[content] evaluation/easy_problems_thinking.jsonl (jsonl)
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?"}], "output": "<think>
Okay, so I have this problem here: \"The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?\" Hmm, let's see. Alright, the question is asking for the probability that it will snow next January 1st, given that the probability is 1/20. Wait, is there more to this? Let me check again.

Wait, maybe I misread. The original probability is 1/20 for snow on January 1st, and they want the probability it will snow on the next January 1st. Hmm. That seems straightforward. But maybe there's a trick here? Like, are they assuming each year is independent? Because if the probability each year is independent, then it's just 1/20 again. But maybe the question is different? Wait, maybe the wording is tricky. Let me parse it again.

\"The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?\" So, \"next\" as in the following year? So next year's January 1st. So if each year's snowfall is independent, then the probability remains 1/20. But maybe there's some hidden assumption here. Wait, sometimes in probability problems, they might be talking about conditional probabilities or something else. But this problem doesn't mention any previous events or dependencies. It just states the probability is 1/20. So if each occurrence is independent, then the probability doesn't change. So next year's probability is also 1/20.

Wait, but let me think again. Maybe the problem is in another language originally, and \"next\" is being translated as \"the next occurrence\" or something? But no, it's about \"next 1st January\". So like, the next time January 1st comes around, which would be the following year. So in terms of probability, assuming each year is an independent event, the probability is the same each year, right? So if the probability of snow on any given January 1st is 1/20, then regardless of previous years, the next one would also be 1/20. 

Alternatively, could it be that the problem is asking for the probability that it will snow next year given that it snowed this year? But the problem doesn't say anything about prior snowfall. It just gives the base rate. So I think the answer is 1/20. Let me confirm.

Another angle: sometimes people confuse \"probability of snow on a specific day\" with \"probability of snow in a year\" but here it's specifically on 1st January each year. So if the probability each year on that date is 1/20, then next year's is also 1/20. Unless there's a pattern, but the problem doesn't indicate that. So I think the answer is 1/20. So I just need to write that in a box. Wait, maybe I need to make sure if the problem is in a different context. For example, maybe it's a different place where the next January 1st is in the southern hemisphere? But that's overcomplicating. The problem is likely testing understanding of independent events. So the answer is 1/20.
</think>

\\boxed{\\dfrac{1}{20}}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

When orange trees are planted to the acre, each tree yields 100 pounds of fruit. For each additional tree over 100, the yield per tree is reduced by 1 pound. How many trees should be planted so that the yield per acre is 4816 pounds?"}], "output": "<think>
Okay, let's try to figure out this problem. So, we're dealing with orange trees planted per acre and their yield. The problem says that when they're planted 100 per acre, each tree gives 100 pounds of fruit. But if you plant more than 100 trees, each additional tree over 100 causes the yield per tree to drop by 1 pound. We need to find how many trees should be planted to get a total yield of 4816 pounds per acre.

Hmm, let's break it down. Let me start by setting up variables. Let's say the number of trees per acre is x. If x is 100, then each tree gives 100 pounds, so total yield is 100*100 = 10,000 pounds. But since we want a yield of 4816, which is much less, maybe we need to plant more trees? Wait, but more trees mean each tree gives less. So there's a trade-off here.

The problem says for each additional tree over 100, the yield per tree decreases by 1 pound. So if we have x trees where x > 100, the number of extra trees is (x - 100). Therefore, the yield per tree would be 100 - (x - 100) pounds. Wait, let me check that. If you have x trees, and the extra trees over 100 are (x - 100), then each tree's yield is 100 minus that number. So, yield per tree = 100 - (x - 100) = 200 - x. Hmm, wait, that might not be right. Let me think again.

Wait, no. Let's say for each tree beyond 100, the yield decreases by 1. So if you plant 101 trees, each tree gives 100 - 1 = 99 pounds. If you plant 102 trees, each gives 98, and so on. So in general, if you plant x trees where x > 100, the yield per tree is 100 - (x - 100). Wait, but 100 - (x - 100) simplifies to 200 - x. So, yield per tree is (200 - x) pounds when x > 100. Then, the total yield would be x * (200 - x). So total yield is x*(200 - x). But we need this total yield to be 4816. So, the equation would be x*(200 - x) = 4816.

Wait, but let me make sure. Let me test with x=100. If x=100, then total yield would be 100*100 = 10,000, which matches the problem statement. If x=150, then yield per tree would be 100 - 50 = 50, so total yield 150*50=7500. That seems right.

So, according to this, the total yield when planting x trees (for x >= 100) is x*(200 - x). Wait, but hold on: when x=100, that gives 100*(200-100)=100*100=10,000, which is correct. For x=101, it's 101*(200 - 101)=101*99=9999, which aligns with each tree producing 1 less pound. So yes, that formula seems correct.

So, we set up the equation:

x*(200 - x) = 4816

But let me check, is 4816 possible? Let's see. The maximum possible yield is at x=100, which is 10,000. So 4816 is less than that. Wait, but when we go beyond 100 trees, each tree's yield decreases, but the number of trees increases. So maybe there's a point where the total yield is 4816. Wait, but maybe there are two such points? Because as x increases beyond 100, the total yield first decreases from 10,000, and maybe reaches a minimum and then... Wait, actually, the equation x*(200 - x) is a quadratic that opens downward, with vertex at x=100, which is the maximum point. Wait, no, x*(200 - x) is equivalent to -x^2 + 200x. The vertex is at x = -b/(2a) = -200/(2*-1) = 100. So yes, that's the maximum point. So as x increases beyond 100, the total yield decreases. Therefore, since 4816 is less than 10,000, there should be two solutions for x: one less than 100 and one greater than 100. But in the context of the problem, since the problem mentions \"additional trees over 100\", we're looking for x >= 100. Wait, but x can't be less than 100 here because the problem states that the reduction starts when you plant more than 100 trees. So if x is less than 100, does the yield stay at 100 pounds per tree? The problem says, \"When orange trees are planted to the acre\" (assuming that's 100 per acre?), \"each tree yields 100 pounds of fruit. For each additional tree over 100, the yield per tree is reduced by 1 pound.\" So if we plant fewer than 100 trees, does the yield per tree stay at 100? The problem doesn't specify, but since the question is about planting \"additional trees over 100\", implying that the base case is 100 trees. So maybe the model is that if you plant up to 100 trees, each gives 100 pounds, and beyond that, each additional tree reduces the yield by 1 per tree. So the total yield function would be:

If x <= 100, total yield = 100x

If x > 100, total yield = x*(200 - x)

But the problem wants the total yield to be 4816 pounds. Let's see if x <= 100 is possible. If x=48, 48*100=4800, which is close. x=49 would be 4900. But 4816 is between 4800 and 4900. But since 4816 isn't a multiple of 100, it can't be achieved with x <= 100. Therefore, the solution must be x > 100.

