The full dataset viewer is not available (click to read why). Only showing a preview of the rows.
Error code: DatasetGenerationError
Exception: CastError
Message: Couldn't cast
steps: int64
elapsed_seconds: double
peak_memory_gb: double
seed: int64
solution: string
problem_id: string
input_ids: list<item: int64>
child 0, item: int64
answer: string
labels: list<item: int64>
child 0, item: int64
problem_source: string
prompt: string
selection_hash: string
source_row_index: int64
id: string
length: int64
target_length: int64
prompt_length: int64
split: string
problem: string
solution_selection_hash: string
to
{'answer': Value('string'), 'id': Value('string'), 'input_ids': List(Value('int64')), 'labels': List(Value('int64')), 'length': Value('int64'), 'problem': Value('string'), 'problem_id': Value('string'), 'problem_source': Value('string'), 'prompt': Value('string'), 'prompt_length': Value('int64'), 'selection_hash': Value('string'), 'solution': Value('string'), 'solution_selection_hash': Value('string'), 'source_row_index': Value('int64'), 'split': Value('string'), 'target_length': Value('int64')}
because column names don't match
Traceback: Traceback (most recent call last):
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1827, in _prepare_split_single
for key, table in generator:
^^^^^^^^^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 613, in wrapped
for item in generator(*args, **kwargs):
~~~~~~~~~^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 343, in _generate_tables
self._cast_table(pa_table, json_field_paths=json_field_paths),
~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 132, in _cast_table
pa_table = table_cast(pa_table, self.info.features.arrow_schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2378, in table_cast
return cast_table_to_schema(table, schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2306, in cast_table_to_schema
raise CastError(
...<3 lines>...
)
datasets.table.CastError: Couldn't cast
steps: int64
elapsed_seconds: double
peak_memory_gb: double
seed: int64
solution: string
problem_id: string
input_ids: list<item: int64>
child 0, item: int64
answer: string
labels: list<item: int64>
child 0, item: int64
problem_source: string
prompt: string
selection_hash: string
source_row_index: int64
id: string
length: int64
target_length: int64
prompt_length: int64
split: string
problem: string
solution_selection_hash: string
to
{'answer': Value('string'), 'id': Value('string'), 'input_ids': List(Value('int64')), 'labels': List(Value('int64')), 'length': Value('int64'), 'problem': Value('string'), 'problem_id': Value('string'), 'problem_source': Value('string'), 'prompt': Value('string'), 'prompt_length': Value('int64'), 'selection_hash': Value('string'), 'solution': Value('string'), 'solution_selection_hash': Value('string'), 'source_row_index': Value('int64'), 'split': Value('string'), 'target_length': Value('int64')}
because column names don't match
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1369, in compute_config_parquet_and_info_response
parquet_operations, partial, estimated_dataset_info = stream_convert_to_parquet(
~~~~~~~~~~~~~~~~~~~~~~~~~^
builder, max_dataset_size_bytes=max_dataset_size_bytes
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
)
^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 948, in stream_convert_to_parquet
builder._prepare_split(split_generator=splits_generators[split], file_format="parquet")
~~~~~~~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1694, in _prepare_split
for job_id, done, content in self._prepare_split_single(
~~~~~~~~~~~~~~~~~~~~~~~~~~^
gen_kwargs=gen_kwargs, job_id=job_id, **_prepare_split_args
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
):
^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1880, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
datasets.exceptions.DatasetGenerationError: An error occurred while generating the datasetNeed help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
answer string | id string | input_ids list | labels list | length int64 | problem string | problem_id string | problem_source string | prompt string | prompt_length int64 | selection_hash string | solution string | solution_selection_hash string | source_row_index int64 | split string | target_length int64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0 | e6db3d64fc795aa252f2703c70cb5b6fb9f24dc8e8736ac2e96da2c1ec53f25c | [
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Find the sum of the remainders when the polynomial $P(x)$ is divided by $x^2+x+1$ and by $x^2-x+1$. | 910b13f86810e0dc4f1c25c94c30fe511033651fb8944b322a5f6d022d872362 | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: A polynomial $P(x)$ is divisible by $x-1$ and the remainder is $R$ when $P(x)$ is divided by $(x-1)^2$.
Find the sum of the remainders when the polynomial $P(x)$ is divided by $x^2+x+1$ and by $x^2-x+1$.
