Explanation:
Okay, so I need to figure out the maximum number of points where at least two of the four distinct circles intersect. Hmm, let me start by recalling what I know about circles intersecting.

First, two circles can intersect in at most two points, right? If they're the same circle, they have infinitely many points, but since the problem says four distinct circles, I guess we don't have to worry about that. So each pair of circles can intersect at two points.

Now, if there are four circles, how many pairs of circles are there? Let me think. The number of ways to choose two circles out of four is given by the combination formula C(n, 2) where n is 4. So that would be 4! / (2! * (4-2)!) = (4*3)/2 = 6. So there are six pairs of circles.

If each pair can intersect at two points, then the maximum number of intersection points would be 6 pairs * 2 points per pair = 12 points. But wait, is there a possibility that some intersections overlap or something? Like, maybe if three circles all intersect at the same point, that would reduce the total number? But the question says "at least two of the circles intersect," so even if three circles intersect at the same point, that point would count as an intersection for each pair. However, the problem is asking for the maximum number of points, so we need to arrange the circles so that no two pairs share an intersection point.

So, if we can arrange the four circles such that every pair intersects at two unique points, not shared with any other pair, then the total number would be 12. But is that possible? Let me think.

Let me start with two circles. They can intersect at two points. Now, add a third circle. To maximize intersections, the third circle should intersect each of the first two circles at two distinct points. So, the third circle would add 2 intersections with the first circle and 2 with the second, so 4 new points. Similarly, adding a fourth circle, it should intersect each of the three existing circles at two distinct points. So that would be 2*3=6 new points.

So total intersections would be 2 (from first pair) + 4 (from third circle) + 6 (from fourth circle) = 12. Wait, but hold on, when you add the third circle, you have to make sure that the intersections with the first two circles don't coincide with any existing intersections. Similarly, when adding the fourth circle, its intersections with the first three circles should not overlap with any previous intersections.

Is this possible? Let me visualize. Let me think of the first two circles intersecting at two points. Now, the third circle can be arranged so that it crosses each of the first two circles at two different points. For example, if the first two circles intersect at points A and B, then the third circle can pass through two new points on the first circle, say C and D, and two new points on the second circle, E and F. Wait, but a circle can't pass through two points on the first circle and two points on the second circle without overlapping? Or maybe it can?

Wait, actually, if you have two circles, the third circle can intersect each of them at two points. So, for example, imagine the first two circles intersecting at two points. Then, the third circle can be arranged so that it intersects the first circle at two points that are not the original intersection points, and similarly for the second circle. So, as long as the third circle is not passing through any of the existing intersection points, then those intersections would be unique.

Similarly, when adding the fourth circle, it can be arranged to intersect each of the three existing circles at two new points, not overlapping with any previous intersections.

Therefore, if all these intersections are unique, then the total number of intersection points would be 2*number of pairs, which is 2*6=12.

But wait, is there a case where three circles can intersect at the same point? If that's possible, then maybe you can have fewer points? But since we want the maximum number, we need to avoid that. So, the maximum occurs when no two pairs share an intersection point.

Therefore, the answer should be 12? But I need to confirm.

Wait, let me check with smaller numbers. For example, with two circles, maximum intersections are 2. For three circles, each pair intersects at two points, so 3 pairs, 6 points. But if you arrange three circles so that each pair intersects at two unique points, then yes, 6 points. Is that possible?

Yes, for example, if you have three circles where each pair intersects at two distinct points, and no three circles intersect at the same point. So, for three circles, maximum is 6.

Similarly, for four circles, it would be 6 pairs, each contributing 2 points, so 12.

But I remember that in some cases, with circles, you can have more intersections? Wait, no. Wait, each pair can only intersect at two points. So, if you have four circles, the maximum number of intersections is 2*C(4,2)=12.

But let me check online... Wait, no, I can't check online. But I need to be sure.

Alternatively, think of it as each circle can intersect with another circle at two points. So, for four circles, each circle can intersect with three others, so 2*3=6 points per circle? Wait, but that would be counting each intersection twice.

Wait, for example, if I have circle A, it intersects with B, C, D. Each intersection is two points, so 2*3=6 points. But each intersection point is shared between two circles. So total number of points would be (number of pairs)*2. Which is again 6*2=12.

So that seems consistent.

But maybe there's a restriction? For example, can four circles all intersect each other in two points without any overlaps?

I think so. For example, if you have four circles arranged such that each pair intersects at two unique points.

But how? Let me try to imagine.

First, take two circles intersecting at two points. Then, add a third circle that intersects each of the first two circles at two new points. Then, add a fourth circle that intersects each of the first three circles at two new points.

But is there a way to arrange the fourth circle so that it doesn't pass through any existing intersection points?

Yes, if you carefully position it. For example, if the first three circles are arranged so that their intersections are all distinct, then the fourth circle can be placed such that it cuts through each of the three existing circles at two points that are not overlapping with any previous intersections.

Therefore, it's possible.

But maybe there's a geometric constraint? For example, with circles, if you have too many intersections, maybe some intersections are forced?

Wait, for example, if you have three circles, each pair intersecting at two points, then you have six points. Now, adding a fourth circle, which needs to intersect each of the three existing circles at two points. So, for each existing circle, the fourth circle can intersect it at two points.

But if the existing circles are arranged in such a way that there's space for the fourth circle to intersect each of them without overlapping previous intersections.

Alternatively, think of circles in general position. In general position, no two circles are tangent, and no three circles meet at a single point. So, in that case, each pair intersects at two distinct points, and no three circles share a common intersection point.

Therefore, with four circles in general position, the number of intersections is 2*C(4,2)=12.

Therefore, the answer should be 12.

But wait, I remember that with lines, the maximum number of intersections is C(n,2), but with circles, it's different. Wait, for lines, each pair intersects at one point, so maximum is C(n,2). For circles, each pair can intersect at two points, so maximum is 2*C(n,2).

So for four circles, 2*C(4,2)=12.

But let me check for n=3. For three circles, maximum intersections would be 6. Which is correct. For example, three circles, each pair intersecting at two points, no three circles intersecting at the same point.

Yes, that's possible. So, for example, take three circles where each pair intersects at two distinct points.

Therefore, for four circles, 12 is the maximum.

But wait, I have a doubt here. Because when you add the fourth circle, it's possible that it can't intersect all three existing circles at two points without overlapping?

Wait, for example, suppose you have three circles arranged so that they are all intersecting each other. Now, can you add a fourth circle that intersects each of them at two points?

I think so. For example, imagine three circles arranged like a Venn diagram, but with each pair intersecting at two points. Then, you can draw a fourth circle that weaves in and out of each of the three circles, intersecting each at two points.

Alternatively, think of each circle as being in a different position.

Alternatively, think of circles with different radii and positions.

But maybe there's a mathematical way to confirm.

Each circle is defined by its center and radius. If we can choose centers and radii such that every pair intersects at two distinct points, and no three circles share an intersection point.

Yes, for example, take four circles with centers at the vertices of a square, and radii large enough so that each circle intersects the others. Wait, but if they are too large, they might overlap too much.

Alternatively, take four circles with centers at different points, each pair of circles having different distances between centers, so that they intersect at two points.

But to ensure that no three circles intersect at the same point, we need to make sure that for any three circles, the two intersection points between each pair are unique.

This seems possible by choosing the circles carefully.

Therefore, I think the maximum number is indeed 12.

But wait, I remember that in some cases, with circles, you can have more intersections? Wait, no. Each pair can only intersect at two points. So, if you have four circles, the maximum number of intersections is 2 times the number of pairs.

Alternatively, maybe there's a different way to count?

Wait, let me think again.

Number of pairs: C(4,2)=6. Each pair contributes up to 2 points. So total is 12.

Yes.

But maybe there's a case where a single point is shared by more than two circles? But if that's the case, then that point would be counted multiple times, but since we are looking for the number of points where at least two circles intersect, even if three circles intersect at the same point, that's still just one point. However, since we want the maximum number of points, we need to avoid such overlaps.

Therefore, the maximum occurs when every intersection is only between two circles, and no three circles meet at a single point.

Therefore, the maximum number is 12.

But let me check with an example.

Suppose we have four circles: C1, C2, C3, C4.

C1 and C2 intersect at points A and B.

C1 and C3 intersect at points C and D.

C1 and C4 intersect at points E and F.

C2 and C3 intersect at points G and H.

C2 and C4 intersect at points I and J.

C3 and C4 intersect at points K and L.

So that's 12 distinct points.

If all these points are distinct, then yes, 12 points.

Therefore, the answer is 12.

But wait, I have a doubt because I remember that with circles, sometimes you can't have too many intersections. For example, if you have too many circles, they might have to overlap in some way.

But with four circles, I think it's possible.

Alternatively, think of each circle as a different size and position. For example, place the first circle anywhere. The second circle intersects it at two points. The third circle intersects the first at two new points and the second at two new points. The fourth circle intersects each of the first three circles at two new points.

Yes, as long as you don't have any coincidences, this should work.

Therefore, I think the answer is 12.

But wait, let me check for n=4 circles, the formula is 2*C(n,2). For n=2, 2*1=2, correct. For n=3, 2*3=6, correct. For n=4, 2*6=12.

Yes, so that seems to be the formula.