So the equation we need to solve is x*(200 - x) = 4816.

Let me write that as:

200x - x^2 = 4816

Rearranged to:

x^2 - 200x + 4816 = 0

Wait, moving everything to the left side:

-x^2 + 200x - 4816 = 0 --> multiply both sides by -1:

x^2 - 200x + 4816 = 0

Now, let's solve this quadratic equation for x.

We can use the quadratic formula:

x = [200 ± sqrt(200^2 - 4*1*4816)] / 2

Compute discriminant D:

D = 40000 - 4*4816 = 40000 - 19264 = 20736

sqrt(20736) = 144 (since 144*144=20736)

So,

x = [200 ± 144]/2

Calculating the two roots:

x = (200 + 144)/2 = 344/2 = 172

x = (200 - 144)/2 = 56/2 = 28

But earlier, we determined that x must be greater than 100, since the yield when x<=100 would be 100x, and 100*28=2800 which is less than 4816, and 100*48=4800, 100*49=4900. So 28 is too low and 172 is over 100. So x=172 is the solution.

Wait, but let me verify with x=172 trees. The yield per tree would be 100 - (172 - 100) = 100 - 72 = 28 pounds per tree. Then total yield is 172 * 28. Let's compute that.

172 * 28: 170*28=4760, 2*28=56, so 4760+56=4816. That's exactly the yield we need. So that works. The other solution x=28 would give 28 trees each at 100 pounds, which is 2800 pounds, which is not the desired 4816. Therefore, the answer is 172 trees.

So the answer should be 172. Let me just check once again. If 172 trees are planted, the yield per tree is 100 - (172 - 100) = 28, total 172*28=4816. Perfect. So yes, that's correct.
</think>

The number of trees to plant is found by solving the quadratic equation derived from the problem's conditions. Let \\( x \\) be the number of trees. For \\( x > 100 \\), the yield per tree is \\( 100 - (x - 100) = 200 - x \\). The total yield equation is:

\\[
x(200 - x) = 4816
\\]

Rewriting this as a quadratic equation:

\\[
x^2 - 200x + 4816 = 0
\\]

Using the quadratic formula \\( x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} \\):

\\[
x = \\frac{200 \\pm \\sqrt{200^2 - 4 \\cdot 1 \\cdot 4816}}{2} = \\frac{200 \\pm 144}{2}
\\]

This yields two solutions: \\( x = 172 \\) and \\( x = 28 \\). Since \\( x \\) must be greater than 100 to affect the yield reduction, the valid solution is:

\\[
\\boxed{172}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

Find the smallest prime number \\( p \\) such that \\( p \\) divides \\( n^2 + 7n + 23 \\)."}], "output": "<think>
Okay, so I need to find the smallest prime number p that divides n² + 7n + 23 for some integer n. Hmm, let me try to figure this out step by step.

First, the problem is asking for the smallest prime p such that there exists an integer n where p divides n² + 7n + 23. That means for that prime p, the equation n² + 7n + 23 ≡ 0 mod p has a solution. So essentially, I need to check the primes starting from the smallest ones (2, 3, 5, 7, 11, etc.) and see if there's an n that makes the quadratic expression divisible by each prime. The first prime for which this is possible will be the answer.

Let me start with the smallest prime, which is 2. Let's check if there exists an n such that n² + 7n + 23 ≡ 0 mod 2. To do this, I can evaluate the expression modulo 2. Let's consider n modulo 2. If n is even (n ≡ 0 mod 2), then n² ≡ 0, 7n ≡ 0, so 0 + 0 + 23 ≡ 1 mod 2. If n is odd (n ≡ 1 mod 2), then n² ≡ 1, 7n ≡ 1, so 1 + 1 + 23 ≡ 1 + 1 + 1 ≡ 3 ≡ 1 mod 2. So in both cases, the expression is congruent to 1 mod 2. Therefore, there's no n for which 2 divides the expression. So 2 is out.

Next prime is 3. Let's check modulo 3. Let's consider n ≡ 0, 1, 2 mod 3.

If n ≡ 0 mod 3:
n² + 7n + 23 ≡ 0 + 0 + 23 ≡ 23 mod 3. 23 ÷ 3 is 7 with remainder 2, so 23 ≡ 2 mod 3.

If n ≡ 1 mod 3:
1 + 7*1 + 23 ≡ 1 + 7 + 23 ≡ 1 + 1 + 2 ≡ 4 ≡ 1 mod 3.

If n ≡ 2 mod 3:
4 + 14 + 23 ≡ 4 + 2 + 2 ≡ 8 ≡ 2 mod 3.

So none of the residues mod 3 give 0. Therefore, 3 doesn't divide the expression for any n. So 3 is also out.

Next prime is 5. Let's check modulo 5. n can be 0,1,2,3,4 mod 5.

n ≡ 0:
0 + 0 + 23 ≡ 23 mod 5 → 23 - 4*5 = 23 -20 = 3 → 3 mod 5.

n ≡ 1:
1 + 7 + 23 = 31 mod 5. 31 ÷5 is 6*5=30, remainder 1. So 1 mod 5.

n ≡ 2:
4 + 14 +23 = 41 mod5. 41-8*5=41-40=1. So 1 mod5.

n ≡3:
9 +21 +23 =53. 53-10*5=53-50=3. 3 mod5.

n ≡4:
16 +28 +23=67. 67-13*5=67-65=2. So 2 mod5.

So none of the residues mod 5 give 0. So 5 is out.

