmost digit:

Units place: 8 + 8 = 16. Write down 6, carryover 1.

Tens place: 3 + 7 = 10 + 1 = 11. Write down 1, carryover 1.

Hundreds place: 7 + 6 = 13 + 1 = 14. Write down 4, carryover 1.

Thousands place: 1 + 5 = 6 + 1 = 7. Write down 7, no carryover.

Ten-thousands place: 7 + 4 = 11. Write down 1, carryover 1.

Hundred-thousands place: 3 + 3 = 6 + 1 = 7. Write down 7, no carryover.

Millions place: 1 + 2 = 3. Write down 3.

So the result is 13,717,416? Wait, let me check again:

Wait, let me write the numbers:

First number: 01,371,738

Second number: 12,345,678

Adding:

Starting from the right:

Digits:

Position: 1 (units): 8 + 8 = 16

Position 2 (tens): 3 + 7 = 10 + 1 = 11

Position 3 (hundreds): 7 + 6 = 13 + 1 = 14

Position 4 (thousands): 1 + 5 = 6 + 1 (carryover) = 7

Position 5 (ten thousands): 7 + 4 = 11

Position 6 (hundred thousands): 3 + 3 = 6

Position 7 (millions): 1 + 2 = 3

Wait, but the first number is 01,371,738 which is 1,371,738. So the digits are:

Millions: 1

Hundred thousands: 3

Ten thousands: 7

Thousands: 1

Hundreds: 7

Tens: 3

Units: 8

Second number is 12,345,678:

Digits:

Ten millions: 1

Millions: 2

Hundred thousands: 3

Ten thousands: 4

Thousands: 5

Hundreds: 6

Tens: 7

Units: 8

So when adding:

Starting from units:

Units: 8 + 8 = 16

Tens: 3 + 7 = 10 + 1 = 11

Hundreds: 7 + 6 = 13 + 1 = 14

Thousands: 1 + 5 = 6 + 1 (carryover) = 7

Ten thousands: 7 + 4 = 11

Hundred thousands: 3 + 3 = 6

Millions: 1 + 2 = 3

Ten millions: 1 (from the second number)

So putting it all together:

Ten millions: 1

Millions: 3

Hundred thousands: 6

Ten thousands: 1 (from ten thousands place, which was 11, so write 1 and carryover 1?)

Wait, hold on, maybe I messed up the positions. Let me try again.

Wait, when adding:

Let me write both numbers with commas for clarity:

First number: 1,371,738

Second number: 12,345,678

Let me write them aligned:

          1,371,738

+       12,345,678

= ?

Let me add them:

Starting from the right:

Units: 8 + 8 = 16. Write 6, carryover 1.

Tens: 3 + 7 = 10 + 1 = 11. Write 1, carryover 1.

Hundreds: 7 + 6 = 13 + 1 = 14. Write 4, carryover 1.

Thousands: 1 + 5 = 6 + 1 = 7. Write 7, no carryover.

Ten thousands: 7 + 4 = 11. Write 1, carryover 1.

Hundred thousands: 3 + 3 = 6 + 1 = 7. Write 7, no carryover.

Millions: 1 + 2 = 3. Write 3, no carryover.

Ten millions: 0 + 1 = 1. Write 1.

So the result is 13,717,416? Wait:

Wait, let me reconstruct the number:

Starting from the left:

Ten millions: 1

Millions: 3

Hundred thousands: 7

Ten thousands: 1

Thousands: 7

Hundreds: 4

Tens: 1

Units: 6

So that's 13,717,416. Wait, but let me check:

Wait, the positions:

After adding, the digits are:

Ten millions: 1

Millions: 3

Hundred thousands: 7

Ten thousands: 1

Thousands: 7

Hundreds: 4

Tens: 1

Units: 6

So the number is 1 (ten millions), 3 (millions), 7 (hundred thousands), 1 (ten thousands), 7 (thousands), 4 (hundreds), 1 (tens), 6 (units). So that's 13,717,416. Let me check with another method.

Alternatively, 1,371,738 + 12,345,678. Let me compute 1,371,738 + 12,345,678.

Let me think of 1,371,738 + 12,345,678 = 1,371,738 + 12,345,678.

Let me add 1,371,738 + 12,345,678:

First, 1,371,738 + 12,345,678 = ?

Well, 1,371,738 + 12,345,678 = 13,717,416. Let me check with another way:

Let me compute 1,371,738 + 12,345,678:

Break down 12,345,678 into 12,000,000 + 345,678.

So 1,371,738 + 12,000,000 = 13,371,738.

Then add 345,678: 13,371,738 + 345,678.

Compute 13,371,738 + 300,000 = 13,671,738.

Then add 45,678: 13,671,738 + 45,678 = 13,717,416. Yes, same result. So total sum is 13,717,416.

Now, need to compute 13,717,416 mod 9. Let me do that.

Again, since 10 ≡ 1 mod 9, the number mod 9 is equal to the sum of its digits. Let me compute the sum of the digits of 13,717,416.

Digits: 1, 3, 7, 1, 7, 4, 1, 6.

Sum: 1 + 3 + 7 + 1 + 7 + 4 + 1 + 6.

Let me compute:

1 + 3 = 4

4 + 7 = 11

11 + 1 = 12

12 + 7 = 19

19 + 4 = 23

23 + 1 = 24

24 + 6 = 30.

So the sum of digits is 30. Then 30 mod 9. 9*3=27, 30-27=3. Therefore, 30 mod 9 is 3. Therefore, the total sum mod 9 is 3. Which matches our previous results. Therefore, n is 3. So this confirms it again.

Therefore, regardless of the method I use, the result is 3. Therefore, the answer is 3. I think that's solid. I don't see any mistakes in my calculations. All methods lead to the same conclusion. Therefore, I can be confident that n is 3.

**Final Answer**
The value of $ n $ is \boxed{3}.

To solve the problem, we are asked to find the value of $ n $ such that:

$$
1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 \equiv n \pmod{9}
$$

with $ 0 \leq n < 9 $.

---

### Step 1: Use the Modulo 9 Rule for Digit Sums

A key property of modulo 9 is that:

$$
n \equiv \text{sum of digits of } n \pmod{9}
$$

This is because $ 10 \equiv 1 \pmod{9} $, so any power of 10 is also congruent to 1 modulo 9. Therefore, the value of a number modulo 9 is the same as the sum of its digits modulo 9.