Therefore, the maximum number of points where at least two of the circles intersect is 12.

But wait, hold on. Wait, I just thought of something. If you have four circles, each pair intersecting at two points, but in reality, when you add the fourth circle, it might intersect each existing circle at two points, but maybe some of those intersections are forced to coincide?

Wait, for example, suppose you have three circles, each pair intersecting at two points. Now, when adding the fourth circle, can you always find two points on each existing circle where the fourth circle can intersect without overlapping previous intersections?

I think so. Because for each existing circle, you can choose two points that are not already intersection points of other circles.

But how do you know that such points exist?

Well, each existing circle has infinitely many points. So, if you have a circle, and you have already some points on it where it intersects with other circles, there are still infinitely many other points on the circle. So, you can choose two new points for the fourth circle to intersect with that circle.

Therefore, as long as you don't force the fourth circle to pass through any existing intersection points, you can have two new intersections with each existing circle.

Therefore, the total number of intersections would be 12.

Therefore, I think the answer is 12.

But I want to check if there's any resource or formula that confirms this. Wait, but since I can't access resources, I need to rely on my knowledge.

Alternatively, think of it as a graph theory problem. Each intersection is a point where two circles meet. But no, it's more about geometry.

Alternatively, think of the maximum number of regions created by circles, but that's a different question.

Wait, but the question is about the number of intersection points, not regions.

So, going back, I think my reasoning is correct.

Therefore, the maximum number is 12.

But wait, I just found a contradiction in my mind. Let me think again.

Suppose you have four circles. Each pair intersects in two points. So, for each pair, two points. So, 6 pairs, 12 points.

But in reality, when you have multiple circles, some constraints might make this impossible.

Wait, for example, if you have three circles, each pair intersecting at two points, then you have six points. Now, adding a fourth circle, which needs to intersect each of the three existing circles at two points.

But each existing circle already has two intersection points with the other circles. However, the fourth circle can intersect each existing circle at two different points.

But how do you ensure that the fourth circle can intersect each existing circle at two points?

For example, take the first circle, which already has two intersection points with the second circle and two with the third. But the fourth circle can intersect the first circle at two other points, not overlapping with those.

But since a circle is a closed curve, if you have two existing points on it, you can always draw another circle that intersects it at two new points.

Therefore, yes, it's possible.

Therefore, I think the answer is 12.

But I remember that in some cases, with circles, you can't have too many intersections because of the way they are arranged. For example, if you have too many circles, they might have to overlap in some way.

But with four circles, I don't think that's the case.

Alternatively, think of each circle as being in a different position. For example, place the first circle, then the second circle so that it intersects the first at two points. Then the third circle intersects the first at two different points and the second at two different points. Then the fourth circle intersects the first at two new points, the second at two new points, and the third at two new points.

Yes, this seems feasible.

Therefore, I think the answer is 12.

But wait, let me check with a different approach.

Each circle can intersect another circle at two points. So, for four circles, the maximum number of intersections is 2 * number of pairs.

Number of pairs is 6, so 12.

Yes.

Alternatively, think of it as each circle can intersect with three others, so 2 intersections per circle, so 2*4=8, but this counts each intersection twice, so divide by 2, getting 4? Wait, no. Wait, that approach is wrong.

Wait, if you think of each circle having 2 intersections with each of the other three circles, that would be 2*3=6 intersections per circle. But each intersection is shared between two circles, so total number of intersections is (4*6)/2=12.

Yes, that's the same as before.

Therefore, that gives 12.

Therefore, I think the answer is 12.

But I want to make sure that there isn't a case where some intersections are not possible.

For example, if you have four circles, can they all intersect each other in two points?

Yes. For example, take four circles with centers at the vertices of a square, and radii such that each circle intersects the other three.

But to ensure that each pair intersects at two points, the distance between centers should be less than the sum of radii and more than the difference of radii.

So, if you have four circles with centers at the square's vertices, and radii larger than half the diagonal of the square, then each circle will intersect the others.

But in this case, how many intersection points would there be?

Each pair of circles would intersect at two points. However, if the square is symmetric, maybe some intersection points coincide?

For example, if you have circles centered at the square's vertices with the same radius, then maybe some intersections coincide?

But if you adjust the radii so that each pair of circles has different distances between centers, then you can have each pair intersecting at two unique points.

Therefore, by choosing different radii and positions, you can avoid overlapping intersections.

Therefore, it's possible.

Therefore, I think the answer is 12.

But I remember that in some problems, the maximum number of intersections for n circles is 2*C(n,2). So for n=4, 12.

Yes, so I think that's the answer.

Therefore, the maximum number of points where at least two of the circles intersect is 12.

**Final Answer**
The maximum number of intersection points is \boxed{12}.

To determine the **maximum number of points where at least two of the four distinct circles intersect**, we begin by analyzing the geometric properties of circle intersections.

---

### Step 1: Understanding Circle Intersections

Two distinct circles in a plane can intersect in **at most two points**. This is a fundamental geometric fact. If the circles are tangent, they intersect at exactly one point, and if they are concentric or non-overlapping, they do not intersect at all. However, to **maximize** the number of intersection points, we must ensure that **each pair of circles intersects at two distinct points**.

---

### Step 2: Counting the Number of Pairs

We are given **four distinct circles**. The number of **unique pairs** of circles that can be formed from four circles is given by the combination formula:

$$
\binom{4}{2} = \frac{4 \times 3}{2} = 6
$$

So, there are **6 unique pairs** of circles.

---

### Step 3: Maximum Intersections per Pair

As established earlier, **each pair of circles can intersect at up to 2 points**. Therefore, the **maximum number of intersection points** is:

$$
6 \text{ pairs} \times 2 \text{ points per pair} = 12 \text{ points}
$$

---

### Step 4: Ensuring All Intersections Are Unique

To achieve this maximum, we must ensure that **no two pairs of circles share the same intersection point**. That is, **no three circles intersect at the same point**, and **no two pairs of circles intersect at the same point**. This is possible by placing the circles in **general position**—that is, ensuring that:

- No two circles are tangent or concentric.
- No three circles intersect at a single point.
- Each pair of circles intersects at two **distinct** points.

This is geometrically feasible by carefully choosing the **centers and radii** of the circles.

---

### Step 5: Final Conclusion

Thus, the **maximum number of points where at least two of the four circles intersect** is:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the maximum number of points where at least two of the four distinct circles intersect. Hmm, let me start by recalling what I know about circles intersecting.

First, two circles can intersect in at most two points, right? If they're the same circle, they have infinitely many points, but since the problem says four distinct circles, I guess we don't have to worry about that. So each pair of circles can intersect at two points.

Now, if there are four circles, how many pairs of circles are there? Let me think. The number of ways to choose two circles out of four is given by the combination formula C(n, 2) where n is 4. So that would be 4! / (2! * (4-2)!) = (4*3)/2 = 6. So there are six pairs of circles.

If each pair can intersect at two points, then the maximum number of intersection points would be 6 pairs * 2 points per pair = 12 points. But wait, is there a possibility that some intersections overlap or something? Like, maybe if three circles all intersect at the same point, that would reduce the total number? But the question says "at least two of the circles intersect," so even if three circles intersect at the same point, that point would count as an intersection for each pair. However, the problem is asking for the maximum number of points, so we need to arrange the circles so that no two pairs share an intersection point.

So, if we can arrange the four circles such that every pair intersects at two unique points, not shared with any other pair, then the total number would be 12. But is that possible? Let me think.

Let me start with two circles. They can intersect at two points. Now, add a third circle. To maximize intersections, the third circle should intersect each of the first two circles at two distinct points. So, the third circle would add 2 intersections with the first circle and 2 with the second, so 4 new points. Similarly, adding a fourth circle, it should intersect each of the three existing circles at two distinct points. So that would be 2*3=6 new points.

So total intersections would be 2 (from first pair) + 4 (from third circle) + 6 (from fourth circle) = 12. Wait, but hold on, when you add the third circle, you have to make sure that the intersections with the first two circles don't coincide with any existing intersections. Similarly, when adding the fourth circle, its intersections with the first three circles should not overlap with any previous intersections.

Is this possible? Let me visualize. Let me think of the first two circles intersecting at two points. Now, the third circle can be arranged so that it crosses each of the first two circles at two different points. For example, if the first two circles intersect at points A and B, then the third circle can pass through two new points on the first circle, say C and D, and two new points on the second circle, E and F. Wait, but a circle can't pass through two points on the first circle and two points on the second circle without overlapping? Or maybe it can?

Wait, actually, if you have two circles, the third circle can intersect each of them at two points. So, for example, imagine the first two circles intersecting at two points. Then, the third circle can be arranged so that it intersects the first circle at two points that are not the original intersection points, and similarly for the second circle. So, as long as the third circle is not passing through any of the existing intersection points, then those intersections would be unique.

Similarly, when adding the fourth circle, it can be arranged to intersect each of the three existing circles at two new points, not overlapping with any previous intersections.

Therefore, if all these intersections are unique, then the total number of intersection points would be 2*number of pairs, which is 2*6=12.

But wait, is there a case where three circles can intersect at the same point? If that's possible, then maybe you can have fewer points? But since we want the maximum number, we need to avoid that. So, the maximum occurs when no two pairs share an intersection point.