Next prime is 7. Let's check modulo7. n from 0 to6.

n=0: 0 +0 +23=23 mod7. 23-3*7=2. 2 mod7.

n=1:1+7+23=31. 31-4*7=31-28=3. 3 mod7.

n=2:4+14+23=41. 41-5*7=41-35=6. 6 mod7.

n=3:9+21+23=53. 53-7*7=53-49=4. 4 mod7.

n=4:16+28+23=67. 67-9*7=67-63=4. 4 mod7.

n=5:25+35+23=83. 83-11*7=83-77=6. 6 mod7.

n=6:36+42+23=101. 101-14*7=101-98=3. 3 mod7.

So none of the residues give 0 mod7. So 7 is out.

Next prime is 11. Wait, but maybe I should check primes in order. The next prime after 7 is 11? Wait, no, primes are 2,3,5,7,11,13,17,19,23,29, etc. So next is 11.

Check modulo11. Let's see:

n can be 0 to 10.

Calculate n² +7n +23 mod11.

First, simplify coefficients modulo11.

7 mod11 is 7.

23 mod11=23-2*11=23-22=1. So expression becomes n² +7n +1 mod11.

Now check each n from 0 to10.

n=0: 0 +0 +1=1 mod11.

n=1:1 +7 +1=9 mod11.

n=2:4 +14 +1=19 mod11=19-11=8.

n=3:9 +21 +1=31 mod11=31-2*11=9.

n=4:16 +28 +1=45 mod11. 45-4*11=45-44=1.

n=5:25 +35 +1=61 mod11. 61-5*11=61-55=6.

n=6:36 +42 +1=79 mod11. 79-7*11=79-77=2.

n=7:49 +49 +1=99 mod11=0. Wait, 49 mod11: 44 is 4*11, so 49=44+5, so 5 mod11. So n=7:

n=7: 7²=49≡5, 7*7=49≡5, so 5 (from n²) + 5 (from 7n) +1 (constant term) =5+5+1=11≡0 mod11. Oh! So when n=7 mod11, the expression is 0 mod11. Therefore, 11 divides the expression when n=7. So 11 is a candidate. But we need to check if there is a smaller prime that works. Since we checked 2,3,5,7 and none worked, the next prime is 11. So the answer should be 11? But wait, maybe I made a mistake here. Let me double-check.

Wait, when n=7, we have n² +7n +23 = 49 + 49 +23 = 121. 121 is 11², so 11 divides 121. So yes, that works. So 11 is the prime here. But since the problem says \"the smallest prime number p such that p divides n² +7n +23\", and 11 is the first prime we found that works, after checking 2,3,5,7. Therefore, 11 is the answer. But wait, just to be thorough, maybe I missed a prime between 7 and 11? No, because 7 is followed by 11, as 8,9,10 aren't primes. So the answer is 11. Hmm. But wait, let me check n=7: 7² +7*7 +23 = 49 +49 +23=121, which is 11 squared. So yes, 11 divides 121. Therefore, the smallest prime is 11. Therefore, the answer should be 11. But let me check one more prime after 11 just to make sure I didn't miss anything. For example, maybe 13 is smaller? Wait, 11 is smaller than 13, so even if 13 works, 11 is the answer. So I think 11 is the answer. But let me check primes up to 11 again. 2,3,5,7 don't work. 11 works. Therefore, the answer is 11. Hmm. Wait, but I just thought of another approach. Maybe using the quadratic formula or discriminant. Let me think.

Alternatively, the equation n² +7n +23 ≡0 mod p. For this equation to have a solution, the discriminant must be a quadratic residue modulo p. The discriminant D = 7² -4*1*23 = 49 -92 = -43. So discriminant is -43. So for the equation to have a solution, -43 must be a quadratic residue modulo p. So (-43|p) =1, where (•|p) is the Legendre symbol. So primes p where -43 is a quadratic residue. Then, we need to find the smallest prime p where (-43|p) =1. Then, the primes in order would be checked. But maybe this is more complicated, but perhaps it can help. But I think the way I checked each prime up to 11 is sufficient. Since 11 works, and smaller primes don't. Therefore, the answer is 11. Wait, but let me verify again. For p=11, does -43 ≡ a square mod 11? Let's compute -43 mod11. 43 divided by 11 is 3*11=33, remainder 10. So 43 ≡10 mod11. So -43 ≡-10 ≡1 mod11. So -43 ≡1 mod11. 1 is a quadratic residue (since 1^2=1). Therefore, the discriminant is 1 mod11, which is a square. Therefore, the equation has solutions. Therefore, 11 is the smallest prime. Thus, the answer is 11. So \\boxed{11} is the answer. I think this is correct.
</think>

The smallest prime number \\( p \\) that divides \\( n^2 + 7n + 23 \\) for some integer \\( n \\) is found by checking primes in ascending order. 

1. **Prime 2**: The expression \\( n^2 + 7n + 23 \\equiv 1 \\mod 2 \\) for all \\( n \\), so 2 is excluded.
2. **Prime 3**: Testing \\( n \\equiv 0, 1, 2 \\mod 3 \\) yields residues 2, 1, and 2 respectively, so 3 is excluded.
3. **Prime 5**: Testing \\( n \\equiv 0, 1, 2, 3, 4 \\mod 5 \\) yields residues 3, 1, 1, 3, and 2 respectively, so 5 is excluded.
4. **Prime 7**: Testing \\( n \\equiv 0, 1, 2, 3, 4, 5, 6 \\mod 7 \\) yields residues 2, 3, 6, 4, 4, 6, and 3 respectively, so 7 is excluded.
5. **Prime 11**: Simplifying the expression modulo 11 gives \\( n^2 + 7n + 1 \\mod 11 \\). Testing \\( n = 7 \\) gives \\( 7^2 + 7 \\cdot 7 + 1 = 121 \\equiv 0 \\mod 11 \\). Thus, 11 divides the expression when \\( n = 7 \\).

Therefore, the smallest prime \\( p \\) is \\(\\boxed{11}\\).", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

Is there a geometric transformation that maps every point to exactly one point such that any triangle can be transformed into another triangle with the same area? (Alternatively, can any triangle be transformed into a regular triangle with the same area?)"}], "output": "<think>
Okay, so I need to figure out if there's a geometric transformation that can map any triangle to another triangle with the same area, or specifically to a regular triangle with the same area. Hmm. Let's start by recalling what geometric transformations preserve areas. Affine transformations, right? Because things like scaling, rotation, translation, shearing—some of these preserve area, but others might change it. Wait, actually, affine transformations can change area unless the determinant of the linear part is 1. So maybe not all affine transformations, but some of them.