This allows us to simplify the problem by computing the **sum of the digits** of each term in the sequence, then summing those digit sums, and finally taking the result modulo 9.

---

### Step 2: Compute the Digit Sums

Let’s compute the digit sum for each term:

- $ 1 $ → $ 1 $
- $ 12 $ → $ 1 + 2 = 3 $
- $ 123 $ → $ 1 + 2 + 3 = 6 $
- $ 1234 $ → $ 1 + 2 + 3 + 4 = 10 $
- $ 12345 $ → $ 1 + 2 + 3 + 4 + 5 = 15 $
- $ 123456 $ → $ 1 + 2 + 3 + 4 + 5 + 6 = 21 $
- $ 1234567 $ → $ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 $
- $ 12345678 $ → $ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 $

Now, sum these digit sums:

$$
1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120
$$

---

### Step 3: Compute $ 120 \mod 9 $

We now compute:

$$
120 \div 9 = 13 \text{ remainder } 3
$$

So,

$$
120 \equiv 3 \pmod{9}
$$

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of n where the sum 1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 is congruent to n modulo 9. And n has to be between 0 and 8. Alright, let me think about how to approach this.

First, I remember that when dealing with modulo 9, there's a trick where you can sum the digits of a number and then take that sum modulo 9. This is because 10 ≡ 1 mod 9, so any power of 10 is also congruent to 1 mod 9. Therefore, a number like 123... is congruent to the sum of its digits mod 9. So maybe I can use this property to simplify each term in the sum before adding them up?

Let me verify that. For example, take the number 12. Its digits are 1 and 2, so 1 + 2 = 3. And 12 mod 9 is 3, which matches. Similarly, 123: 1 + 2 + 3 = 6, and 123 divided by 9 is 13*9=117, 123-117=6. Yep, that works. So this seems like a valid approach.

Therefore, instead of calculating each term's actual value and then adding them up, which might be tedious and prone to errors, I can just compute the sum of the digits of each term, then add those sums together, and then take that total modulo 9. That should give me the same result as the original sum modulo 9. That sounds manageable.

So let me list out each term and compute the sum of their digits:

1. The first term is 1. Its digit sum is 1.
2. The second term is 12. Digits are 1 and 2, sum is 1 + 2 = 3.
3. The third term is 123. Digits are 1, 2, 3. Sum is 1 + 2 + 3 = 6.
4. The fourth term is 1234. Digits are 1, 2, 3, 4. Sum is 1 + 2 + 3 + 4 = 10.
5. The fifth term is 12345. Digits are 1, 2, 3, 4, 5. Sum is 1 + 2 + 3 + 4 + 5 = 15.
6. The sixth term is 123456. Digits are 1, 2, 3, 4, 5, 6. Sum is 1 + 2 + 3 + 4 + 5 + 6 = 21.
7. The seventh term is 1234567. Digits are 1, 2, 3, 4, 5, 6, 7. Sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
8. The eighth term is 12345678. Digits are 1, 2, 3, 4, 5, 6, 7, 8. Sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.

Okay, so now I have the digit sums for each term. Let me write them down again:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 10

Term 5: 15

Term 6: 21

Term 7: 28

Term 8: 36

Now, I need to add all these numbers together and then take the result modulo 9. Let me compute the total sum first.

Let me add them step by step:

Start with 1 (from term 1).

Add term 2: 1 + 3 = 4.

Add term 3: 4 + 6 = 10.

Add term 4: 10 + 10 = 20.

Add term 5: 20 + 15 = 35.

Add term 6: 35 + 21 = 56.

Add term 7: 56 + 28 = 84.

Add term 8: 84 + 36 = 120.

So the total sum of the digit sums is 120. Now, I need to compute 120 mod 9.

But wait, maybe I can simplify this even further. Since each digit sum was already congruent to the original number mod 9, then adding them up is equivalent to adding the original numbers mod 9. Therefore, the total sum mod 9 is equal to 120 mod 9. Let me compute that.

Alternatively, maybe I can compute each digit sum mod 9 first, then add those up mod 9. Let me try that approach as well to cross-verify.

Let me compute each digit sum mod 9:

Term 1: 1 mod 9 = 1

Term 2: 3 mod 9 = 3

Term 3: 6 mod 9 = 6

Term 4: 10 mod 9 = 1 (since 9*1=9, 10-9=1)

Term 5: 15 mod 9. 9*1=9, 15-9=6. So 6.

Term 6: 21 mod 9. 9*2=18, 21-18=3. So 3.

Term 7: 28 mod 9. Let's see, 9*3=27, 28-27=1. So 1.

Term 8: 36 mod 9. Well, 36 is 4*9, so 0.

So now, adding these up:

1 + 3 + 6 + 1 + 6 + 3 + 1 + 0.

Let me compute that step by step:

Start with 1.

Add 3: 1 + 3 = 4.

Add 6: 4 + 6 = 10.

Add 1: 10 + 1 = 11.

Add 6: 11 + 6 = 17.

Add 3: 17 + 3 = 20.

Add 1: 20 + 1 = 21.

Add 0: 21 + 0 = 21.

So the total is 21. Now, 21 mod 9 is... 9*2=18, 21-18=3. Therefore, 21 mod 9 is 3. Therefore, the total sum mod 9 is 3. Therefore, n is 3?

Wait, but earlier when I added all the digit sums to get 120, then 120 mod 9. Let me check that as well. 120 divided by 9: 9*13=117, 120-117=3. So 120 mod 9 is also 3. So both methods give me 3. Therefore, n is 3. So the answer is 3?

But let me check again, just to make sure I didn't make any mistakes in my calculations. It's easy to make an arithmetic error here.

First, let me recheck the digit sums:

Term 1: 1. Correct.

Term 2: 1+2=3. Correct.

Term 3: 1+2+3=6. Correct.

Term 4: 1+2+3+4=10. Correct.

Term 5: 1+2+3+4+5=15. Correct.

Term 6: 1+2+3+4+5+6=21. Correct.

Term 7: 1+2+3+4+5+6+7=28. Correct.

Term 8: 1+2+3+4+5+6+7+8=36. Correct.

So the digit sums are correct. Then adding them up:

1 + 3 = 4; 4 + 6 = 10; 10 + 10 = 20; 20 + 15 = 35; 35 + 21 = 56; 56 + 28 = 84; 84 + 36 = 120. Correct. 120 mod 9 is 3. Correct.