Solution:
| 98 | 001e76ebf9ff7dc84aba53b98429b18d2177656d106f099f062caf4f43b199bb | Since $P(x)$ is divisible by $x-1$, we can write $P(x) = (x-1)Q(x)$ for some polynomial $Q(x)$.
Given that the remainder is $R$ when $P(x)$ is divided by $(x-1)^2$, we can write:
\[ P(x) = (x-1)^2S(x) + R \]
for some polynomial $S(x)$.
Notice that $P(1) = 0$ since $P(x)$ is divisible by $x-1$.
Also, the derivative ... | 707314cda32febe22b5244281f49bc0c4268c66b33c306ff83ee96d90d0c9ee2 | 576,508 | train | 595 |
100\sqrt{3} | 737d0b775f3428d49a89d2fea1507a7d6ec293a1c81ca7596cea5edff1262711 | [
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Problem: A regular triangle and a square share a common side. The perimeter of the triangle is 60 cm and that of the square is 52 cm. What is the area of the triangle?
Solution:
| 63 | 001e9738fd79b6fb5b8d4b7d1413b99095cb282c0b6005682eece9923ee819e9 | Let's call the side length of the square $s$. Then the perimeter of the square is $4s$, and we know this is equal to 52 cm:
\[ 4s = 52 \Rightarrow s = 52 : 4 \Rightarrow s = 13 \text{ cm} \]
Since the triangle and square share a common side, the length of this shared side is also $s = 13$ cm.
Let's call the side len... | 6cde6b77a13fdf562e53f76482acbcccea6efb12cf17423132ad037087810cbf | 578,552 | train | 363 |
\frac{15}{4} | 3cd10fd26442dc280bb44b50baed03511fb4481e3bc2a283d7852bb1a09c7d2b | [
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Problem: Find the radius of a sphere inscribed in a pyramid with a square base of side length 15 cm, if the height of the pyramid is 10 cm. Express your answer in terms of $\pi$.
Solution:
| 66 | 001eedadb4de588b9b75bb2561f596dc93f9e4c30219055eb23741ff23fe5091 | To solve this problem, we can use the fact that the radius of the inscribed sphere is equal to the ratio of the volume of the pyramid to the surface area of the pyramid.
The volume of the pyramid is given by the formula:
\[ V = \frac{1}{3} Bh \]
where $B$ is the area of the base, and $h$ is the height of the pyramid.
... | c950aa313b80f72f1aad893ef2ad203ddbbf39dea6629af0fa3086b5d516374f | 876,098 | train | 726 |
5 | 36bdc49edee89d52603d78ad652ae36848eea11d1cfe630b87d4773529b018d8 | [
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Problem: A certain company produces a set of identical-looking cylindrical containers in three different sizes. The dimensions of the containers are such that the volume of the largest one is 4 times the volume of the smallest one, ... | 138 | 001f114ddccb29e49f2e9593fc18d64c97a7605cb9d6268009df988dacae98a2 | To find the number of different combinations of three containers that can be selected such that the total volume of the selected containers is equal to the volume of exactly 6 middle-sized containers, let's first denote the volumes of the small, middle, and large containers as $S$, $M$, and $L$ respectively.
From the ... | 72f335d02206a73a6e02943effcb042c95bf077596ecbc9de4a0fe55c9c6442f | 200,569 | train | 584 |
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Problem: What is the largest positive integer $m$ such that the solutions to the equation $x^3 - x + 1 = 0$ are not $m^{\text{th}}$ roots of any integer?
Solution:
| 67 | 001f1df94cf2fe5ca2c0942038fb21459276b22125570400d22f3e37825acc05 | Suppose there exists an integer $n$ such that $x^m = n$. Let $x_1$, $x_2$, and $x_3$ be the roots of the given cubic equation.
\[ x^3 - x + 1 = 0 \Rightarrow (x - x_1)(x - x_2)(x - x_3) = 0 \]
Using Vieta's formulas for the sum and product of the roots of a cubic equation:
\[ x_1 + x_2 + x_3 = 0 \quad (1) \]
\[ x_1x_... | cbe048fa9102fd6609651d4bee95a0bb8e64df645e8733848aa3bf321aea3d97 | 531,592 | train | 452 |
360^\circ | d665a5b8b61957608e6c2b2a79f6771a59f93cddb6dd218ad617861beaeff962 | [
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$$\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-12=0.$$
Express the solutions in the form $x=e^{i\theta}$, where $0^\circ \leq \theta < 360^\circ$. What is the sum of all values of $\theta$? | c2bbe20f56902be984ac4a28d03d45276fe713bfafe3548d6dced5be1fed972b | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: Solve the equation
$$\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-12=0.$$
Express the solutions in the form $x=e^{i\theta}$, where $0^\circ \leq \theta < 360^\circ$. What is the sum of all values of $\theta$?