Wait, the question is asking if there's a transformation that can map any triangle to another triangle with the same area. So perhaps an affine transformation? Because affine transformations can take any triangle to any other triangle. But do they preserve area? Hmm. Let me think. An affine transformation is a combination of linear transformation (like scaling, rotation, shearing) followed by a translation. The area scaling factor of an affine transformation is the absolute value of the determinant of the linear part. So if I can adjust the linear part so that the determinant is 1, then the area would be preserved. But wait, if we have two triangles with the same area, can we find an affine transformation that maps one to the other and has determinant 1?

Alternatively, maybe the answer is yes, because given two triangles with the same area, there exists an affine transformation that maps one to the other. Let me verify. Suppose I have triangle ABC and triangle DEF, both with area S. Since affine transformations preserve ratios of areas, but not the absolute areas. Wait, but if the determinant is adjusted so that the area scaling factor is 1, then it would preserve area. So maybe if I compose a scaling transformation that adjusts the area appropriately. Wait, but if the two triangles have the same area, then maybe the linear transformation that maps one to the other can have determinant 1. Hmm.

Wait, let's think of triangles as affine equivalent. In affine geometry, any triangle can be mapped to any other triangle via an affine transformation. But affine transformations don't necessarily preserve area. However, if we can compose an affine transformation that maps triangle A to triangle B, and then scale appropriately to adjust the area. Wait, but the problem states that the transformation should map every point to exactly one point (bijection?) and that any triangle can be transformed into another triangle with the same area. So maybe if we can adjust the transformation so that the determinant is scaled to correct the area. But since we want to map any triangle to another with the same area, maybe such a transformation exists. 

Alternatively, the question mentions \"can any triangle be transformed into a regular triangle with the same area?\" So maybe instead of mapping between any two triangles, just mapping to a regular triangle. Let's consider that. Let's say we have a triangle with area S. We want to find a transformation that turns it into a regular triangle with area S. How?

First, take any triangle. We can apply an affine transformation to make it regular. But affine transformations change the area by the determinant. So if the original triangle has area S, then after the affine transformation, the area would be S * |det A|, where A is the linear part. To get a regular triangle with area S, we need |det A| = 1. So can we find an affine transformation that maps the original triangle to a regular triangle and has determinant 1? Maybe, but I'm not sure.

Alternatively, perhaps using a combination of transformations. Let's think step by step. Let's say we have triangle ABC. We can translate it so that one vertex is at the origin. Then, we can apply a linear transformation that maps the triangle to a regular triangle. Suppose the original triangle has vertices at (0,0), (a,b), (c,d). The area is (ad - bc)/2. We want to map this to a regular triangle with the same area. A regular triangle with area S has side length (4S)/(√3))^(1/2), but maybe coordinates can be set as, say, (0,0), (s, 0), (s/2, (s√3)/2), where the area is (√3/4)s² = S, so s = √(4S/√3). But maybe using coordinates would complicate.

Alternatively, we can use an affine transformation. Let's suppose the original triangle has vertices A, B, C. Let's translate vertex A to the origin. Then we can apply a linear transformation that maps vectors AB and AC to vectors (1,0) and (0,1). The area of the original triangle would be half the determinant of the matrix formed by AB and AC. If we apply a linear transformation that maps AB and AC to standard basis vectors, then the area of the transformed triangle would be 1/2. But the original area is S, so the determinant of the transformation matrix is 1/(2S). But then if we want to map it to a regular triangle with area S, maybe we need another scaling.

Wait, maybe here's the idea: first, apply an affine transformation that maps the original triangle to a standard regular triangle, but scaled such that the area is adjusted. Then, if necessary, apply a scaling to correct the area. But scaling would change the area, so if we adjust the scaling factor so that the area becomes S, then the composition of these transformations would have the desired effect. However, the problem is whether such a transformation can be done without knowing the original area, or is it possible for any triangle. Hmm.

Alternatively, maybe using a shear transformation. Shearing preserves area, right? Because the determinant of a shear matrix is 1. So perhaps we can use shears and rotations to transform a triangle into a regular one. But can shears and rotations (which are area-preserving) turn any triangle into a regular triangle? I don't think so. For example, a very skinny triangle can't be made into a regular triangle just by shearing and rotating, since shearing doesn't change the base or height, only slides one direction. Wait, actually, shearing does change the shape but preserves area. For instance, a shear can turn a rectangle into a parallelogram with the same base and height. But maybe combining shears and rotations can make a triangle regular?

Wait, consider a triangle with base b and height h. Its area is (1/2)bh. A regular triangle with area (1/2)bh would have sides of length s where (√3/4)s² = (1/2)bh, so s² = (2/√3)bh. So if we can transform the original triangle into a regular triangle with side s, but how? Maybe using a combination of scaling (to adjust the base and height) and shearing to adjust the angles. But scaling would change the area. But if the problem allows for scaling as long as the area is preserved... Wait, but scaling changes the area unless the scaling factor is 1. So maybe if we first scale the triangle to have the correct dimensions for a regular triangle, but that scaling would change the area. However, since we want the final area to be the same as the original, that would require the scaling factor to be 1, which would do nothing. Hmm, this is confusing.

Wait, the key here is probably affine transformations. Since affine transformations can map any triangle to any other triangle. So if two triangles have the same area, then there exists an affine transformation between them with determinant preserving the area. Wait, but affine transformations can scale the area. So if we have two triangles with area S, can we find an affine transformation that maps one to the other with determinant 1 (thus preserving area)? Because the determinant would be the ratio of the areas. Wait, but if both areas are S, then the determinant would need to be 1. Wait, no. Let me think. If you have an affine transformation that maps triangle T1 to T2, then the area scaling factor is |det L|, where L is the linear part. If T1 and T2 have the same area, then we need |det L| = 1. So yes, if such a linear transformation exists. But does it?

Yes, because any two triangles are affinely equivalent. So given two triangles with the same area, there exists an affine transformation with determinant ±1 that maps one to the other. Wait, but is that true? Let me check with coordinates.