Therefore, the answer should be 12? But I need to confirm.

Wait, let me check with smaller numbers. For example, with two circles, maximum intersections are 2. For three circles, each pair intersects at two points, so 3 pairs, 6 points. But if you arrange three circles so that each pair intersects at two unique points, then yes, 6 points. Is that possible?

Yes, for example, if you have three circles where each pair intersects at two distinct points, and no three circles intersect at the same point. So, for three circles, maximum is 6.

Similarly, for four circles, it would be 6 pairs, each contributing 2 points, so 12.

But I remember that in some cases, with circles, you can have more intersections? Wait, no. Wait, each pair can only intersect at two points. So, if you have four circles, the maximum number of intersections is 2*C(4,2)=12.

But let me check online... Wait, no, I can't check online. But I need to be sure.

Alternatively, think of it as each circle can intersect with another circle at two points. So, for four circles, each circle can intersect with three others, so 2*3=6 points per circle? Wait, but that would be counting each intersection twice.

Wait, for example, if I have circle A, it intersects with B, C, D. Each intersection is two points, so 2*3=6 points. But each intersection point is shared between two circles. So total number of points would be (number of pairs)*2. Which is again 6*2=12.

So that seems consistent.

But maybe there's a restriction? For example, can four circles all intersect each other in two points without any overlaps?

I think so. For example, if you have four circles arranged such that each pair intersects at two unique points.

But how? Let me try to imagine.

First, take two circles intersecting at two points. Then, add a third circle that intersects each of the first two circles at two new points. Then, add a fourth circle that intersects each of the first three circles at two new points.

But is there a way to arrange the fourth circle so that it doesn't pass through any existing intersection points?

Yes, if you carefully position it. For example, if the first three circles are arranged so that their intersections are all distinct, then the fourth circle can be placed such that it cuts through each of the three existing circles at two points that are not overlapping with any previous intersections.

Therefore, it's possible.

But maybe there's a geometric constraint? For example, with circles, if you have too many intersections, maybe some intersections are forced?

Wait, for example, if you have three circles, each pair intersecting at two points, then you have six points. Now, adding a fourth circle, which needs to intersect each of the three existing circles at two points. So, for each existing circle, the fourth circle can intersect it at two points.

But if the existing circles are arranged in such a way that there's space for the fourth circle to intersect each of them without overlapping previous intersections.

Alternatively, think of circles in general position. In general position, no two circles are tangent, and no three circles meet at a single point. So, in that case, each pair intersects at two distinct points, and no three circles share a common intersection point.

Therefore, with four circles in general position, the number of intersections is 2*C(4,2)=12.

Therefore, the answer should be 12.

But wait, I remember that with lines, the maximum number of intersections is C(n,2), but with circles, it's different. Wait, for lines, each pair intersects at one point, so maximum is C(n,2). For circles, each pair can intersect at two points, so maximum is 2*C(n,2).

So for four circles, 2*C(4,2)=12.

But let me check for n=3. For three circles, maximum intersections would be 6. Which is correct. For example, three circles, each pair intersecting at two points, no three circles intersecting at the same point.

Yes, that's possible. So, for example, take three circles where each pair intersects at two distinct points.

Therefore, for four circles, 12 is the maximum.

But wait, I have a doubt here. Because when you add the fourth circle, it's possible that it can't intersect all three existing circles at two points without overlapping?

Wait, for example, suppose you have three circles arranged so that they are all intersecting each other. Now, can you add a fourth circle that intersects each of them at two points?

I think so. For example, imagine three circles arranged like a Venn diagram, but with each pair intersecting at two points. Then, you can draw a fourth circle that weaves in and out of each of the three circles, intersecting each at two points.

Alternatively, think of each circle as being in a different position.

Alternatively, think of circles with different radii and positions.

But maybe there's a mathematical way to confirm.

Each circle is defined by its center and radius. If we can choose centers and radii such that every pair intersects at two distinct points, and no three circles share an intersection point.

Yes, for example, take four circles with centers at the vertices of a square, and radii large enough so that each circle intersects the others. Wait, but if they are too large, they might overlap too much.

Alternatively, take four circles with centers at different points, each pair of circles having different distances between centers, so that they intersect at two points.

But to ensure that no three circles intersect at the same point, we need to make sure that for any three circles, the two intersection points between each pair are unique.

This seems possible by choosing the circles carefully.

Therefore, I think the maximum number is indeed 12.

But wait, I remember that in some cases, with circles, you can have more intersections? Wait, no. Each pair can only intersect at two points. So, if you have four circles, the maximum number of intersections is 2 times the number of pairs.

Alternatively, maybe there's a different way to count?

Wait, let me think again.

Number of pairs: C(4,2)=6. Each pair contributes up to 2 points. So total is 12.

Yes.

But maybe there's a case where a single point is shared by more than two circles? But if that's the case, then that point would be counted multiple times, but since we are looking for the number of points where at least two circles intersect, even if three circles intersect at the same point, that's still just one point. However, since we want the maximum number of points, we need to avoid such overlaps.

Therefore, the maximum occurs when every intersection is only between two circles, and no three circles meet at a single point.

Therefore, the maximum number is 12.

But let me check with an example.

Suppose we have four circles: C1, C2, C3, C4.

C1 and C2 intersect at points A and B.

C1 and C3 intersect at points C and D.

C1 and C4 intersect at points E and F.

C2 and C3 intersect at points G and H.

C2 and C4 intersect at points I and J.

C3 and C4 intersect at points K and L.

So that's 12 distinct points.

If all these points are distinct, then yes, 12 points.

Therefore, the answer is 12.

But wait, I have a doubt because I remember that with circles, sometimes you can't have too many intersections. For example, if you have too many circles, they might have to overlap in some way.

But with four circles, I think it's possible.

Alternatively, think of each circle as a different size and position. For example, place the first circle anywhere. The second circle intersects it at two points. The third circle intersects the first at two new points and the second at two new points. The fourth circle intersects each of the first three circles at two new points.

Yes, as long as you don't have any coincidences, this should work.

Therefore, I think the answer is 12.

But wait, let me check for n=4 circles, the formula is 2*C(n,2). For n=2, 2*1=2, correct. For n=3, 2*3=6, correct. For n=4, 2*6=12.

Yes, so that seems to be the formula.

Therefore, the maximum number of points where at least two of the circles intersect is 12.

But wait, hold on. Wait, I just thought of something. If you have four circles, each pair intersecting at two points, but in reality, when you add the fourth circle, it might intersect each existing circle at two points, but maybe some of those intersections are forced to coincide?

Wait, for example, suppose you have three circles, each pair intersecting at two points. Now, when adding the fourth circle, can you always find two points on each existing circle where the fourth circle can intersect without overlapping previous intersections?

I think so. Because for each existing circle, you can choose two points that are not already intersection points of other circles.

But how do you know that such points exist?

Well, each existing circle has infinitely many points. So, if you have a circle, and you have already some points on it where it intersects with other circles, there are still infinitely many other points on the circle. So, you can choose two new points for the fourth circle to intersect with that circle.

Therefore, as long as you don't force the fourth circle to pass through any existing intersection points, you can have two new intersections with each existing circle.

Therefore, the total number of intersections would be 12.

Therefore, I think the answer is 12.

But I want to check if there's any resource or formula that confirms this. Wait, but since I can't access resources, I need to rely on my knowledge.

Alternatively, think of it as a graph theory problem. Each intersection is a point where two circles meet. But no, it's more about geometry.

Alternatively, think of the maximum number of regions created by circles, but that's a different question.

Wait, but the question is about the number of intersection points, not regions.

So, going back, I think my reasoning is correct.

Therefore, the maximum number is 12.

But wait, I just found a contradiction in my mind. Let me think again.

Suppose you have four circles. Each pair intersects in two points. So, for each pair, two points. So, 6 pairs, 12 points.

But in reality, when you have multiple circles, some constraints might make this impossible.

Wait, for example, if you have three circles, each pair intersecting at two points, then you have six points. Now, adding a fourth circle, which needs to intersect each of the three existing circles at two points.

But each existing circle already has two intersection points with the other circles. However, the fourth circle can intersect each existing circle at two different points.

But how do you ensure that the fourth circle can intersect each existing circle at two points?

For example, take the first circle, which already has two intersection points with the second circle and two with the third. But the fourth circle can intersect the first circle at two other points, not overlapping with those.

But since a circle is a closed curve, if you have two existing points on it, you can always draw another circle that intersects it at two new points.

Therefore, yes, it's possible.

Therefore, I think the answer is 12.

But I remember that in some cases, with circles, you can't have too many intersections because of the way they are arranged. For example, if you have too many circles, they might have to overlap in some way.

But with four circles, I don't think that's the case.

Alternatively, think of each circle as being in a different position. For example, place the first circle, then the second circle so that it intersects the first at two points. Then the third circle intersects the first at two different points and the second at two different points. Then the fourth circle intersects the first at two new points, the second at two new points, and the third at two new points.

Yes, this seems feasible.

Therefore, I think the answer is 12.

But wait, let me check with a different approach.

Each circle can intersect another circle at two points. So, for four circles, the maximum number of intersections is 2 * number of pairs.

Number of pairs is 6, so 12.