Alternatively, when I took each digit sum mod 9:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 10 mod 9 is 1

Term 5: 15 mod 9 is 6

Term 6: 21 mod 9 is 3

Term 7: 28 mod 9 is 1

Term 8: 36 mod 9 is 0

Adding those: 1 + 3 + 6 + 1 + 6 + 3 + 1 + 0. Let me add again:

1 + 3 = 4; 4 + 6 = 10; 10 + 1 = 11; 11 + 6 = 17; 17 + 3 = 20; 20 + 1 = 21; 21 + 0 = 21. 21 mod 9 is 3. Correct.

So both methods give me 3. Therefore, n is 3. Therefore, the answer is 3. But let me check once more by another approach, just to be thorough.

Another way: instead of calculating digit sums, maybe compute each term mod 9 directly, then add them up. Let me try that.

First, let me recall that for any number, its value mod 9 is equal to the sum of its digits mod 9. Therefore, each term mod 9 is equal to the sum of its digits mod 9. Therefore, this is the same as the previous approach. So, if I compute each term mod 9, then add them up, I should get the same result.

Alternatively, maybe I can compute each term mod 9 directly. Let me try that.

First term: 1 mod 9 = 1.

Second term: 12 mod 9. 12 divided by 9 is 1 with remainder 3. So 3.

Third term: 123 mod 9. Let me compute 123 divided by 9. 9*13=117, 123-117=6. So 6.

Fourth term: 1234 mod 9. Let me compute this. Alternatively, since 1234 is 123*10 + 4. Since 123 mod 9 is 6, then 123*10 mod 9 is 6*10 mod 9. 10 mod 9 is 1, so 6*1 = 6. Then add 4: 6 + 4 = 10. 10 mod 9 is 1. So 1234 mod 9 is 1. Which matches the previous result where the digit sum was 10, which mod 9 is 1. So that's consistent.

Fifth term: 12345 mod 9. Let me compute this. Alternatively, since 1234 mod 9 is 1, then 12345 is 1234*10 + 5. So 1*10 + 5 = 10 + 5 = 15. 15 mod 9 is 6. Which matches the digit sum of 15 mod 9 is 6. Correct.

Sixth term: 123456 mod 9. Let me compute. 12345 mod 9 is 6. Then 123456 is 12345*10 + 6. So 6*10 + 6 = 60 + 6 = 66. 66 mod 9. 9*7=63, 66-63=3. So 3. Which matches the digit sum of 21 mod 9 is 3. Correct.

Seventh term: 1234567 mod 9. Let me compute. 123456 mod 9 is 3. Then 1234567 is 123456*10 + 7. So 3*10 + 7 = 30 + 7 = 37. 37 mod 9. 9*4=36, 37-36=1. So 1. Which matches the digit sum of 28 mod 9 is 1. Correct.

Eighth term: 12345678 mod 9. Let me compute. 1234567 mod 9 is 1. Then 12345678 is 1234567*10 + 8. So 1*10 + 8 = 10 + 8 = 18. 18 mod 9 is 0. Which matches the digit sum of 36 mod 9 is 0. Correct.

So now, adding up all these mod 9 results:

Term 1: 1

Term 2: 3

Term 3: 6

Term 4: 1

Term 5: 6

Term 6: 3

Term 7: 1

Term 8: 0

Adding them up: 1 + 3 + 6 + 1 + 6 + 3 + 1 + 0. Which is the same as before, 21. 21 mod 9 is 3. So same result. Therefore, n is 3. So seems like 3 is the answer.

But just to be absolutely thorough, let me compute the original sum and then take mod 9. Although the numbers are large, maybe I can compute them step by step.

Original sum: 1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678.

Let me compute this step by step:

Start with 1.

Add 12: 1 + 12 = 13.

Add 123: 13 + 123 = 136.

Add 1234: 136 + 1234 = 1370.

Add 12345: 1370 + 12345 = let's compute 1370 + 12345. 12345 + 1000 = 13345, then + 370 = 13345 + 370 = 13715.

Wait, let me check again:

Wait 1370 + 12345. Let me do it properly:

12345 + 1370:

Breakdown:

12345 + 1000 = 13345

13345 + 300 = 13645

13645 + 70 = 13715. Yes, that's correct.

So now total is 13715.

Next term: add 123456. So 13715 + 123456.

Let me compute that:

123456 + 13715. Let me think:

123456 + 10000 = 133456

133456 + 3000 = 136456

136456 + 700 = 137156

137156 + 15 = 137171. So total is 137171.

Next term: add 1234567. So 137171 + 1234567.

Let me compute:

1234567 + 137171. Let me break it down:

1234567 + 100000 = 1334567

1334567 + 30000 = 1364567

1364567 + 7000 = 1371567

1371567 + 171 = 1371738. So total is 1,371,738.

Next term: add 12345678. So 1,371,738 + 12,345,678.

Let me compute that:

1,371,738 + 12,345,678. Let me add them:

First, 1,371,738 + 12,345,678.

Adding the numbers:

Let me write them aligned:

1,371,738

+12,345,678

= ?

Adding:

Starting from the right:

8 + 8 = 16

3 + 7 = 10, plus 1 = 11

7 + 6 = 13, plus 1 = 14

1 + 5 = 6, plus 1 = 7

7 + 4 = 11, plus 1 = 12

3 + 3 = 6, plus 1 = 7

1 + 2 = 3

Wait, maybe I should do it more carefully:

Let me write the numbers:

1,371,738

+12,345,678

Let me convert them to the same number of digits:

1,371,738 is 01,371,738

Adding to 12,345,678:

So:

01,371,738

+12,345,678

= ?

Adding digit by digit:

Starting from the rightmost digit:

Units place: 8 + 8 = 16. Write down 6, carryover 1.

Tens place: 3 + 7 = 10 + 1 = 11. Write down 1, carryover 1.

Hundreds place: 7 + 6 = 13 + 1 = 14. Write down 4, carryover 1.

Thousands place: 1 + 5 = 6 + 1 = 7. Write down 7, no carryover.

Ten-thousands place: 7 + 4 = 11. Write down 1, carryover 1.

Hundred-thousands place: 3 + 3 = 6 + 1 = 7. Write down 7, no carryover.

Millions place: 1 + 2 = 3. Write down 3.