Suppose we have triangle T1 with vertices at (0,0), (1,0), (0,1). Its area is 0.5. Suppose we want to map it to another triangle T2 with area 0.5. Let's say T2 is a right triangle with vertices (0,0), (a,0), (0,b), where ab/2 = 0.5, so ab=1. We need a linear transformation that maps (1,0) to (a,0) and (0,1) to (0,b). The matrix would be [[a, 0], [0, b]]. The determinant is ab=1, as required. So in this case, such a transformation exists. Therefore, for right triangles with the same area, we can find such a transformation.

Similarly, for any triangle, we can first map it to a right triangle, then to another triangle. Wait, but affine transformations can map any triangle to any other triangle. So if two triangles have the same area, then there's an affine transformation between them with determinant 1. Therefore, the answer should be yes. Because given any two triangles with the same area, you can apply an affine transformation (linear transformation with determinant 1 and translation) to map one to the other. Therefore, such a transformation exists.

Alternatively, if the problem is asking specifically about transforming any triangle into a regular triangle with the same area, then similarly, since a regular triangle is just another triangle with the same area, we can use an affine transformation to do so. Therefore, the answer is yes.
</think>

\\boxed{Yes}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

How many natural numbers greater than 6 but less than 60 are relatively prime to 15?"}], "output": "<think>
Okay, so I need to find how many natural numbers greater than 6 and less than 60 are relatively prime to 15. Hmm, let's break this down step by step. 

First, let me recall what \"relatively prime\" means. Two numbers are relatively prime if their greatest common divisor (GCD) is 1. So, I need to count all numbers between 7 and 59 (since it's greater than 6 and less than 60) that don't share any common factors with 15 except 1. 

15 factors into prime factors as 3 × 5. So any number that is not divisible by 3 or 5 will be relatively prime to 15. That means I need to exclude numbers in that range that are multiples of 3 or 5. 

First, let's figure out the total number of numbers we're considering. The numbers start from 7 and go up to 59. To find the count: 59 - 7 + 1 = 53 numbers. Wait, is that right? Let me check. If you subtract 6 from 59, you get 53, but since we start at 7, which is 6 + 1, yeah, the count is 53. So there are 53 numbers in total.

Now, I need to subtract the numbers that are multiples of 3 or 5 in this range. But I have to be careful not to double subtract those that are multiples of both 3 and 5 (i.e., multiples of 15). So this sounds like using the inclusion-exclusion principle.

First, find the number of multiples of 3 between 7 and 59. Then find the number of multiples of 5 in the same range. Then subtract those, but add back the multiples of 15 that were subtracted twice.

Let me start with multiples of 3. The smallest multiple of 3 greater than 6 is 3×3=9, and the largest multiple of 3 less than 60 is 59 ÷ 3 = 19.666..., so 19×3=57. Now the number of multiples is from 9 to 57, stepping by 3. The formula for the number of terms is ((last - first)/step) +1. So (57 -9)/3 +1 = (48/3)+1 =16 +1=17. So 17 numbers divisible by 3.

Next, multiples of 5. The smallest multiple of 5 greater than 6 is 10, and the largest less than 60 is 55. The numbers are 10,15,...,55. Number of terms: (55 -10)/5 +1 = (45/5)+1=9+1=10. So 10 numbers divisible by 5.

Now, multiples of 15 (since LCM(3,5)=15). The smallest multiple of 15 greater than 6 is 15, and the largest less than 60 is 60 -15=45? Wait, 15×3=45, 15×4=60, but 60 is not less than 60, so the largest is 45. So from 15 to 45, stepping by 15. Let's see: 15,30,45. How many? (45 -15)/15 +1 = (30/15)+1=2+1=3. So 3 numbers divisible by 15.

Applying inclusion-exclusion: the number of numbers divisible by 3 or 5 is (17 +10) -3=24. Therefore, the numbers not divisible by 3 or 5 (i.e., relatively prime to 15) would be total numbers (53) minus 24, which is 53 -24=29.

Wait, but wait a minute. Let me double-check the count of the multiples. Maybe I made a mistake there.

Starting with multiples of 3:

First multiple of 3 ≥7 is 9 (since 6 is excluded). Last multiple of 3 <60 is 57. The sequence is 9,12,...,57. The number of terms can be calculated by ((57 -9)/3) +1 = (48/3)+1=16 +1=17. So that's correct.

Multiples of 5:

First multiple of 5 >6 is 10. Last multiple of 5 <60 is 55. So numbers: 10,15,...,55. The count is ((55 -10)/5)+1=(45/5)+1=9+1=10. That's correct.

Multiples of 15:

First multiple of 15 >6 is 15. Last multiple of 15 <60 is 45. So numbers:15,30,45. So 3 terms. Correct.

So inclusion-exclusion gives 17+10-3=24. Total numbers 53. So 53-24=29.

Wait, but let's verify with another approach. Maybe list the numbers and count, but that's tedious. Alternatively, use Euler's totient function. Wait, Euler's totient function φ(n) gives the count of numbers less than n that are coprime to n. But here, we have numbers from 7 to 59. So the problem isn't exactly φ(15), but let's see.

Alternatively, if we think of the numbers from 1 to 59, then subtract the numbers from 1 to 6. But maybe this complicates.

Alternatively, φ(15) = 8, since numbers 1,2,4,7,8,11,13,14 are coprime to 15. So every 15 numbers, there are 8 coprime numbers. So perhaps the density is 8/15. But our range is 53 numbers. However, 53 divided by 15 is 3*15=45, so 3 full cycles of 15 and then 8 more numbers. Let me see if that's a useful approach.

Wait, but 7 to 59 is 53 numbers. Let's see, 15*3=45, so 45 numbers in 3 cycles. Then 53-45=8. So perhaps there's 3 full cycles (each contributing 8 coprime numbers) and then 8 numbers. Let's check that.

But wait, the first number in the range is 7, which is 7 mod 15. So the cycle might not align. For example, the first 15 numbers from 1-15, next 16-30, etc. But our range is 7-59, which spans from 7-15 (9 numbers), 16-30 (15 numbers), 31-45 (15 numbers), 46-59 (14 numbers). Hmm, perhaps splitting the range into these intervals.