Yes.

Alternatively, think of it as each circle can intersect with three others, so 2 intersections per circle, so 2*4=8, but this counts each intersection twice, so divide by 2, getting 4? Wait, no. Wait, that approach is wrong.

Wait, if you think of each circle having 2 intersections with each of the other three circles, that would be 2*3=6 intersections per circle. But each intersection is shared between two circles, so total number of intersections is (4*6)/2=12.

Yes, that's the same as before.

Therefore, that gives 12.

Therefore, I think the answer is 12.

But I want to make sure that there isn't a case where some intersections are not possible.

For example, if you have four circles, can they all intersect each other in two points?

Yes. For example, take four circles with centers at the vertices of a square, and radii such that each circle intersects the other three.

But to ensure that each pair intersects at two points, the distance between centers should be less than the sum of radii and more than the difference of radii.

So, if you have four circles with centers at the square's vertices, and radii larger than half the diagonal of the square, then each circle will intersect the others.

But in this case, how many intersection points would there be?

Each pair of circles would intersect at two points. However, if the square is symmetric, maybe some intersection points coincide?

For example, if you have circles centered at the square's vertices with the same radius, then maybe some intersections coincide?

But if you adjust the radii so that each pair of circles has different distances between centers, then you can have each pair intersecting at two unique points.

Therefore, by choosing different radii and positions, you can avoid overlapping intersections.

Therefore, it's possible.

Therefore, I think the answer is 12.

But I remember that in some problems, the maximum number of intersections for n circles is 2*C(n,2). So for n=4, 12.

Yes, so I think that's the answer.

Therefore, the maximum number of points where at least two of the circles intersect is 12.

**Final Answer**
The maximum number of intersection points is \boxed{12}.

To determine the **maximum number of points where at least two of the four distinct circles intersect**, we begin by analyzing the geometric properties of circle intersections.

---

### Step 1: Understanding Circle Intersections

Two distinct circles in a plane can intersect in **at most two points**. This is a fundamental geometric fact. If the circles are tangent, they intersect at exactly one point, and if they are concentric or non-overlapping, they do not intersect at all. However, to **maximize** the number of intersection points, we must ensure that **each pair of circles intersects at two distinct points**.

---

### Step 2: Counting the Number of Pairs

We are given **four distinct circles**. The number of **unique pairs** of circles that can be formed from four circles is given by the combination formula:

$$
\binom{4}{2} = \frac{4 \times 3}{2} = 6
$$

So, there are **6 unique pairs** of circles.

---

### Step 3: Maximum Intersections per Pair

As established earlier, **each pair of circles can intersect at up to 2 points**. Therefore, the **maximum number of intersection points** is:

$$
6 \text{ pairs} \times 2 \text{ points per pair} = 12 \text{ points}
$$

---

### Step 4: Ensuring All Intersections Are Unique

To achieve this maximum, we must ensure that **no two pairs of circles share the same intersection point**. That is, **no three circles intersect at the same point**, and **no two pairs of circles intersect at the same point**. This is possible by placing the circles in **general position**—that is, ensuring that:

- No two circles are tangent or concentric.
- No three circles intersect at a single point.
- Each pair of circles intersects at two **distinct** points.

This is geometrically feasible by carefully choosing the **centers and radii** of the circles.

---

### Step 5: Final Conclusion

Thus, the **maximum number of points where at least two of the four circles intersect** is:

$$
\boxed{12}
$$

Solution:
Each pair of circles has at most two intersection points. There are $\binom{4}{2} = 6$ pairs of circles, so there are at most $6\times 2 = 12$ points of intersection. The following configuration shows that $\boxed{12}$ points of intersection are indeed possible:

[asy]
draw(Circle((0,0),2));
draw(Circle((1,1.3),2));
draw(Circle((-1,1.3),2));
draw(Circle((0,0.7),2));
[/asy]


Explanation:
Okay, so I need to figure out how many digits repeat in the decimal expansion of 31 divided by 11111. Let me start by recalling that when you divide a number by another, if the denominator has prime factors other than 2 and 5, the decimal will repeat. Since 11111 is the denominator here, I should check its prime factors to understand the repeating pattern.

First, let me try dividing 31 by 11111 directly. But maybe there's a smarter way than long division. Let me think. The denominator is 11111. Hmm, 11111... that number looks familiar. Wait, 11111 is a number with five 1s. Maybe it's related to 10^5 - 1 over 9? Let me check:

10^5 is 100000. So 10^5 - 1 is 99999. Then 99999 divided by 9 is 11111. Yes! So 11111 = (10^5 - 1)/9. Therefore, 1/11111 is equal to 9/(10^5 - 1). So 1/11111 = 9/99999. Therefore, 31/11111 would be 31*9 / 99999. Let me compute that numerator: 31*9 is 279. So 279/99999.

Now, 279 divided by 99999. Since 99999 is 10^5 - 1, which is 99999, so the decimal expansion of 279/99999 would be a repeating decimal with a period equal to the length of the repeating cycle of 1/99999. Wait, but 1/99999 is 0.000010000100001..., repeating every 5 digits? Wait, no. Wait, 1/99999 is 0.\overline{00001}, right? Because 1/99999 = 0.000010000100001..., so the repeating part is 00001, which is 5 digits. But then multiplying by 279 would give a decimal that's 279 times that. So maybe the decimal expansion of 279/99999 is 0.002790027900279..., repeating every 5 digits? Wait, but 279 is 279, so maybe the decimal is 0.002790027900279...? Let me check.

Alternatively, maybe I can think of 279/99999 as 0.002790027900279..., so the repeating part is "00279"? Wait, but that's 5 digits. Wait, but 279 is three digits. Let me think again. If I have 1/99999 = 0.0000100001..., so multiplying by 279 would give 0.00001 * 279 = 0.00279. Then, since the decimal repeats every 5 digits, the decimal expansion of 279/99999 would be 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the repeating decimal has a cycle of 5 digits. Therefore, the answer would be 5?

But wait, let me verify this with another approach. Maybe I should perform the division 31 divided by 11111 manually to check.

So, 31 divided by 11111. Since 11111 is larger than 31, the decimal will start with 0. Let me write it as 31.000000...

So, 11111 goes into 310000 how many times? Let me compute 11111 * 2 = 22222. 11111 * 3 = 33333. But 33333 is larger than 310000? Wait, no. Wait, 11111 * 2 = 22222. So 22222 is less than 31000. Wait, but 31000 is the first few digits. Wait, maybe I need to adjust.

Wait, actually, when doing decimal division, we need to consider how many decimal places. Let me try again.

Dividing 31 by 11111:

First, 11111 goes into 31 zero times. So we write 0., and then consider 310 divided by 11111, still zero. Then 3100 divided by 11111, still zero. Then 31000 divided by 11111. Let me compute how many times 11111 fits into 31000.

11111 * 2 = 22222. 11111 * 3 = 33333. 33333 is larger than 31000, so it's 2 times. So 2 * 11111 = 22222. Subtract that from 31000: 31000 - 22222 = 8778. Bring down a zero: 87780.

Now, how many times does 11111 go into 87780? Let me compute 11111 * 7 = 77777. 11111 * 8 = 88888. 88888 is larger than 87780, so it's 7 times. 7 * 11111 = 77777. Subtract from 87780: 87780 - 77777 = 10003. Bring down a zero: 100030.

Now, how many times does 11111 go into 100030? Let me compute 11111 * 9 = 99999. That's very close. 99999 is less than 100030. So 9 times. Subtract 99999 from 100030: 100030 - 99999 = 31. Bring down a zero: 310.

Wait, now we have 310 again, which is where we started after the decimal point. So the decimal repeats from here. Let me write down the decimal digits we have so far:

First, after the decimal, we had 0. Then:

First division step: 2 (from 31000 / 11111 = 2), then 7 (from 87780 / 11111 = 7), then 9 (from 100030 / 11111 = 9), then we had a remainder of 31, which is the same as the original numerator. So the decimal expansion is 0.00279... and then it repeats. Wait, let me check the digits:

Wait, let me track the decimal places step by step.

Starting with 31.000000...

First, 11111 into 31.0 is 0. So first decimal digit: 0.

Then 11111 into 310 is 0. Second decimal digit: 0.

Then 11111 into 3100 is 0. Third decimal digit: 0.

Then 11111 into 31000 is 2 (as 2*11111=22222). So fourth decimal digit: 2.

Subtract 22222 from 31000 gives 8778. Bring down a zero: 87780.

Now, 11111 into 87780 is 7 (7*11111=77777). Fifth decimal digit: 7.

Subtract 77777 from 87780 gives 10003. Bring down a zero: 100030.

11111 into 100030 is 9 (9*11111=99999). Sixth decimal digit: 9.

Subtract 99999 from 100030 gives 31. Bring down a zero: 310.

Now, we have 310, which is the same as after the decimal point initially. So the decimal repeats from here. Therefore, the decimal expansion is 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the number of repeating digits is 5.

Wait, but let me check again. The decimal expansion is 0.002790027900279..., so the repeating cycle is "00279", which is five digits. So the answer is 5?