S... | 102 | 001f22110e138bd957c4abd77996d5f2c76d63aa76d51adcdcbfcaa121b931e9 | Let $y = x + \frac{1}{x}$. Then we have the quadratic equation:
\[ y^2 + y - 12 = 0 \]
Factor the quadratic equation:
\[ (y - 3)(y + 4) = 0 \]
This gives us two possible values for $y$:
\[ y = 3 \quad \text{or} \quad y = -4 \]
Now, we solve the equations $x + \frac{1}{x} = 3$ and $x + \frac{1}{x} = -4$ for $x$.
1. ... | f2bf7b8176b26faaf8219da750d25f8e3e0fb852474babffeed75a58a8698277 | 432,528 | train | 450 |
(-\infty, 1) \cup (1, \infty) | 0d7c5e023a853f58c7990408de79ebf463091f340edbbb52baf6a35a265b828d | [
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Problem: Find the range of the function $\frac{x^2-4}{x^2+x-2}$, and determine the intervals on which the function is increasing or decreasing.
Solution:
| 60 | 001f3673e2dc28dd03b4a527d7e40d9407535ffd271c9ec929aa10f520c51ab2 | To find the range of the function $\frac{x^2-4}{x^2+x-2}$, we can simplify it by factoring the numerator and the denominator.
The numerator can be factored as:
\[ x^2 - 4 = (x + 2)(x - 2) \]
The denominator can be factored as:
\[ x^2 + x - 2 = (x + 2)(x - 1) \]
So, the function can be rewritten as:
\[ f(x) = \frac{(... | 156bec518504e8f5b425deba0c80f92cba1ccda3356c6004d0745839c70df987 | 108,311 | train | 685 |
20 | f50a612d98864e7cfb01afae6f61be2349df110e690f4e3473406510b4f8061c | [
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$$22x - 15x = 47.60$$
Simplifying the equation, we get:
$$7x = 47.60$$
Dividing both ... | 570b2c1314709b9fae0323184b9bf3334185b02bda915061887134be6b0082fa | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: Let's say the hourly wage is $x$ dollars per hour. Then, in the first week, Xenia earned $15x$ dollars, and in the second week, she earned $22x$ dollars. Since she earned $47.60 more in the second week, we can set up the eq... | 327 | 001f6584a77ac02a07d9af58668daa1890832745e579ec14d3080f8d0cddcb13 | To find out how many weeks it will take Tom to save enough money, we need to calculate how much he saves each week and how much more he needs to save.
Tom gets $5 per week but spends $2 per week, so he saves $5 - $2 = $3 per week.
Tom needs $180 - $120 = $60 more to buy the bike.
To find out how many weeks it will t... | a53a46b850f0a3e29171dba8f708b45f3d190f2fe530df3860fed42ea39bd541 | 793,300 | train | 152 |
17 | 846c56c97d660b0d053655b359ff5c753c7ec2b53a30dd17e6063b27ebfb6000 | [
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Problem: A square array of numbers has a first row consisting of the even integers $2,4,6,\ldots,100$ in increasing order. Each row below the first has one more entry than the row above it, and the bottom row has $50$ entries. Each ... | 118 | 001f8f3e8eb6c7cb1c76878bc0d548c0a60cf121766472fb0a0dce1bb33a79f8 | To find the number of entries that are multiples of 17, we can examine the pattern of the numbers in the array.
First row:
The numbers are 2, 4, 6,..., 100.
Second row:
The numbers are 2 + 4 = 6, 4 + 6 = 10, 6 + 8 = 14,..., 98 + 100 = 198.
Third row:
The numbers are 6 + 10 = 16, 10 + 14 = 24, 14 + 18 = 32,..., 192 +... | b3f120879f5cefd8d5507d43db0b7311833ccc5dedfe8b31fca979b4a10f4640 | 685,050 | train | 489 |
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$$\omega^n=1$$
and
$$\omega^{2n}+\omega^n+1=0$$
where $n$ is a positive integer.