But maybe this complicates. Alternatively, using inclusion-exclusion as before, the answer is 29.

But let's verify by considering the counts.

Total numbers:53.

Numbers divisible by 3:17. Numbers divisible by 5:10. Numbers divisible by 15:3.

So total numbers not divisible by 3 or 5:53 -17 -10 +3=29. That's the inclusion-exclusion result. So seems correct.

But let's check a small example to see if inclusion-exclusion works. Suppose the problem was numbers greater than 6 and less than 16. So numbers from 7 to 15. Total numbers:9. Numbers coprime to 15: numbers not divisible by 3 or 5.

Numbers in 7-15:7,8,9,10,11,12,13,14,15.

Which of these are divisible by 3:9,12,15.

Divisible by 5:10,15.

Numbers to exclude:9,12,15,10. Wait, 15 is in both. So total excluded numbers:3 (multiples of 3) +2 (multiples of 5) -1 (multiple of 15)=4. So 9 total numbers -4=5 coprime numbers. Let's list them:7,8,11,13,14. Yes, those are 5 numbers, which is 9 -4=5. So 5 numbers. φ(15)=8, but since we started at 7, which is after 1-6, which includes 1,2,4,7 (but wait, φ(15) is 8 numbers from 1-14. Wait, 1-14: φ(15)=8. But in our small example, 7-15 is 9 numbers, but 15 is excluded. Wait, no, original small example was up to 15, which is included. Wait, in my hypothetical example: numbers greater than 6 and less than 16: so 7 to 15 inclusive. 15 is included? Wait, if the upper limit is less than 16, then 15 is included. Wait, the original problem is less than 60, so 59. So in the small example, numbers greater than 6 and less than 16:7-15. So 9 numbers. And among them, 15 is included and is divisible by both 3 and 5. So the count for that is 5 numbers coprime to 15. That works with inclusion-exclusion. So that method works. So our original method of inclusion-exclusion should be correct.

Therefore, for the original problem, 29 numbers. So the answer is 29. Let me just check once more.

Total numbers: from 7 to 59 inclusive. 59-7+1=53.

Multiples of 3: starting at 9 (3×3) up to 57 (3×19). Number of terms: (57-9)/3 +1=48/3 +1=16 +1=17. Okay.

Multiples of 5: starting at 10 (5×2) up to 55 (5×11). (55-10)/5 +1=45/5 +1=9+1=10. Okay.

Multiples of 15:15,30,45. That's three numbers. So inclusion-exclusion:17+10-3=24.

So 53-24=29. Seems correct.

I think that's the answer. So the number of natural numbers greater than 6 and less than 60 that are relatively prime to 15 is 29.
</think>

The natural numbers greater than 6 and less than 60 are from 7 to 59, inclusive. There are \\(59 - 7 + 1 = 53\\) numbers in total. We need to count numbers that are relatively prime to 15, which means they are not divisible by 3 or 5.

1. **Multiples of 3**: The smallest multiple of 3 greater than 6 is 9, and the largest less than 60 is 57. The sequence is \\(9, 12, \\ldots, 57\\), which has \\(\\frac{57 - 9}{3} + 1 = 17\\) terms.
2. **Multiples of 5**: The smallest multiple of 5 greater than 6 is 10, and the largest less than 60 is 55. The sequence is \\(10, 15, \\ldots, 55\\), which has \\(\\frac{55 - 10}{5} + 1 = 10\\) terms.
3. **Multiples of 15 (LCM of 3 and 5)**: The smallest multiple of 15 greater than 6 is 15, and the largest less than 60 is 45. The sequence is \\(15, 30, 45\\), which has \\(\\frac{45 - 15}{15} + 1 = 3\\) terms.

Using the inclusion-exclusion principle, the number of numbers divisible by 3 or 5 is \\(17 + 10 - 3 = 24\\). Subtracting these from the total gives \\(53 - 24 = 29\\).

\\boxed{29}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

Resolve the expression $\\frac{x^5 + x^4 - x^3 + 2x^2 + x}{(x-1)^2 (x^2 + 1)^2}$."}], "output": "<think>
Okay, so I need to solve this expression: (x^5 + x^4 - x^3 + 2x^2 + x) divided by (x-1)^2 (x^2 + 1)^2. The problem says to resolve it, which I think means to perform partial fraction decomposition. Alright, let's start by recalling how partial fractions work. 

First, the denominator is already factored into (x-1)^2 and (x^2 +1)^2. Since both of these are repeated irreducible factors, the partial fraction decomposition will have terms for each power up to the exponent. 

So for (x-1)^2, we'll have constants over (x-1) and (x-1)^2. Similarly, for (x^2 +1)^2, we'll have linear terms over (x^2 +1) and (x^2 +1)^2. So the general form should be something like:

(A)/(x-1) + (B)/(x-1)^2 + (Cx + D)/(x^2 +1) + (Ex + F)/(x^2 +1)^2

Now, the numerator of the original fraction is x^5 + x^4 - x^3 + 2x^2 + x. So when we express the original expression as the sum of these partial fractions, we can multiply both sides by the denominator (x-1)^2 (x^2 +1)^2 and equate coefficients.

Let me write that equation out:

x^5 + x^4 - x^3 + 2x^2 + x = A(x-1)(x^2 +1)^2 + B(x^2 +1)^2 + (Cx + D)(x-1)^2 (x^2 +1) + (Ex + F)(x-1)^2

Hmm, that seems a bit complicated, but let's expand each term step by step.

First, let's handle term by term.