But let me check with another approach. Since 11111 is (10^5 - 1)/9, so 1/11111 = 9/99999. Therefore, 31/11111 = 31*9 / 99999 = 279 / 99999. So 279 divided by 99999. Since 99999 is 10^5 - 1, the decimal expansion of 279/99999 is 0.\overline{00279}, which is a repeating cycle of 5 digits. Therefore, the answer is 5.

But let me check with another method. Let me think about the decimal expansion of fractions with denominator 99999. Since 99999 is 10^5 - 1, the decimal expansion of 1/99999 is 0.000010000100001..., repeating every 5 digits. Therefore, multiplying by 279 would give 0.002790027900279..., so the repeating part is 00279, which is 5 digits. Therefore, the answer is 5.

But let me check if there is a shorter period. For example, sometimes the period can be shorter than the denominator's length. For example, 1/3 = 0.333..., period 1, even though denominator is 3. But in this case, since 11111 is a factor of 99999, which is 10^5 - 1. Let me check if 11111 divides 10^k - 1 for some smaller k. If so, then the period would be smaller.

So, let me check if 11111 divides 10^k - 1 for some k < 5. Let me check k=1: 10^1 -1 = 9, which is not divisible by 11111. k=2: 99, nope. k=3: 999, nope. k=4: 9999. Let me divide 9999 by 11111. Well, 9999 is less than 11111, so no. Therefore, the minimal k is 5. Therefore, the period is 5. Therefore, the decimal expansion of 1/11111 has period 5, so multiplying by 31 would not change the period. Therefore, the decimal expansion of 31/11111 also has period 5. Therefore, the answer is 5.

But let me check with another example. Let me take 1/11111. If I compute 1 divided by 11111, what is the decimal expansion? Let me try to compute it.

1 divided by 11111. Let me do the division:

1.000000...

11111 goes into 100000 (after adding decimal and zeros) how many times? Let me compute 11111 * 9 = 99999. So 9 times. So 9*11111=99999. Subtract from 100000: 1. Bring down a zero: 10. Then 11111 goes into 10 zero times. Bring down another zero: 100. Still zero. Bring down another zero: 1000. Still zero. Bring down another zero: 10000. Still zero. Bring down another zero: 100000. Now we are back to where we started. So the decimal expansion is 0.000090000900009..., so repeating every 5 digits: "00009". Therefore, the period is 5. Therefore, multiplying by 31 would give 0.0027900279..., which is the same as before. Therefore, the repeating cycle is 5 digits. Therefore, the answer is 5.

But let me check if there's a possibility that the decimal repeats with a shorter period. For example, maybe 31/11111 has a shorter period? How can I check that?

The period of a fraction a/b is equal to the multiplicative order of 10 modulo b', where b' is the denominator after removing factors of 2 and 5. In this case, the denominator is 11111. Let me check if 11111 has factors other than 1. Let me factorize 11111.

Let me try dividing 11111 by small primes. Let me check divisibility by 3: 1+1+1+1+1=5, which is not divisible by 3. Divisible by 7? Let me check 11111 divided by 7. 7*1587 = 11109, remainder 2. Not divisible by 7. Divided by 11? Let me use the divisibility rule for 11: difference between sum of digits in odd positions and even positions. For 11111: positions are 1 (position 1), 1 (position 2), 1 (position 3), 1 (position 4), 1 (position 5). Sum of odd positions: 1 + 1 + 1 = 3. Sum of even positions: 1 + 1 = 2. Difference is 3 - 2 = 1, not divisible by 11. Divided by 13? Let me try 13*854 = 11102, remainder 9. Not divisible. Divided by 17? Let me try 17*653 = 11101, remainder 10. Not divisible. Divided by 19? 19*584 = 11096, remainder 15. Not divisible. Divided by 23? 23*483 = 11109, remainder 2. Not divisible. Divided by 29? Let me try 29*383 = 11107, remainder 4. Not divisible. Divided by 31? 31*358 = 111, 31*358 = 31*(300 + 58) = 9300 + 1798 = 11098. Remainder 13. Not divisible. Divided by 37? Let me try 37*300 = 11100. 11111 - 11100 = 11. Not divisible. Divided by 41? 41*271 = 11111? Let me check 41*270 = 11070, plus 41 is 11111. Yes! Wait, 41*271 = 11111? Let me compute 41*271:

First, 40*271 = 10840, plus 1*271 = 271. Total is 10840 + 271 = 11111. Yes! So 11111 factors into 41 and 271. Let me check if 271 is prime. Let me check divisibility of 271. Divided by 2? No. Divided by 3? 2+7+1=10, not divisible by 3. Divided by 5? No. Divided by 7? 7*38=266, remainder 5. Divided by 11? 11*24=264, remainder 7. Divided by 13? 13*20=260, remainder 11. Divided by 17? 17*15=255, remainder 16. Divided by 19? 19*14=266, remainder 5. Divided by 23? 23*11=253, remainder 18. So 271 is prime. Therefore, 11111 factors into 41*271. Both primes.

Therefore, the denominator is 41*271. Now, to find the period of 31/11111, we need to find the multiplicative order of 10 modulo 11111, but since 11111 is co-prime with 10 (since it's 41*271, neither of which are 2 or 5). Wait, but the multiplicative order of 10 modulo 11111 is the least common multiple of the multiplicative orders modulo 41 and 271. Because 41 and 271 are coprime.

So, first, find the multiplicative order of 10 modulo 41. Then find the multiplicative order of 10 modulo 271. Then take the LCM of those two numbers.

Let me start with modulo 41.

What is the multiplicative order of 10 modulo 41?

We need to find the smallest positive integer k such that 10^k ≡ 1 mod 41.

First, φ(41) = 40, so the order divides 40. The divisors of 40 are 1, 2, 4, 5, 8, 10, 20, 40.

Check 10^1 mod 41 = 10 ≠ 1.

10^2 = 100 mod 41. 41*2=82, 100-82=18. 10^2 ≡ 18 ≠ 1.

10^4 = (10^2)^2 ≡ 18^2 = 324 mod 41. 41*7=287, 324 - 287 = 37. 10^4 ≡ 37 ≠ 1.

10^5 = 10^4 * 10 ≡ 37*10 = 370 mod 41. 41*9=369, so 370 - 369 = 1. So 10^5 ≡ 1 mod 41. Therefore, the multiplicative order of 10 modulo 41 is 5.

Now, check modulo 271.

Find the multiplicative order of 10 modulo 271.

First, φ(271) = 270, since 271 is prime. So the order divides 270. The divisors of 270 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270.

We need to find the smallest k such that 10^k ≡ 1 mod 271.

Let me check the divisors in order.

First, check k=1: 10^1=10≠1.

k=2: 10^2=100≠1.

k=3: 10^3=1000 mod 271. Let me compute 271*3=813. 1000 - 813 = 187. 187≠1.

k=5: Let me check k=5. Maybe compute 10^5 mod 271. Alternatively, check if 10^k ≡1 for smaller divisors.

Alternatively, since 270 factors into 2*3^3*5. So possible orders are divisors of 270.

Let me check k=9: 10^9 mod 271. Maybe compute step by step.

Alternatively, use exponentiation by squaring.

First, compute 10^1 mod 271 = 10

10^2 = 100

10^4 = (10^2)^2 = 100^2 = 10000 mod 271.

Compute 271*37 = 271*(30 + 7) = 8130 + 1897 = 9, 8130 + 1897 = 10027. Wait, 271*37: Let me compute 271*30 = 8130, 271*7=1897, so total 8130 + 1897 = 10027. Then 10000 - 10027 = -27 mod 271. So 10^4 ≡ -27 mod 271.

Then 10^5 = 10^4 * 10 ≡ (-27)*10 = -270 ≡ 1 mod 271? Wait, -270 mod 271 is 1, because 271 - 270 = 1. So 10^5 ≡ 1 mod 271? Wait, that would mean the order is 5? But 5 divides 270? Yes, 270 divided by 5 is 54. But let me check again.

Wait, 10^5 mod 271: Let me verify my calculation.

Earlier, I said 10^4 ≡ -27 mod 271. Then 10^5 = 10^4 * 10 ≡ (-27)*10 = -270. Then -270 + 271 = 1. Therefore, 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is 5? But 5 divides φ(271)=270? Yes, since 270 = 5*54. Therefore, the order is 5? Wait, but that seems surprising. Let me check with another exponent.

Wait, if 10^5 ≡ 1 mod 271, then 10^5 - 1 is divisible by 271. Let me check 10^5 - 1 = 100000 - 1 = 99999. Is 99999 divisible by 271? Let me divide 99999 by 271.

Compute 271*300 = 81300. 99999 - 81300 = 18699.

271*60 = 16260. 18699 - 16260 = 2439.

271*9 = 2439. So total is 300 + 60 + 9 = 369. Therefore, 271*369 = 99999. Therefore, yes, 10^5 - 1 = 99999 = 271*369. Therefore, 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is indeed 5.

Therefore, the multiplicative order of 10 modulo 41 is 5, and modulo 271 is also 5. Therefore, the multiplicative order modulo 11111 is LCM(5,5)=5. Therefore, the period of the decimal expansion of 31/11111 is 5. Therefore, the number of repeating digits is 5.

Therefore, the answer is 5. But let me check once more with the decimal expansion I did earlier. When I did the long division, I saw that after 0.00279, the remainder came back to 31, which was the original numerator, so the cycle repeats every 5 digits. Therefore, the repeating part is 5 digits. Therefore, the answer is 5.