So the result is 13,717,416? Wait, let me check again:

Wait, let me write the numbers:

First number: 01,371,738

Second number: 12,345,678

Adding:

Starting from the right:

Digits:

Position: 1 (units): 8 + 8 = 16

Position 2 (tens): 3 + 7 = 10 + 1 = 11

Position 3 (hundreds): 7 + 6 = 13 + 1 = 14

Position 4 (thousands): 1 + 5 = 6 + 1 (carryover) = 7

Position 5 (ten thousands): 7 + 4 = 11

Position 6 (hundred thousands): 3 + 3 = 6

Position 7 (millions): 1 + 2 = 3

Wait, but the first number is 01,371,738 which is 1,371,738. So the digits are:

Millions: 1

Hundred thousands: 3

Ten thousands: 7

Thousands: 1

Hundreds: 7

Tens: 3

Units: 8

Second number is 12,345,678:

Digits:

Ten millions: 1

Millions: 2

Hundred thousands: 3

Ten thousands: 4

Thousands: 5

Hundreds: 6

Tens: 7

Units: 8

So when adding:

Starting from units:

Units: 8 + 8 = 16

Tens: 3 + 7 = 10 + 1 = 11

Hundreds: 7 + 6 = 13 + 1 = 14

Thousands: 1 + 5 = 6 + 1 (carryover) = 7

Ten thousands: 7 + 4 = 11

Hundred thousands: 3 + 3 = 6

Millions: 1 + 2 = 3

Ten millions: 1 (from the second number)

So putting it all together:

Ten millions: 1

Millions: 3

Hundred thousands: 6

Ten thousands: 1 (from ten thousands place, which was 11, so write 1 and carryover 1?)

Wait, hold on, maybe I messed up the positions. Let me try again.

Wait, when adding:

Let me write both numbers with commas for clarity:

First number: 1,371,738

Second number: 12,345,678

Let me write them aligned:

          1,371,738

+       12,345,678

= ?

Let me add them:

Starting from the right:

Units: 8 + 8 = 16. Write 6, carryover 1.

Tens: 3 + 7 = 10 + 1 = 11. Write 1, carryover 1.

Hundreds: 7 + 6 = 13 + 1 = 14. Write 4, carryover 1.

Thousands: 1 + 5 = 6 + 1 = 7. Write 7, no carryover.

Ten thousands: 7 + 4 = 11. Write 1, carryover 1.

Hundred thousands: 3 + 3 = 6 + 1 = 7. Write 7, no carryover.

Millions: 1 + 2 = 3. Write 3, no carryover.

Ten millions: 0 + 1 = 1. Write 1.

So the result is 13,717,416? Wait:

Wait, let me reconstruct the number:

Starting from the left:

Ten millions: 1

Millions: 3

Hundred thousands: 7

Ten thousands: 1

Thousands: 7

Hundreds: 4

Tens: 1

Units: 6

So that's 13,717,416. Wait, but let me check:

Wait, the positions:

After adding, the digits are:

Ten millions: 1

Millions: 3

Hundred thousands: 7

Ten thousands: 1

Thousands: 7

Hundreds: 4

Tens: 1

Units: 6

So the number is 1 (ten millions), 3 (millions), 7 (hundred thousands), 1 (ten thousands), 7 (thousands), 4 (hundreds), 1 (tens), 6 (units). So that's 13,717,416. Let me check with another method.

Alternatively, 1,371,738 + 12,345,678. Let me compute 1,371,738 + 12,345,678.

Let me think of 1,371,738 + 12,345,678 = 1,371,738 + 12,345,678.

Let me add 1,371,738 + 12,345,678:

First, 1,371,738 + 12,345,678 = ?

Well, 1,371,738 + 12,345,678 = 13,717,416. Let me check with another way:

Let me compute 1,371,738 + 12,345,678:

Break down 12,345,678 into 12,000,000 + 345,678.

So 1,371,738 + 12,000,000 = 13,371,738.

Then add 345,678: 13,371,738 + 345,678.

Compute 13,371,738 + 300,000 = 13,671,738.

Then add 45,678: 13,671,738 + 45,678 = 13,717,416. Yes, same result. So total sum is 13,717,416.

Now, need to compute 13,717,416 mod 9. Let me do that.

Again, since 10 ≡ 1 mod 9, the number mod 9 is equal to the sum of its digits. Let me compute the sum of the digits of 13,717,416.

Digits: 1, 3, 7, 1, 7, 4, 1, 6.

Sum: 1 + 3 + 7 + 1 + 7 + 4 + 1 + 6.

Let me compute:

1 + 3 = 4

4 + 7 = 11

11 + 1 = 12

12 + 7 = 19

19 + 4 = 23

23 + 1 = 24

24 + 6 = 30.

So the sum of digits is 30. Then 30 mod 9. 9*3=27, 30-27=3. Therefore, 30 mod 9 is 3. Therefore, the total sum mod 9 is 3. Which matches our previous results. Therefore, n is 3. So this confirms it again.

Therefore, regardless of the method I use, the result is 3. Therefore, the answer is 3. I think that's solid. I don't see any mistakes in my calculations. All methods lead to the same conclusion. Therefore, I can be confident that n is 3.

**Final Answer**
The value of $ n $ is \boxed{3}.

To solve the problem, we are asked to find the value of $ n $ such that:

$$
1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 \equiv n \pmod{9}
$$

with $ 0 \leq n < 9 $.

---

### Step 1: Use the Modulo 9 Rule for Digit Sums

A key property of modulo 9 is that:

$$
n \equiv \text{sum of digits of } n \pmod{9}
$$

This is because $ 10 \equiv 1 \pmod{9} $, so any power of 10 is also congruent to 1 modulo 9. Therefore, the value of a number modulo 9 is the same as the sum of its digits modulo 9.

This allows us to simplify the problem by computing the **sum of the digits** of each term in the sequence, then summing those digit sums, and finally taking the result modulo 9.