Starting with A(x-1)(x^2 +1)^2:

First, expand (x^2 +1)^2: that's x^4 + 2x^2 +1. Then multiply by (x-1):

(x-1)(x^4 + 2x^2 +1) = x(x^4 + 2x^2 +1) -1(x^4 + 2x^2 +1)
= x^5 + 2x^3 + x - x^4 - 2x^2 -1
= x^5 - x^4 + 2x^3 -2x^2 + x -1

Multiply by A: A(x^5 -x^4 +2x^3 -2x^2 +x -1)

Next, term B(x^2 +1)^2:

(x^2 +1)^2 is x^4 +2x^2 +1, so B(x^4 +2x^2 +1) = Bx^4 + 2Bx^2 + B

Then, term (Cx + D)(x-1)^2 (x^2 +1):

First expand (x-1)^2: x^2 -2x +1

Multiply by (x^2 +1): (x^2 -2x +1)(x^2 +1) 

Multiply term by term:

x^2*(x^2 +1) = x^4 +x^2

-2x*(x^2 +1) = -2x^3 -2x

1*(x^2 +1) = x^2 +1

Combine those: x^4 +x^2 -2x^3 -2x +x^2 +1 = x^4 -2x^3 +2x^2 -2x +1

Now multiply by (Cx + D):

(Cx + D)(x^4 -2x^3 +2x^2 -2x +1) 

Let's distribute:

Cx*(x^4) = Cx^5

Cx*(-2x^3) = -2Cx^4

Cx*(2x^2) = 2Cx^3

Cx*(-2x) = -2Cx^2

Cx*(1) = Cx

D*(x^4) = Dx^4

D*(-2x^3) = -2Dx^3

D*(2x^2) = 2Dx^2

D*(-2x) = -2Dx

D*(1) = D

So combining all terms:

Cx^5 -2Cx^4 +2Cx^3 -2Cx^2 +Cx +Dx^4 -2Dx^3 +2Dx^2 -2Dx +D

Group like terms:

x^5: C

x^4: (-2C + D)

x^3: (2C - 2D)

x^2: (-2C + 2D)

x: (C - 2D)

constant: D

So this term becomes:

Cx^5 + (-2C + D)x^4 + (2C - 2D)x^3 + (-2C + 2D)x^2 + (C - 2D)x + D

Next, the term (Ex + F)(x-1)^2:

First, expand (x-1)^2 = x^2 - 2x +1

Multiply by (Ex + F):

Ex*(x^2 -2x +1) = Ex^3 -2Ex^2 +Ex

F*(x^2 -2x +1) = Fx^2 -2Fx +F

Combine:

Ex^3 -2Ex^2 +Ex + Fx^2 -2Fx +F

Group terms:

x^3: E

x^2: (-2E + F)

x: (E -2F)

constant: F

So this term is:

Ex^3 + (-2E + F)x^2 + (E -2F)x + F

Now, adding all the terms together (A's term, B's term, Cx+D term, Ex+F term):

Left side: x^5 +x^4 -x^3 +2x^2 +x

Right side:

From A: A x^5 -A x^4 +2A x^3 -2A x^2 +A x -A

From B: B x^4 +2B x^2 +B

From Cx+D: C x^5 + (-2C + D)x^4 + (2C - 2D)x^3 + (-2C + 2D)x^2 + (C - 2D)x + D

From Ex+F: E x^3 + (-2E + F)x^2 + (E -2F)x + F

Now, combine all these terms on the right side:

x^5 terms: A + C

x^4 terms: (-A + B -2C + D)

x^3 terms: (2A + 2C - 2D + E)

x^2 terms: (-2A + 2B -2C + 2D -2E + F)

x terms: (A + C -2D + E -2F)

constant terms: (-A + B + D + F)

Now, equate coefficients with the left side. The left side is:

x^5 +x^4 -x^3 +2x^2 +x +0

So setting up equations:

For x^5: A + C = 1

For x^4: (-A + B -2C + D) = 1

For x^3: (2A + 2C -2D + E) = -1

For x^2: (-2A + 2B -2C + 2D -2E + F) = 2

For x: (A + C -2D + E -2F) = 1

For constants: (-A + B + D + F) = 0

So we have a system of 6 equations:

1. A + C = 1

2. -A + B -2C + D = 1

3. 2A + 2C -2D + E = -1

4. -2A + 2B -2C + 2D -2E + F = 2

5. A + C -2D + E -2F = 1

6. -A + B + D + F = 0

Let's solve these step by step.

From equation 1: A + C =1, so C =1 - A.

Let's substitute C in other equations.

Equation 2: -A + B -2(1 - A) + D =1

Simplify: -A + B -2 + 2A + D =1

Which is (A) + B + D = 3 --> equation 2a: A + B + D =3

Equation 3: 2A + 2(1 - A) -2D + E = -1

Simplify: 2A +2 -2A -2D + E = -1 --> 0A +2 -2D + E = -1

Thus: -2D + E = -3 --> equation 3a: E = 2D -3

Equation 4: -2A + 2B -2(1 - A) +2D -2E + F =2

Simplify: -2A +2B -2 +2A +2D -2E + F =2

Simplify terms: (-2A +2A)=0, so 2B -2 +2D -2E + F =2

Which becomes 2B +2D -2E + F =4 --> equation 4a: 2B +2D -2E + F =4

Equation 5: A + (1 - A) -2D + E -2F =1

Simplify: 1 -2D + E -2F =1

Thus: -2D + E -2F =0 --> equation 5a: -2D + E -2F =0

Equation 6: -A + B + D + F =0 --> equation 6 remains as is.

Now, let's note from equation 3a: E =2D -3

Let's substitute E into equations 4a and 5a.

Equation 4a: 2B +2D -2(2D -3) + F =4

Simplify: 2B +2D -4D +6 + F =4 --> 2B -2D + F = -2 --> equation 4b: 2B -2D + F = -2

Equation 5a: -2D + (2D -3) -2F =0

Simplify: -2D +2D -3 -2F =0 --> -3 -2F=0 --> -2F=3 --> F= -3/2

So F= -3/2.

Now, substitute F into equation 6: -A + B + D + (-3/2)=0

So, -A + B + D = 3/2 --> equation 6a: -A + B + D = 3/2

From equation 2a: A + B + D =3

Now, subtract equation 6a from equation 2a:

(A + B + D) - (-A + B + D) =3 - 3/2

Left side: A + B + D + A - B - D = 2A

Right side: 3/2

Thus, 2A = 3/2 --> A= 3/4

Since A=3/4, then from equation1: C=1 - A=1 - 3/4=1/4

So C=1/4

Now, equation 3a: E=2D -3

We need to find D. Let's use equation 2a: A + B + D =3, with A=3/4.

So 3/4 + B + D =3 --> B + D= 3 - 3/4=9/4 --> equation 2b: B + D=9/4

From equation 6a: -A + B + D=3/2, with A=3/4:

-3/4 + B + D=3/2 --> B + D=3/2 +3/4= (6/4 +3/4)=9/4, which is consistent with equation 2b.