But just to be thorough, let me check if there's a shorter cycle. Suppose the decimal repeats every 1, 2, 3, or 4 digits. Let me check.

Suppose the decimal expansion is 0.0027900279..., so the repeating part is "00279". Let me check if "00279" can be broken into a shorter cycle. For example, if it was "00279" repeating, but maybe there's a shorter cycle. However, since the multiplicative order is 5, which is prime, the period cannot be shorter. Because if the order is 5, then the period is exactly 5. Therefore, there is no shorter period. Therefore, the answer is 5.

Therefore, I can be confident that the number of repeating digits is 5.

**Final Answer**
The number of repeating digits is \boxed{5}.

To determine how many digits repeat in the decimal expansion of the fraction $\frac{31}{11111}$, we begin by analyzing the structure of the denominator.

---

### Step 1: Understanding the Denominator

The denominator is $11111$. This number is not a power of 10, and it is not divisible by 2 or 5, so the decimal expansion of $\frac{31}{11111}$ will be a **repeating decimal**.

We can express $11111$ in terms of powers of 10:

$$
11111 = \frac{10^5 - 1}{9}
$$

This is because:

$$
\frac{10^5 - 1}{9} = \frac{99999}{9} = 11111
$$

So, we can rewrite the original fraction as:

$$
\frac{31}{11111} = \frac{31 \cdot 9}{99999} = \frac{279}{99999}
$$

---

### Step 2: Decimal Expansion of $\frac{279}{99999}$

The number $99999 = 10^5 - 1$, and the decimal expansion of any fraction of the form $\frac{a}{10^n - 1}$ has a **repeating cycle of length $n$**. This is because the denominator is one less than a power of 10, and the decimal repeats every $n$ digits.

So, the decimal expansion of $\frac{279}{99999}$ will repeat every **5 digits**.

To confirm, we can perform the long division of $279 \div 99999$, or observe that:

$$
\frac{279}{99999} = 0.\overline{00279}
$$

This means the repeating part is **"00279"**, which is **5 digits long**.

---

### Step 3: Confirming the Period Using Multiplicative Order

To further verify, we can use the concept of **multiplicative order**. The period of the decimal expansion of a fraction $\frac{a}{b}$ (where $b$ is co-prime to 10) is equal to the **multiplicative order of 10 modulo $b$**.

We factor $11111 = 41 \times 271$, both of which are prime numbers. The multiplicative order of 10 modulo 41 and 271 is 5 in both cases. Therefore, the least common multiple of these orders is:

$$
\text{lcm}(5, 5) = 5
$$

This confirms that the decimal expansion of $\frac{31}{11111}$ has a **repeating cycle of 5 digits**.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many digits repeat in the decimal expansion of 31 divided by 11111. Let me start by recalling that when you divide a number by another, if the denominator has prime factors other than 2 and 5, the decimal will repeat. Since 11111 is the denominator here, I should check its prime factors to understand the repeating pattern.

First, let me try dividing 31 by 11111 directly. But maybe there's a smarter way than long division. Let me think. The denominator is 11111. Hmm, 11111... that number looks familiar. Wait, 11111 is a number with five 1s. Maybe it's related to 10^5 - 1 over 9? Let me check:

10^5 is 100000. So 10^5 - 1 is 99999. Then 99999 divided by 9 is 11111. Yes! So 11111 = (10^5 - 1)/9. Therefore, 1/11111 is equal to 9/(10^5 - 1). So 1/11111 = 9/99999. Therefore, 31/11111 would be 31*9 / 99999. Let me compute that numerator: 31*9 is 279. So 279/99999.

Now, 279 divided by 99999. Since 99999 is 10^5 - 1, which is 99999, so the decimal expansion of 279/99999 would be a repeating decimal with a period equal to the length of the repeating cycle of 1/99999. Wait, but 1/99999 is 0.000010000100001..., repeating every 5 digits? Wait, no. Wait, 1/99999 is 0.\overline{00001}, right? Because 1/99999 = 0.000010000100001..., so the repeating part is 00001, which is 5 digits. But then multiplying by 279 would give a decimal that's 279 times that. So maybe the decimal expansion of 279/99999 is 0.002790027900279..., repeating every 5 digits? Wait, but 279 is 279, so maybe the decimal is 0.002790027900279...? Let me check.

Alternatively, maybe I can think of 279/99999 as 0.002790027900279..., so the repeating part is "00279"? Wait, but that's 5 digits. Wait, but 279 is three digits. Let me think again. If I have 1/99999 = 0.0000100001..., so multiplying by 279 would give 0.00001 * 279 = 0.00279. Then, since the decimal repeats every 5 digits, the decimal expansion of 279/99999 would be 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the repeating decimal has a cycle of 5 digits. Therefore, the answer would be 5?

But wait, let me verify this with another approach. Maybe I should perform the division 31 divided by 11111 manually to check.

So, 31 divided by 11111. Since 11111 is larger than 31, the decimal will start with 0. Let me write it as 31.000000...

So, 11111 goes into 310000 how many times? Let me compute 11111 * 2 = 22222. 11111 * 3 = 33333. But 33333 is larger than 310000? Wait, no. Wait, 11111 * 2 = 22222. So 22222 is less than 31000. Wait, but 31000 is the first few digits. Wait, maybe I need to adjust.

Wait, actually, when doing decimal division, we need to consider how many decimal places. Let me try again.

Dividing 31 by 11111:

First, 11111 goes into 31 zero times. So we write 0., and then consider 310 divided by 11111, still zero. Then 3100 divided by 11111, still zero. Then 31000 divided by 11111. Let me compute how many times 11111 fits into 31000.

11111 * 2 = 22222. 11111 * 3 = 33333. 33333 is larger than 31000, so it's 2 times. So 2 * 11111 = 22222. Subtract that from 31000: 31000 - 22222 = 8778. Bring down a zero: 87780.

Now, how many times does 11111 go into 87780? Let me compute 11111 * 7 = 77777. 11111 * 8 = 88888. 88888 is larger than 87780, so it's 7 times. 7 * 11111 = 77777. Subtract from 87780: 87780 - 77777 = 10003. Bring down a zero: 100030.

Now, how many times does 11111 go into 100030? Let me compute 11111 * 9 = 99999. That's very close. 99999 is less than 100030. So 9 times. Subtract 99999 from 100030: 100030 - 99999 = 31. Bring down a zero: 310.

Wait, now we have 310 again, which is where we started after the decimal point. So the decimal repeats from here. Let me write down the decimal digits we have so far:

First, after the decimal, we had 0. Then:

First division step: 2 (from 31000 / 11111 = 2), then 7 (from 87780 / 11111 = 7), then 9 (from 100030 / 11111 = 9), then we had a remainder of 31, which is the same as the original numerator. So the decimal expansion is 0.00279... and then it repeats. Wait, let me check the digits:

Wait, let me track the decimal places step by step.

Starting with 31.000000...

First, 11111 into 31.0 is 0. So first decimal digit: 0.

Then 11111 into 310 is 0. Second decimal digit: 0.

Then 11111 into 3100 is 0. Third decimal digit: 0.

Then 11111 into 31000 is 2 (as 2*11111=22222). So fourth decimal digit: 2.

Subtract 22222 from 31000 gives 8778. Bring down a zero: 87780.

Now, 11111 into 87780 is 7 (7*11111=77777). Fifth decimal digit: 7.

Subtract 77777 from 87780 gives 10003. Bring down a zero: 100030.

11111 into 100030 is 9 (9*11111=99999). Sixth decimal digit: 9.

Subtract 99999 from 100030 gives 31. Bring down a zero: 310.

Now, we have 310, which is the same as after the decimal point initially. So the decimal repeats from here. Therefore, the decimal expansion is 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the number of repeating digits is 5.

Wait, but let me check again. The decimal expansion is 0.002790027900279..., so the repeating cycle is "00279", which is five digits. So the answer is 5?

But let me check with another approach. Since 11111 is (10^5 - 1)/9, so 1/11111 = 9/99999. Therefore, 31/11111 = 31*9 / 99999 = 279 / 99999. So 279 divided by 99999. Since 99999 is 10^5 - 1, the decimal expansion of 279/99999 is 0.\overline{00279}, which is a repeating cycle of 5 digits. Therefore, the answer is 5.

But let me check with another method. Let me think about the decimal expansion of fractions with denominator 99999. Since 99999 is 10^5 - 1, the decimal expansion of 1/99999 is 0.000010000100001..., repeating every 5 digits. Therefore, multiplying by 279 would give 0.002790027900279..., so the repeating part is 00279, which is 5 digits. Therefore, the answer is 5.

But let me check if there is a shorter period. For example, sometimes the period can be shorter than the denominator's length. For example, 1/3 = 0.333..., period 1, even though denominator is 3. But in this case, since 11111 is a factor of 99999, which is 10^5 - 1. Let me check if 11111 divides 10^k - 1 for some smaller k. If so, then the period would be smaller.