Determine $n$ and $\omega$. | f47ded81535855dd27cb4abe19a134d18082ad912d347023f82ac44865ebefe5 | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: Let $\omega$ be the complex number such that
$$\omega^n=1$$
and
$$\omega^{2n}+\omega^n+1=0$$
where $n$ is a positive integer.
Determine $n$ and $\omega$.
Solution:
| 76 | 001fe20c70eb1e639f9b25492a27882dc28e08e47123ff14c074dee381174878 | We are given that $\omega^n=1$ and $\omega^{2n}+\omega^n+1=0$. Since $\omega^{2n} = (\omega^n)^2 = 1^2 = 1$, we can substitute this into the second equation:
\[ 1 + \omega^n + 1 = 0 \Rightarrow 2 + \omega^n = 0 \Rightarrow \omega^n = -2 \]
This contradicts the first equation $\omega^n = 1$ unless we consider the poss... | 94c4f60cde14876aa2a6abb6925d9eb0e9dcf9315c4db66f1516982e29742686 | 391,229 | train | 552 |
98\% | 31e6d22f62dd6d93b2a191020f3897f6fb1d67239b0875c2871b9fe581a35e74 | [
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Problem: A candy bar weighing 30 grams has its weight reduced to 70% after losing some water. Then, the candy is left outside for a day and its water content increases by 40%. What percentage of the original weight is the new weight... | 78 | 001ff16a5d65f07589aa0f1dc630e9640f9545999f9b2cfc0254447a0ef4460e | The candy bar initially weighs 30 grams. Its weight is reduced to 70% of the original weight, which is:
\[ 30 \cdot 0.7 = 21 \text{ grams} \]
Then, the candy's water content increases by 40%. This means the weight increases by 40% of the current weight:
\[ 21 \cdot 0.4 = 8.4 \text{ grams} \]
Add the weight increase t... | c8b0cd13fe7d8d78d192bce1500308780bab8fefa917f440283b1df9ec08723e | 625,634 | train | 185 |
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Problem: Determine the remainder when the polynomial $x^{12}+2x^{11}+3x^{10}+...+10x+11$ is divided by $x^4-x^2+1$.
Solution:
| 66 | 002000eafc78f75e0f27dfd469b0a6fc8db130ecbfdb6bdf6fb58dde415f4122 | We know that:
\[ x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1) \]
So, $x^6 \equiv -1 \pmod{x^4-x^2+1}$.
Now, we can express the given polynomial in terms of $x^6$:
\[ x^{12}+2x^{11}+3x^{10}+...+10x+11 = \]
\[ (x^{12} + x^6) + (2x^{11} + 2x^5) + (3x^{10} + 3x^4) + (4x^9 + 4x^3) + (5x^8 + 5x^2) + (6x^7 + 6x) + 11 \]
Using the fa... | 4f18a43034e2035f81ded54ebe9e45e164e9f3bf0bcaef4fe53c3236cbdcc907 | 380,061 | train | 409 |
r^2 \arccos \left( \sqrt{\frac{r^2 - 1}{2r^2}} \right) | cefd1f8a0ba999ff57ec12579e90fc3c40ea8f53905219c4753710fdc0ad4702 | [
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\[\cos \frac{\theta}{2} = \sqrt{\frac{r^2 - 1}{2r^2}}.\]Express the area of the circle sector with central angle $\theta$ in terms of $r.$ | 5bd7d98ac89aa6cfb8f187139ac8fbe064ee205111fd85db7888a723db2430f1 | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: Let $r$ be the radius of a circle, and let
\[\cos \frac{\theta}{2} = \sqrt{\frac{r^2 - 1}{2r^2}}.\]Express the area of the circle sector with central angle $\theta$ in terms of $r.$
Solution:
| 88 | 002011d22af84230935f6ae1f1c1c70ba89c1db09c6e122945650e6fd1837abd | To find the area of the circle sector with central angle $\theta$, we first need to express $\theta$ in terms of $r$.