---

### Step 2: Compute the Digit Sums

Let’s compute the digit sum for each term:

- $ 1 $ → $ 1 $
- $ 12 $ → $ 1 + 2 = 3 $
- $ 123 $ → $ 1 + 2 + 3 = 6 $
- $ 1234 $ → $ 1 + 2 + 3 + 4 = 10 $
- $ 12345 $ → $ 1 + 2 + 3 + 4 + 5 = 15 $
- $ 123456 $ → $ 1 + 2 + 3 + 4 + 5 + 6 = 21 $
- $ 1234567 $ → $ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 $
- $ 12345678 $ → $ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 $

Now, sum these digit sums:

$$
1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120
$$

---

### Step 3: Compute $ 120 \mod 9 $

We now compute:

$$
120 \div 9 = 13 \text{ remainder } 3
$$

So,

$$
120 \equiv 3 \pmod{9}
$$

---

### Final Answer

$$
\boxed{3}
$$

Solution:
A number is congruent to the sum of its digits $\pmod 9$. Thus, \begin{align*}
&1+12+123+1234+12345+123456\\
&\qquad+1234567+12345678\\
&\quad\equiv 1+3+6+10+15+21+28+36\pmod 9 \\
&\quad\equiv 1+3+6+1+6+3+1+9\pmod 9 \\
&\quad\equiv 30\pmod 9 \\
&\quad\equiv \boxed{3}\pmod 9.
\end{align*}


Explanation:
Okay, so I need to find positive integers a, b, and 2009, where a < b < 2009, and they form a geometric sequence with an integer ratio. The question is asking for the value of a. Let me think about how to approach this.

First, let me recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio. Since the ratio is an integer, that means the ratio r is an integer. So, if the three terms are a, b, 2009, then:

b = a * r

and

2009 = b * r = a * r^2

So, 2009 = a * r^2. Therefore, a = 2009 / r^2. Since a must be a positive integer, r^2 must divide 2009. Also, since a < b < 2009, let's think about the constraints on r.

First, since a < b, and b = a * r, then a < a * r implies that r > 1. Similarly, since b < 2009, then a * r < 2009. But since 2009 = a * r^2, substituting that in, we have a * r < a * r^2, which simplifies to r < r^2, which is true for r > 1. So that's consistent.

Therefore, the key is that r^2 divides 2009, and r is an integer greater than 1. So, first, I need to factorize 2009 to find its divisors, then find possible values of r such that r^2 divides 2009.

Let me factorize 2009. Let me start by checking divisibility by small primes.

First, check if 2009 is even? No, it's odd. Divisible by 3? Let's check: 2 + 0 + 0 + 9 = 11, which is not divisible by 3. Next, 5? Ends with 9, so no. Next, 7? Let me do 2009 divided by 7. 7*287 is 2009? Let me check 7*280 = 1960, 7*287 = 1960 + 49 = 2009. Yes! So 2009 = 7 * 287.

Now factorize 287. Let's check divisibility by 7 again: 7*41 is 287? 7*40 = 280, so 7*41 = 287. Yes! So 2009 = 7 * 7 * 41. Therefore, prime factorization of 2009 is 7² * 41¹.

So, 2009 = 7² × 41.

Now, since r² divides 2009, then r² must be a divisor of 7² × 41. Let me think about the exponents in the prime factorization. For r² to divide 2009, the exponents of primes in r² must be less than or equal to those in 2009.

Let me denote r as a product of primes. Since 2009's prime factors are 7 and 41, then r must be of the form 7^x * 41^y, where x and y are non-negative integers. Then, r² would be 7^(2x) * 41^(2y). For this to divide 2009 = 7² * 41¹, we need:

2x ≤ 2 and 2y ≤ 1.

So, solving these inequalities:

For 2x ≤ 2: x ≤ 1.

For 2y ≤ 1: y ≤ 0.5. Since y is an integer (as r is an integer), y must be 0.

Therefore, the only possible exponents are x = 0 or 1, and y = 0. Therefore, possible values of r are:

If x=0, y=0: r = 7^0 * 41^0 = 1. But earlier we said that r must be greater than 1 because a < b. So r=1 is invalid.

If x=1, y=0: r = 7^1 * 41^0 = 7. So r=7.

Are there any other possibilities? Let me check. Since y must be 0, and x can be 0 or 1. So only r=1 and r=7. Since r=1 is invalid, only r=7 is possible?

Wait, but hold on. Let me check if there are other factors. Wait, 2009 is 7² × 41. So, the divisors of 2009 are of the form 7^a * 41^b where a=0,1,2 and b=0,1. Therefore, the square factors would be those divisors which are squares. So, for a divisor to be a square, all exponents in its prime factorization must be even. So, for 7^a * 41^b, we need a even and b even. Since a can be 0,1,2 and b can be 0,1. Therefore, possible exponents:

For a: 0 or 2 (since 0 is even, 2 is even, 1 is odd). For b: 0 (since 1 is odd). Therefore, the square divisors are 7^0 * 41^0 = 1, 7^2 * 41^0 = 49, and 7^0 * 41^0 = 1 again. Wait, so only 1 and 49?

Wait, so the square divisors of 2009 are 1 and 49? Because 41^2 is 1681, which is larger than 2009, so it can't be. Therefore, the square divisors are 1 and 49. Therefore, r² must be 1 or 49. Since r is integer greater than 1, then r² must be 49, so r=7. Therefore, only possible ratio is 7. Therefore, a = 2009 / r² = 2009 / 49. Let me compute that.

2009 divided by 49. Well, 49*40 = 1960. 2009 - 1960 = 49. Therefore, 49*41 = 2009. Therefore, a = 41. Then b = a*r = 41*7 = 287. Then check if a < b < 2009: 41 < 287 < 2009. Yes, that works. So is that the answer? 41?

Wait, but let me check if there are other possibilities. Wait, earlier I thought that the square divisors are only 1 and 49. But let me check again. Suppose someone says, maybe r is not necessarily composed of primes in 7 and 41? But since 2009 only has primes 7 and 41, then any divisor of 2009 must be of the form 7^a * 41^b. Therefore, if r² divides 2009, then r must be of the form 7^c * 41^d, where 2c ≤ 2 and 2d ≤ 1. Which gives c ≤ 1 and d ≤ 0. So, d=0, c=0 or 1. Therefore, r is 1 or 7. Therefore, only possible ratio is 7. Therefore, only possible a is 2009 / 7² = 41. Therefore, answer is 41?

But let me check again. Wait, maybe there's another way? Let me think. Suppose that the geometric sequence is a, b, 2009. So, the ratio could be negative? But the problem says positive integers. So if the ratio is negative, then b would be negative if a is positive, which is not allowed. Therefore, ratio must be positive integer. So, ratio is positive integer greater than 1. So, only possible ratio is 7. Therefore, a=41. Therefore, answer is 41?