So, let me check if 11111 divides 10^k - 1 for some k < 5. Let me check k=1: 10^1 -1 = 9, which is not divisible by 11111. k=2: 99, nope. k=3: 999, nope. k=4: 9999. Let me divide 9999 by 11111. Well, 9999 is less than 11111, so no. Therefore, the minimal k is 5. Therefore, the period is 5. Therefore, the decimal expansion of 1/11111 has period 5, so multiplying by 31 would not change the period. Therefore, the decimal expansion of 31/11111 also has period 5. Therefore, the answer is 5.

But let me check with another example. Let me take 1/11111. If I compute 1 divided by 11111, what is the decimal expansion? Let me try to compute it.

1 divided by 11111. Let me do the division:

1.000000...

11111 goes into 100000 (after adding decimal and zeros) how many times? Let me compute 11111 * 9 = 99999. So 9 times. So 9*11111=99999. Subtract from 100000: 1. Bring down a zero: 10. Then 11111 goes into 10 zero times. Bring down another zero: 100. Still zero. Bring down another zero: 1000. Still zero. Bring down another zero: 10000. Still zero. Bring down another zero: 100000. Now we are back to where we started. So the decimal expansion is 0.000090000900009..., so repeating every 5 digits: "00009". Therefore, the period is 5. Therefore, multiplying by 31 would give 0.0027900279..., which is the same as before. Therefore, the repeating cycle is 5 digits. Therefore, the answer is 5.

But let me check if there's a possibility that the decimal repeats with a shorter period. For example, maybe 31/11111 has a shorter period? How can I check that?

The period of a fraction a/b is equal to the multiplicative order of 10 modulo b', where b' is the denominator after removing factors of 2 and 5. In this case, the denominator is 11111. Let me check if 11111 has factors other than 1. Let me factorize 11111.

Let me try dividing 11111 by small primes. Let me check divisibility by 3: 1+1+1+1+1=5, which is not divisible by 3. Divisible by 7? Let me check 11111 divided by 7. 7*1587 = 11109, remainder 2. Not divisible by 7. Divided by 11? Let me use the divisibility rule for 11: difference between sum of digits in odd positions and even positions. For 11111: positions are 1 (position 1), 1 (position 2), 1 (position 3), 1 (position 4), 1 (position 5). Sum of odd positions: 1 + 1 + 1 = 3. Sum of even positions: 1 + 1 = 2. Difference is 3 - 2 = 1, not divisible by 11. Divided by 13? Let me try 13*854 = 11102, remainder 9. Not divisible. Divided by 17? Let me try 17*653 = 11101, remainder 10. Not divisible. Divided by 19? 19*584 = 11096, remainder 15. Not divisible. Divided by 23? 23*483 = 11109, remainder 2. Not divisible. Divided by 29? Let me try 29*383 = 11107, remainder 4. Not divisible. Divided by 31? 31*358 = 111, 31*358 = 31*(300 + 58) = 9300 + 1798 = 11098. Remainder 13. Not divisible. Divided by 37? Let me try 37*300 = 11100. 11111 - 11100 = 11. Not divisible. Divided by 41? 41*271 = 11111? Let me check 41*270 = 11070, plus 41 is 11111. Yes! Wait, 41*271 = 11111? Let me compute 41*271:

First, 40*271 = 10840, plus 1*271 = 271. Total is 10840 + 271 = 11111. Yes! So 11111 factors into 41 and 271. Let me check if 271 is prime. Let me check divisibility of 271. Divided by 2? No. Divided by 3? 2+7+1=10, not divisible by 3. Divided by 5? No. Divided by 7? 7*38=266, remainder 5. Divided by 11? 11*24=264, remainder 7. Divided by 13? 13*20=260, remainder 11. Divided by 17? 17*15=255, remainder 16. Divided by 19? 19*14=266, remainder 5. Divided by 23? 23*11=253, remainder 18. So 271 is prime. Therefore, 11111 factors into 41*271. Both primes.

Therefore, the denominator is 41*271. Now, to find the period of 31/11111, we need to find the multiplicative order of 10 modulo 11111, but since 11111 is co-prime with 10 (since it's 41*271, neither of which are 2 or 5). Wait, but the multiplicative order of 10 modulo 11111 is the least common multiple of the multiplicative orders modulo 41 and 271. Because 41 and 271 are coprime.

So, first, find the multiplicative order of 10 modulo 41. Then find the multiplicative order of 10 modulo 271. Then take the LCM of those two numbers.

Let me start with modulo 41.

What is the multiplicative order of 10 modulo 41?

We need to find the smallest positive integer k such that 10^k ≡ 1 mod 41.

First, φ(41) = 40, so the order divides 40. The divisors of 40 are 1, 2, 4, 5, 8, 10, 20, 40.

Check 10^1 mod 41 = 10 ≠ 1.

10^2 = 100 mod 41. 41*2=82, 100-82=18. 10^2 ≡ 18 ≠ 1.

10^4 = (10^2)^2 ≡ 18^2 = 324 mod 41. 41*7=287, 324 - 287 = 37. 10^4 ≡ 37 ≠ 1.

10^5 = 10^4 * 10 ≡ 37*10 = 370 mod 41. 41*9=369, so 370 - 369 = 1. So 10^5 ≡ 1 mod 41. Therefore, the multiplicative order of 10 modulo 41 is 5.

Now, check modulo 271.

Find the multiplicative order of 10 modulo 271.

First, φ(271) = 270, since 271 is prime. So the order divides 270. The divisors of 270 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270.

We need to find the smallest k such that 10^k ≡ 1 mod 271.

Let me check the divisors in order.

First, check k=1: 10^1=10≠1.

k=2: 10^2=100≠1.

k=3: 10^3=1000 mod 271. Let me compute 271*3=813. 1000 - 813 = 187. 187≠1.

k=5: Let me check k=5. Maybe compute 10^5 mod 271. Alternatively, check if 10^k ≡1 for smaller divisors.

Alternatively, since 270 factors into 2*3^3*5. So possible orders are divisors of 270.

Let me check k=9: 10^9 mod 271. Maybe compute step by step.

Alternatively, use exponentiation by squaring.

First, compute 10^1 mod 271 = 10

10^2 = 100

10^4 = (10^2)^2 = 100^2 = 10000 mod 271.

Compute 271*37 = 271*(30 + 7) = 8130 + 1897 = 9, 8130 + 1897 = 10027. Wait, 271*37: Let me compute 271*30 = 8130, 271*7=1897, so total 8130 + 1897 = 10027. Then 10000 - 10027 = -27 mod 271. So 10^4 ≡ -27 mod 271.

Then 10^5 = 10^4 * 10 ≡ (-27)*10 = -270 ≡ 1 mod 271? Wait, -270 mod 271 is 1, because 271 - 270 = 1. So 10^5 ≡ 1 mod 271? Wait, that would mean the order is 5? But 5 divides 270? Yes, 270 divided by 5 is 54. But let me check again.

Wait, 10^5 mod 271: Let me verify my calculation.

Earlier, I said 10^4 ≡ -27 mod 271. Then 10^5 = 10^4 * 10 ≡ (-27)*10 = -270. Then -270 + 271 = 1. Therefore, 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is 5? But 5 divides φ(271)=270? Yes, since 270 = 5*54. Therefore, the order is 5? Wait, but that seems surprising. Let me check with another exponent.

Wait, if 10^5 ≡ 1 mod 271, then 10^5 - 1 is divisible by 271. Let me check 10^5 - 1 = 100000 - 1 = 99999. Is 99999 divisible by 271? Let me divide 99999 by 271.

Compute 271*300 = 81300. 99999 - 81300 = 18699.

271*60 = 16260. 18699 - 16260 = 2439.

271*9 = 2439. So total is 300 + 60 + 9 = 369. Therefore, 271*369 = 99999. Therefore, yes, 10^5 - 1 = 99999 = 271*369. Therefore, 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is indeed 5.

Therefore, the multiplicative order of 10 modulo 41 is 5, and modulo 271 is also 5. Therefore, the multiplicative order modulo 11111 is LCM(5,5)=5. Therefore, the period of the decimal expansion of 31/11111 is 5. Therefore, the number of repeating digits is 5.

Therefore, the answer is 5. But let me check once more with the decimal expansion I did earlier. When I did the long division, I saw that after 0.00279, the remainder came back to 31, which was the original numerator, so the cycle repeats every 5 digits. Therefore, the repeating part is 5 digits. Therefore, the answer is 5.

But just to be thorough, let me check if there's a shorter cycle. Suppose the decimal repeats every 1, 2, 3, or 4 digits. Let me check.

Suppose the decimal expansion is 0.0027900279..., so the repeating part is "00279". Let me check if "00279" can be broken into a shorter cycle. For example, if it was "00279" repeating, but maybe there's a shorter cycle. However, since the multiplicative order is 5, which is prime, the period cannot be shorter. Because if the order is 5, then the period is exactly 5. Therefore, there is no shorter period. Therefore, the answer is 5.

Therefore, I can be confident that the number of repeating digits is 5.

**Final Answer**
The number of repeating digits is \boxed{5}.

To determine how many digits repeat in the decimal expansion of the fraction $\frac{31}{11111}$, we begin by analyzing the structure of the denominator.

---

### Step 1: Understanding the Denominator

The denominator is $11111$. This number is not a power of 10, and it is not divisible by 2 or 5, so the decimal expansion of $\frac{31}{11111}$ will be a **repeating decimal**.