So, so far, we have:

A=3/4, C=1/4, F=-3/2

Now, equation 4b:2B -2D + F = -2

F=-3/2, so:

2B -2D -3/2 = -2

Multiply both sides by 2 to eliminate fractions:

4B -4D -3= -4

Thus,4B -4D= -1 --> divide by 4: B -D= -1/4 --> equation 4c: B= D -1/4

From equation 2b: B + D=9/4

Substitute B= D -1/4 into equation 2b:

(D -1/4) + D =9/4

2D -1/4=9/4

2D=10/4=5/2

D=5/4

Then, B=5/4 -1/4=4/4=1

So B=1, D=5/4

Now, E=2D -3=2*(5/4) -3=10/4 -12/4= -2/4= -1/2

So E= -1/2

Now, check equation5a: already used in substitution, but let's verify:

Equation5a: -2D + E -2F=0

-2*(5/4) + (-1/2) -2*(-3/2)= (-10/4 -1/2 +3)= (-5/2 -1/2 +3)= (-3 +3)=0. Correct.

Now, check equation3: already used.

Now, check equation4: Let me check equation4a: 2B +2D -2E + F=4?

But 2B=2*1=2, 2D=2*(5/4)=5/2, -2E= -2*(-1/2)=1, F= -3/2

So sum:2 +5/2 +1 -3/2=2 + (5/2 -3/2) +1=2 +1 +1=4. Correct.

Now, all equations are satisfied.

So now, we have all constants:

A=3/4

B=1

C=1/4

D=5/4

E= -1/2

F= -3/2

So the partial fractions decomposition is:

(A)/(x-1) + (B)/(x-1)^2 + (Cx + D)/(x^2 +1) + (Ex + F)/(x^2 +1)^2

Which is:

(3/4)/(x-1) + 1/(x-1)^2 + ( (1/4)x +5/4 )/(x^2 +1) + ( (-1/2)x -3/2 )/(x^2 +1)^2

We can write this with denominators as:

3/(4(x-1)) + 1/(x-1)^2 + (x +5)/(4(x^2 +1)) + (-x -3)/(2(x^2 +1)^2)

Alternatively, factoring out the constants:

= \\frac{3}{4(x-1)} + \\frac{1}{(x-1)^2} + \\frac{x + 5}{4(x^2 +1)} + \\frac{ -x -3 }{2(x^2 +1)^2}

To check, maybe we can substitute a value for x and see if both sides match.

Let's pick x=0. Compute original expression and the decomposition at x=0.

Original expression numerator at x=0:0 +0 -0 +0 +0=0. Denominator: (-1)^2*(0+1)^2=1*1=1. So expression is 0/1=0.

Decomposition at x=0:

3/(4*(-1)) + 1/((-1)^2) + (0 +5)/(4*(1)) + (0 -3)/(2*(1)^2)

= -3/4 +1 +5/4 -3/2

Convert to quarters:

-3/4 +4/4 +5/4 -6/4= ( -3 +4 +5 -6 )/4=0/4=0. Correct.

Another test, x=1. But x=1 makes denominator zero, so skip. Let's try x=2.

Original expression:

Numerator: 32 +16 -8 +8 +2= 50. Denominator: (1)^2*(4 +1)^2=1*25=25. So 50/25=2.

Decomposition at x=2:

3/(4*(1)) +1/(1)^2 + (2 +5)/(4*(5)) + (-2 -3)/(2*(25))

= 3/4 +1 +7/20 + (-5)/50

Convert to 100 denominators:

3/4=75/100, 1=100/100, 7/20=35/100, -5/50= -10/100.

Sum:75 +100 +35 -10=200. 200/100=2. Correct.

Another check x= -1.

Original expression numerator: -1 +1 +1 +2 +(-1)=2. Denominator: (-2)^2*(1 +1)^2=4*4=16. So 2/16=1/8.

Decomposition at x=-1:

3/(4*(-2)) +1/((-2)^2) + (-1 +5)/(4*(2)) + (1 -3)/(2*(2)^2)

= 3/(-8) +1/4 +4/8 + (-2)/8

= -3/8 +2/8 +4/8 -2/8= ( -3 +2 +4 -2 )/8=1/8. Correct.

All checks passed. So the partial fractions decomposition is correct.
</think>

The partial fraction decomposition of the given expression is:

\\boxed{\\frac{3}{4(x-1)} + \\frac{1}{(x-1)^2} + \\frac{x + 5}{4(x^2 + 1)} - \\frac{x + 3}{2(x^2 + 1)^2}}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

What is the total volume in cubic feet of three boxes if each box is a cube with edge length 4 feet?"}], "output": "<think>
Okay, so I need to find the total volume of three boxes, and each box is a cube with an edge length of 4 feet. Hmm, let's start by recalling the formula for the volume of a cube. I think the volume of a cube is calculated by edge length cubed, right? So if each edge is 4 feet, then the volume of one box would be 4 multiplied by 4 multiplied by 4. Let me check that again.

Yes, volume of a cube is side³. So, 4 feet * 4 feet * 4 feet. Let me compute that. First, 4*4 is 16, and then 16*4 is 64. So each cube has a volume of 64 cubic feet. Now, since there are three boxes, I need to multiply the volume of one box by 3. That should give the total volume. 

So, 64 cubic feet per box times 3 boxes. Let's do 64*3. Breaking it down: 60*3 is 180, and 4*3 is 12. Adding them together, 180 +12 is 192. So, the total volume for all three boxes should be 192 cubic feet. 

Wait, let me make sure I didn't make a mistake in the multiplication. 64*3... 64+64=128, and 128+64=192. Yep, that's correct. So the individual volumes are 64 each, three of them add up to 192. I think that's it. No, wait, maybe I should check the edge length again. The problem says each box is a cube with edge length 4 feet. So the calculation is right. Edge length is 4, so 4^3 is 64. Three boxes, so 3*64=192. Yep, that seems right. I don't think I made any errors here.
</think>

The volume of one cube is calculated as edge length cubed: \\(4^3 = 64\\) cubic feet. For three such boxes, the total volume is \\(3 \	imes 64 = 192\\) cubic feet.

\\boxed{192}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}