But let me check if there are other possible ratios. Wait, suppose that the ratio is a fraction? But the problem says integer ratio. So ratio must be integer. Therefore, no. So, seems like only possible ratio is 7. Therefore, answer is 41. But let me check again.

Wait, let me check if 2009 can be expressed as a * r^2 with a and r integers, and a < b = a*r < 2009. So, if r=7, then a=41, b=287, which is less than 2009. If there is another ratio?

Wait, suppose that r is something else. For example, suppose that r is 41? Then, a = 2009 / (41)^2. Let me compute that. 41 squared is 1681. 2009 divided by 1681 is approximately 1.195... which is not integer. Therefore, not possible. Similarly, if r= something else?

Wait, but according to our previous analysis, the only square divisors are 1 and 49. Therefore, only possible r is 7. Therefore, answer is 41. But let me check if there is another way. Maybe the geometric sequence is not in the order a, b, 2009? Wait, the problem says "form a geometric sequence with an integer ratio". So, the three terms are a, b, 2009, with a < b < 2009. So, the order is important. So, the ratio is positive integer, so the sequence is increasing. Therefore, the ratio is greater than 1. So, only possible ratio is 7. Therefore, answer is 41. But let me check if there are other factors. Wait, 2009 is 7²×41. So, if we think of possible r, which is integer, then r must divide sqrt(2009). But sqrt(2009) is approx 44.8. So, possible integer ratios less than 44.8. But since r² divides 2009, then r must be such that r² divides 2009. Which as we saw, only 1 and 7. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is the answer.

But let me check again. Let me think if there is another possible ratio. For example, suppose that the ratio is  something else. Let me think. Suppose that the ratio is  something like  sqrt(7*41). But that is irrational. So, no. Or maybe if there is another way to factor?

Wait, maybe the problem is not that the ratio is integer, but that the ratio is an integer. So, the ratio is integer, but maybe the terms are not necessarily in the order a, b, 2009? Wait, no, the problem says "form a geometric sequence with an integer ratio", and the terms are a, b, 2009 with a < b < 2009. So, the order is fixed. So, the three terms are in geometric progression, so the ratio is consistent. Therefore, the ratio is r, so b = a*r, 2009 = b*r = a*r². Therefore, 2009 = a*r². Therefore, a must be 2009 divided by r squared. So, since a must be integer, r squared divides 2009. Which as we saw, only possible r is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there is another possible value. For example, if r is  something else. Let me check if 2009 can be divided by some square number other than 1 and 49. For example, 2009 divided by 49 is 41, which is prime. What about 2009 divided by 1 is 2009. So, if r squared is 1, then r=1, but that gives a=2009, but then b = a*r = 2009, which is not less than 2009. So invalid. If r squared is 49, then a=41, which is valid. What about other squares? For example, 2009 divided by 7² is 41. What about 2009 divided by something else? For example, if r squared is 7*41=287? But 287 is not a square. So, 287 is 7*41, which is not a square. Therefore, if r squared is 287, then r would be sqrt(287), which is not integer. Therefore, invalid. Similarly, if r squared is 7^2*41^2, which is way larger than 2009. Therefore, no. Therefore, only possible square divisors are 1 and 49. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there is another way. Suppose that the geometric sequence is not a, b, 2009, but maybe 2009 is the first term? But the problem says a, b, 2009 with a < b < 2009. So 2009 is the third term. Therefore, the order is fixed. Therefore, the ratio is positive integer greater than 1. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is the answer.

But let me check with actual numbers. If a=41, ratio=7, then b=41*7=287, then next term is 287*7=2009. Yes, that works. And 41 < 287 < 2009. So that's valid. Is there another possible sequence? Let me think. Suppose that there is another ratio. For example, if ratio is  something else. Let me think. Suppose that ratio is  something like  41? Then, a would be 2009 / (41)^2 = 2009 / 1681 ≈ 1.195, which is not integer. If ratio is  something else, like  7* something? Wait, but ratio must be integer. Let me think. Suppose that ratio is  something else. For example, if ratio is  1/ something? But ratio must be integer, so no. Therefore, no. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there are other factors. Wait, 2009 is 7²×41. So, if we think of possible r, which is integer, then r must divide sqrt(2009). But sqrt(2009) is irrational. Therefore, the only way for r² to divide 2009 is if r is 7 or 1. Therefore, only possible. Therefore, answer is 41. Therefore, I think that is correct.

Therefore, the answer is \boxed{41}.

**Final Answer**
\boxed{41}

To solve the problem, we are given that three positive integers $ a $, $ b $, and $ 2009 $ form a geometric sequence with an **integer ratio**, and that $ a < b < 2009 $. We are to find the value of $ a $.

---

### Step 1: Understanding the Geometric Sequence

In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio $ r $. So, if the sequence is $ a, b, 2009 $, then:

$$
b = a \cdot r \quad \text{and} \quad 2009 = b \cdot r = a \cdot r^2
$$

From this, we can express $ a $ in terms of $ r $:

$$
a = \frac{2009}{r^2}
$$

Since $ a $ must be a **positive integer**, $ r^2 $ must divide 2009.

---

### Step 2: Prime Factorization of 2009

We factorize 2009:

$$
2009 = 7^2 \times 41
$$

This tells us that the only prime factors of 2009 are 7 and 41.

---

### Step 3: Determine Possible Integer Values of $ r $

We are looking for integer values of $ r $ such that $ r^2 $ divides 2009. That is, $ r^2 \mid 2009 $.

Let’s consider the square divisors of 2009. Since the prime factorization is $ 7^2 \times 41 $, the square divisors are:

- $ 1 = 7^0 \times 41^0 $
- $ 49 = 7^2 \times 41^0 $

These are the only square divisors of 2009.

So, the possible values of $ r^2 $ are 1 and 49. Since $ r > 1 $ (because $ a < b $), we discard $ r = 1 $, and take:

$$
r^2 = 49 \Rightarrow r = 7
$$

---

### Step 4: Compute $ a $

Now that we know $ r = 7 $, we can compute:

$$
a = \frac{2009}{r^2} = \frac{2009}{49} = 41
$$

---

### Step 5: Verify the Sequence

Let’s verify that the sequence $ a = 41 $, $ b = a \cdot r = 41 \cdot 7 = 287 $, and $ 2009 $ forms a geometric sequence with integer ratio:

- $ b = 41 \cdot 7 = 287 $
- $ 2009 = 287 \cdot 7 $

So the sequence is $ 41, 287, 2009 $, with common ratio $ r = 7 $, and the condition $ a < b < 2009 $ is satisfied.