We can express $11111$ in terms of powers of 10:

$$
11111 = \frac{10^5 - 1}{9}
$$

This is because:

$$
\frac{10^5 - 1}{9} = \frac{99999}{9} = 11111
$$

So, we can rewrite the original fraction as:

$$
\frac{31}{11111} = \frac{31 \cdot 9}{99999} = \frac{279}{99999}
$$

---

### Step 2: Decimal Expansion of $\frac{279}{99999}$

The number $99999 = 10^5 - 1$, and the decimal expansion of any fraction of the form $\frac{a}{10^n - 1}$ has a **repeating cycle of length $n$**. This is because the denominator is one less than a power of 10, and the decimal repeats every $n$ digits.

So, the decimal expansion of $\frac{279}{99999}$ will repeat every **5 digits**.

To confirm, we can perform the long division of $279 \div 99999$, or observe that:

$$
\frac{279}{99999} = 0.\overline{00279}
$$

This means the repeating part is **"00279"**, which is **5 digits long**.

---

### Step 3: Confirming the Period Using Multiplicative Order

To further verify, we can use the concept of **multiplicative order**. The period of the decimal expansion of a fraction $\frac{a}{b}$ (where $b$ is co-prime to 10) is equal to the **multiplicative order of 10 modulo $b$**.

We factor $11111 = 41 \times 271$, both of which are prime numbers. The multiplicative order of 10 modulo 41 and 271 is 5 in both cases. Therefore, the least common multiple of these orders is:

$$
\text{lcm}(5, 5) = 5
$$

This confirms that the decimal expansion of $\frac{31}{11111}$ has a **repeating cycle of 5 digits**.

---

### Final Answer

$$
\boxed{5}
$$

Solution:
We begin by noting that    $$\frac{31}{11111} = \frac{31 \times 9}{11111 \times 9} = \frac{279}{99999}.$$We will show that   $$\frac{279}{99999} = 0.\overline{00279},$$so our final answer is $\boxed{5}.$

Proof that $279/99999 = 0.\overline{00279}$:

Let $s = 0.\overline{00279}$. Then multiplying both sides $10^5$ gives   $$10^5 s = 279.\overline{00279}.$$Subtracting the left side by $s$ and the right side by $0.\overline{00279}$ gives   $$99999s = 279,$$so $s = 279/99999$. It follows that $0.\overline{00279} = 279 / 99999,$ as desired.


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. The square roots can be combined:

√(18/19) / √(8/19) = √[(18/19) / (8/19)] = √(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 denominators can't be zero, so 3x - 1 ≠ 0 => x ≠ 1/3. Also, the expressions under the square roots must be non-negative:

For √(2x): 2x ≥ 0 => x ≥ 0

For √(3x - 1): 3x - 1 ≥ 0 => x ≥ 1/3

Therefore, the domain is x ≥ 1/3. Now, x = 9/19 is approximately 0.473, which is greater than 1/3 (which is approximately 0.333). So that's okay. Therefore, x = 9/19 is within the domain and satisfies the original equation. So that's the solution.

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, the algebra led me directly to x = 9/19, and since the equation after squaring was linear, there's only one solution. Therefore, I think that's the only solution.

Alternatively, maybe I can approach the problem by rationalizing the denominator or another method? Let me try another way to see if I get the same result. Let me write the original equation again:

√(2x)/√(3x - 1) = 3/2

Another approach could be to write this as √(2x) = (3/2)√(3x - 1). Then square both sides:

(√(2x))² = (3/2)²*(√(3x - 1))² => 2x = (9/4)*(3x - 1). Which is the same equation as before: 2x = (27x - 9)/4. Multiply both sides by 4: 8x = 27x - 9, leading to the same result. So same steps, same answer. So that's consistent.

Alternatively, maybe I can write the original equation as √(2x)/√(3x - 1) = 3/2, and then combine the square roots into a single square root: √[(2x)/(3x - 1)] = 3/2. Then square both sides: (2x)/(3x - 1) = 9/4. Which again is the same equation. So regardless of the method, I end up with the same equation. Therefore, solving that gives x = 9/19, which checks out.

Therefore, I think that's the correct answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

Original equation: √(2x)/√(3x - 1) = 3/2.

Square both sides: 2x/(3x - 1) = 9/4.

Cross-multiplying: 4*2x = 9*(3x - 1) => 8x = 27x - 9.

Subtract 8x: 0 = 19x - 9 => 19x = 9 => x = 9/19.

Check domain: x ≥ 1/3. 9/19 ≈ 0.473, which is greater than 1/3 ≈ 0.333. So valid.

Check in original equation: √(2*(9/19))/√(3*(9/19) - 1) = √(18/19)/√(27/19 - 19/19) = √(18/19)/√(8/19) = √(18/8) = √(9/4) = 3/2. Correct.

Therefore, x = 9/19 is the solution. I think that's solid. I don't see any mistakes here. So I think that's the answer.

**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. This is valid because both sides of the equation are positive (since square roots are non-negative and the right-hand side is positive), so squaring will not introduce extraneous solutions in this case.

---

### 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 for $ x $

Subtract $ 8x $ from both sides:

$$
0 = 19x - 9
$$

$$
19x = 9
$$

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

---

### Step 4: Verify the Solution

We must ensure that the solution satisfies the original equation and that the expressions under the square roots are valid (i.e., non-negative).

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

- Since $ \frac{9}{19} \approx 0.473 $, which is greater than $ \frac{1}{3} \approx 0.333 $, the solution is valid.

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

- $ \sqrt{2x} = \sqrt{\frac{18}{19}} $
- $ \sqrt{3x - 1} = \sqrt{\frac{27}{19} - \frac{19}{19}} = \sqrt{\frac{8}{19}} $

So the left-hand side becomes:

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

This matches the right-hand side, confirming the solution is correct.

---

### 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. The square roots can be combined:

√(18/19) / √(8/19) = √[(18/19) / (8/19)] = √(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 denominators can't be zero, so 3x - 1 ≠ 0 => x ≠ 1/3. Also, the expressions under the square roots must be non-negative:

For √(2x): 2x ≥ 0 => x ≥ 0

For √(3x - 1): 3x - 1 ≥ 0 => x ≥ 1/3

Therefore, the domain is x ≥ 1/3. Now, x = 9/19 is approximately 0.473, which is greater than 1/3 (which is approximately 0.333). So that's okay. Therefore, x = 9/19 is within the domain and satisfies the original equation. So that's the solution.

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, the algebra led me directly to x = 9/19, and since the equation after squaring was linear, there's only one solution. Therefore, I think that's the only solution.

Alternatively, maybe I can approach the problem by rationalizing the denominator or another method? Let me try another way to see if I get the same result. Let me write the original equation again:

√(2x)/√(3x - 1) = 3/2

Another approach could be to write this as √(2x) = (3/2)√(3x - 1). Then square both sides:

(√(2x))² = (3/2)²*(√(3x - 1))² => 2x = (9/4)*(3x - 1). Which is the same equation as before: 2x = (27x - 9)/4. Multiply both sides by 4: 8x = 27x - 9, leading to the same result. So same steps, same answer. So that's consistent.

Alternatively, maybe I can write the original equation as √(2x)/√(3x - 1) = 3/2, and then combine the square roots into a single square root: √[(2x)/(3x - 1)] = 3/2. Then square both sides: (2x)/(3x - 1) = 9/4. Which again is the same equation. So regardless of the method, I end up with the same equation. Therefore, solving that gives x = 9/19, which checks out.

Therefore, I think that's the correct answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

Original equation: √(2x)/√(3x - 1) = 3/2.

Square both sides: 2x/(3x - 1) = 9/4.

Cross-multiplying: 4*2x = 9*(3x - 1) => 8x = 27x - 9.

Subtract 8x: 0 = 19x - 9 => 19x = 9 => x = 9/19.

Check domain: x ≥ 1/3. 9/19 ≈ 0.473, which is greater than 1/3 ≈ 0.333. So valid.

Check in original equation: √(2*(9/19))/√(3*(9/19) - 1) = √(18/19)/√(27/19 - 19/19) = √(18/19)/√(8/19) = √(18/8) = √(9/4) = 3/2. Correct.

Therefore, x = 9/19 is the solution. I think that's solid. I don't see any mistakes here. So I think that's the answer.

**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. This is valid because both sides of the equation are positive (since square roots are non-negative and the right-hand side is positive), so squaring will not introduce extraneous solutions in this case.

---

### 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 for $ x $

Subtract $ 8x $ from both sides:

$$
0 = 19x - 9
$$

$$
19x = 9
$$

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

---

### Step 4: Verify the Solution

We must ensure that the solution satisfies the original equation and that the expressions under the square roots are valid (i.e., non-negative).

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

- Since $ \frac{9}{19} \approx 0.473 $, which is greater than $ \frac{1}{3} \approx 0.333 $, the solution is valid.

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

- $ \sqrt{2x} = \sqrt{\frac{18}{19}} $
- $ \sqrt{3x - 1} = \sqrt{\frac{27}{19} - \frac{19}{19}} = \sqrt{\frac{8}{19}} $

So the left-hand side becomes:

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

This matches the right-hand side, confirming the solution is correct.

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### 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.