---

### Final Answer

$$
\boxed{41}
$$

[Assistant]
Explanation:
Okay, so I need to find positive integers a, b, and 2009, where a < b < 2009, and they form a geometric sequence with an integer ratio. The question is asking for the value of a. Let me think about how to approach this.

First, let me recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio. Since the ratio is an integer, that means the ratio r is an integer. So, if the three terms are a, b, 2009, then:

b = a * r

and

2009 = b * r = a * r^2

So, 2009 = a * r^2. Therefore, a = 2009 / r^2. Since a must be a positive integer, r^2 must divide 2009. Also, since a < b < 2009, let's think about the constraints on r.

First, since a < b, and b = a * r, then a < a * r implies that r > 1. Similarly, since b < 2009, then a * r < 2009. But since 2009 = a * r^2, substituting that in, we have a * r < a * r^2, which simplifies to r < r^2, which is true for r > 1. So that's consistent.

Therefore, the key is that r^2 divides 2009, and r is an integer greater than 1. So, first, I need to factorize 2009 to find its divisors, then find possible values of r such that r^2 divides 2009.

Let me factorize 2009. Let me start by checking divisibility by small primes.

First, check if 2009 is even? No, it's odd. Divisible by 3? Let's check: 2 + 0 + 0 + 9 = 11, which is not divisible by 3. Next, 5? Ends with 9, so no. Next, 7? Let me do 2009 divided by 7. 7*287 is 2009? Let me check 7*280 = 1960, 7*287 = 1960 + 49 = 2009. Yes! So 2009 = 7 * 287.

Now factorize 287. Let's check divisibility by 7 again: 7*41 is 287? 7*40 = 280, so 7*41 = 287. Yes! So 2009 = 7 * 7 * 41. Therefore, prime factorization of 2009 is 7² * 41¹.

So, 2009 = 7² × 41.

Now, since r² divides 2009, then r² must be a divisor of 7² × 41. Let me think about the exponents in the prime factorization. For r² to divide 2009, the exponents of primes in r² must be less than or equal to those in 2009.

Let me denote r as a product of primes. Since 2009's prime factors are 7 and 41, then r must be of the form 7^x * 41^y, where x and y are non-negative integers. Then, r² would be 7^(2x) * 41^(2y). For this to divide 2009 = 7² * 41¹, we need:

2x ≤ 2 and 2y ≤ 1.

So, solving these inequalities:

For 2x ≤ 2: x ≤ 1.

For 2y ≤ 1: y ≤ 0.5. Since y is an integer (as r is an integer), y must be 0.

Therefore, the only possible exponents are x = 0 or 1, and y = 0. Therefore, possible values of r are:

If x=0, y=0: r = 7^0 * 41^0 = 1. But earlier we said that r must be greater than 1 because a < b. So r=1 is invalid.

If x=1, y=0: r = 7^1 * 41^0 = 7. So r=7.

Are there any other possibilities? Let me check. Since y must be 0, and x can be 0 or 1. So only r=1 and r=7. Since r=1 is invalid, only r=7 is possible?

Wait, but hold on. Let me check if there are other factors. Wait, 2009 is 7² × 41. So, the divisors of 2009 are of the form 7^a * 41^b where a=0,1,2 and b=0,1. Therefore, the square factors would be those divisors which are squares. So, for a divisor to be a square, all exponents in its prime factorization must be even. So, for 7^a * 41^b, we need a even and b even. Since a can be 0,1,2 and b can be 0,1. Therefore, possible exponents:

For a: 0 or 2 (since 0 is even, 2 is even, 1 is odd). For b: 0 (since 1 is odd). Therefore, the square divisors are 7^0 * 41^0 = 1, 7^2 * 41^0 = 49, and 7^0 * 41^0 = 1 again. Wait, so only 1 and 49?

Wait, so the square divisors of 2009 are 1 and 49? Because 41^2 is 1681, which is larger than 2009, so it can't be. Therefore, the square divisors are 1 and 49. Therefore, r² must be 1 or 49. Since r is integer greater than 1, then r² must be 49, so r=7. Therefore, only possible ratio is 7. Therefore, a = 2009 / r² = 2009 / 49. Let me compute that.

2009 divided by 49. Well, 49*40 = 1960. 2009 - 1960 = 49. Therefore, 49*41 = 2009. Therefore, a = 41. Then b = a*r = 41*7 = 287. Then check if a < b < 2009: 41 < 287 < 2009. Yes, that works. So is that the answer? 41?

Wait, but let me check if there are other possibilities. Wait, earlier I thought that the square divisors are only 1 and 49. But let me check again. Suppose someone says, maybe r is not necessarily composed of primes in 7 and 41? But since 2009 only has primes 7 and 41, then any divisor of 2009 must be of the form 7^a * 41^b. Therefore, if r² divides 2009, then r must be of the form 7^c * 41^d, where 2c ≤ 2 and 2d ≤ 1. Which gives c ≤ 1 and d ≤ 0. So, d=0, c=0 or 1. Therefore, r is 1 or 7. Therefore, only possible ratio is 7. Therefore, only possible a is 2009 / 7² = 41. Therefore, answer is 41?

But let me check again. Wait, maybe there's another way? Let me think. Suppose that the geometric sequence is a, b, 2009. So, the ratio could be negative? But the problem says positive integers. So if the ratio is negative, then b would be negative if a is positive, which is not allowed. Therefore, ratio must be positive integer. So, ratio is positive integer greater than 1. So, only possible ratio is 7. Therefore, a=41. Therefore, answer is 41?

But let me check if there are other possible ratios. Wait, suppose that the ratio is a fraction? But the problem says integer ratio. So ratio must be integer. Therefore, no. So, seems like only possible ratio is 7. Therefore, answer is 41. But let me check again.

Wait, let me check if 2009 can be expressed as a * r^2 with a and r integers, and a < b = a*r < 2009. So, if r=7, then a=41, b=287, which is less than 2009. If there is another ratio?

Wait, suppose that r is something else. For example, suppose that r is 41? Then, a = 2009 / (41)^2. Let me compute that. 41 squared is 1681. 2009 divided by 1681 is approximately 1.195... which is not integer. Therefore, not possible. Similarly, if r= something else?

Wait, but according to our previous analysis, the only square divisors are 1 and 49. Therefore, only possible r is 7. Therefore, answer is 41. But let me check if there is another way. Maybe the geometric sequence is not in the order a, b, 2009? Wait, the problem says "form a geometric sequence with an integer ratio". So, the three terms are a, b, 2009, with a < b < 2009. So, the order is important. So, the ratio is positive integer, so the sequence is increasing. Therefore, the ratio is greater than 1. So, only possible ratio is 7. Therefore, answer is 41. But let me check if there are other factors. Wait, 2009 is 7²×41. So, if we think of possible r, which is integer, then r must divide sqrt(2009). But sqrt(2009) is approx 44.8. So, possible integer ratios less than 44.8. But since r² divides 2009, then r must be such that r² divides 2009. Which as we saw, only 1 and 7. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is the answer.

But let me check again. Let me think if there is another possible ratio. For example, suppose that the ratio is  something else. Let me think. Suppose that the ratio is  something like  sqrt(7*41). But that is irrational. So, no. Or maybe if there is another way to factor?

Wait, maybe the problem is not that the ratio is integer, but that the ratio is an integer. So, the ratio is integer, but maybe the terms are not necessarily in the order a, b, 2009? Wait, no, the problem says "form a geometric sequence with an integer ratio", and the terms are a, b, 2009 with a < b < 2009. So, the order is fixed. So, the three terms are in geometric progression, so the ratio is consistent. Therefore, the ratio is r, so b = a*r, 2009 = b*r = a*r². Therefore, 2009 = a*r². Therefore, a must be 2009 divided by r squared. So, since a must be integer, r squared divides 2009. Which as we saw, only possible r is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there is another possible value. For example, if r is  something else. Let me check if 2009 can be divided by some square number other than 1 and 49. For example, 2009 divided by 49 is 41, which is prime. What about 2009 divided by 1 is 2009. So, if r squared is 1, then r=1, but that gives a=2009, but then b = a*r = 2009, which is not less than 2009. So invalid. If r squared is 49, then a=41, which is valid. What about other squares? For example, 2009 divided by 7² is 41. What about 2009 divided by something else? For example, if r squared is 7*41=287? But 287 is not a square. So, 287 is 7*41, which is not a square. Therefore, if r squared is 287, then r would be sqrt(287), which is not integer. Therefore, invalid. Similarly, if r squared is 7^2*41^2, which is way larger than 2009. Therefore, no. Therefore, only possible square divisors are 1 and 49. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there is another way. Suppose that the geometric sequence is not a, b, 2009, but maybe 2009 is the first term? But the problem says a, b, 2009 with a < b < 2009. So 2009 is the third term. Therefore, the order is fixed. Therefore, the ratio is positive integer greater than 1. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is the answer.

But let me check with actual numbers. If a=41, ratio=7, then b=41*7=287, then next term is 287*7=2009. Yes, that works. And 41 < 287 < 2009. So that's valid. Is there another possible sequence? Let me think. Suppose that there is another ratio. For example, if ratio is  something else. Let me think. Suppose that ratio is  something like  41? Then, a would be 2009 / (41)^2 = 2009 / 1681 ≈ 1.195, which is not integer. If ratio is  something else, like  7* something? Wait, but ratio must be integer. Let me think. Suppose that ratio is  something else. For example, if ratio is  1/ something? But ratio must be integer, so no. Therefore, no. Therefore, only possible ratio is 7. Therefore, answer is 41. Therefore, I think that is correct.

But let me check if there are other factors. Wait, 2009 is 7²×41. So, if we think of possible r, which is integer, then r must divide sqrt(2009). But sqrt(2009) is irrational. Therefore, the only way for r² to divide 2009 is if r is 7 or 1. Therefore, only possible. Therefore, answer is 41. Therefore, I think that is correct.

Therefore, the answer is \boxed{41}.

**Final Answer**
\boxed{41}

To solve the problem, we are given that three positive integers $ a $, $ b $, and $ 2009 $ form a geometric sequence with an **integer ratio**, and that $ a < b < 2009 $. We are to find the value of $ a $.

---

### Step 1: Understanding the Geometric Sequence

In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio $ r $. So, if the sequence is $ a, b, 2009 $, then:

$$
b = a \cdot r \quad \text{and} \quad 2009 = b \cdot r = a \cdot r^2
$$

From this, we can express $ a $ in terms of $ r $:

$$
a = \frac{2009}{r^2}
$$

Since $ a $ must be a **positive integer**, $ r^2 $ must divide 2009.

---

### Step 2: Prime Factorization of 2009

We factorize 2009:

$$
2009 = 7^2 \times 41
$$

This tells us that the only prime factors of 2009 are 7 and 41.

---

### Step 3: Determine Possible Integer Values of $ r $

We are looking for integer values of $ r $ such that $ r^2 $ divides 2009. That is, $ r^2 \mid 2009 $.

Let’s consider the square divisors of 2009. Since the prime factorization is $ 7^2 \times 41 $, the square divisors are:

- $ 1 = 7^0 \times 41^0 $
- $ 49 = 7^2 \times 41^0 $

These are the only square divisors of 2009.

So, the possible values of $ r^2 $ are 1 and 49. Since $ r > 1 $ (because $ a < b $), we discard $ r = 1 $, and take:

$$
r^2 = 49 \Rightarrow r = 7
$$

---

### Step 4: Compute $ a $

Now that we know $ r = 7 $, we can compute:

$$
a = \frac{2009}{r^2} = \frac{2009}{49} = 41
$$

---

### Step 5: Verify the Sequence

Let’s verify that the sequence $ a = 41 $, $ b = a \cdot r = 41 \cdot 7 = 287 $, and $ 2009 $ forms a geometric sequence with integer ratio:

- $ b = 41 \cdot 7 = 287 $
- $ 2009 = 287 \cdot 7 $

So the sequence is $ 41, 287, 2009 $, with common ratio $ r = 7 $, and the condition $ a < b < 2009 $ is satisfied.

---

### Final Answer

$$
\boxed{41}
$$

Solution: The prime factorization of $2009$ is $2009 = 7\cdot 7\cdot 41$. As $a<b<2009$, the ratio must be positive and larger than $1$. Hence, there is only one possibility: the ratio must be $7$, and thus $b=7\cdot 41$ and $a=\boxed{41}$.