We are given:
\[ \cos \frac{\theta}{2} = \sqrt{\frac{r^2 - 1}{2r^2}} \]
To find $\theta$, we can take the inverse cosine of both sides:
\[ \frac{\theta}{2} = \arccos \left( \sqrt{\frac{r^2 - 1}{2r^2}}... | 2cc372d88194392ab46e75c9d1308ce77887f82fe04633b2d5a7f8c39ea384f4 | 709,819 | train | 344 |
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Problem: A sequence is defined as follows: $b_1 = 2$, $b_2 = 3$, and for all positive integers $n$, $b_{n+2} = b_{n+1}b_n + 1$. Given that $b_{15} = 2752556$ and $b_{16} = 1567645923$, find the remainder when $\displaystyle \sum_{k=... | 135 | 002053cd53004c0e68681f765d8416dbc93a7df2ce5456a8813b36b2ed117845 | The problem provides the recursive sequence $b_{n+2} = b_{n+1}b_n + 1$ and the initial terms $b_1 = 2$, $b_2 = 3$. We are tasked with finding the remainder when $\displaystyle \sum_{k=1}^{15} \frac{1}{b_k}$ is divided by $\frac{1}{100}$.
Given the recursive definition, we can derive a relation between consecutive term... | 407cb0fc8941c1d49c9b43f940b66d26f254e27ed908ceaae2b3ce026b1fd498 | 520,873 | train | 704 |
1225 | 56a50af30c644399629cf064906ba29f616a034fe801f937f77a61189c866880 | [
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Problem: Find the smallest positive integer $n$ for which $\lfloor \sqrt{n} \rfloor$ is divisible by 5 and $\lceil \sqrt[3]{n} \rceil$ is divisible by 7.
Solution:
| 74 | 00208dfb688b25af0b6f4a84784a1e5d8ac360161a70ea61621d4f3ffcf9f9e6 | ## Step 1: Understand the conditions for n
We need to find the smallest positive integer $n$ such that $\lfloor \sqrt{n} \rfloor$ is divisible by 5 and $\lceil \sqrt[3]{n} \rceil$ is divisible by 7.
## Step 2: Analyze the first condition
For $\lfloor \sqrt{n} \rfloor$ to be divisible by 5, $\sqrt{n}$ must be greater t... | 11699148b55111799b7e3c241549fbb19cb2c0c459aa5a690d4f6950078d08fa | 224,653 | train | 697 |
270 | ad6e3a4ce3aa10216d49cdf9dde99bd2f4c1f1d4dfead45b061ab9834eb011cf | [
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Problem: A computer science club has 5 members who know C++, 4 members who know Java, and 3 members who know Python. The club is planning to send a team of 4 members to a coding competition, where at least one member should know eac... | 90 | 0020917254f375fc7e4437dbd78675779c78fc6017e80d7dcde443c6c13dfb12 | To solve this problem, we need to count the number of ways to select a team of 4 members that includes at least one member who knows C++, at least one member who knows Java, and at least one member who knows Python.
First, we find the total number of ways to choose 4 members from the club without any restrictions:
\[ ... | 0ce688ac9ff7f0e5da4db5d0049327c48341a3257b0ef6970969385a36d72d5a | 749,087 | train | 527 |
10 | 22e66e0c28e467c058f5de5e651658770564cdd248c57ef1029929797719b978 | [
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Problem: Find the largest number of positive, consecutive perfect squares whose sum is less than or equal to 400.
Solution:
| 46 | 0020a5d3d6a08d6db4ad0d4b381ca0735050444dfdcca1f90a209d04c14c54f5 | We can write a sum of consecutive perfect squares starting from $1^2$ as follows:
\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 \]
This sum can be expressed using the formula:
\[ \frac{n(n + 1)(2n + 1)}{6} \]
We want to find the largest value of $n$ such that the sum is less than or equal to 400. We can solve this inequality:
\[... | 3ce0f3e02be3c48c0dd5c9d02d18d65fe892cc88cbacc381ba11b2668bd1c34c | 991,213 | train | 850 |
17 | ab4b32011c9db21d3ce5b1a7cb09f618b2d76e68fc378619b8d124a5f4805d7b | [
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Problem: Let $z$ be a complex number such that $z = a + bi,$ where $a$ and $b$ are positive integers. If the imaginary part of $z^2$ is 8 times the real part, find $z \times \overline{z}.$
Solution:
| 83 | 0020a6df4551e862e6c1c2c30ef8b1a8967630be98d488928b16115502a674f5 | Given that $z = a + bi$, we can square $z$ to find $z^2$.
First, let's square $z$:
\[ z^2 = (a + bi)^2 = a^2 + 2abi + b^2 i^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + 2abi \]
The real part of $z^2$ is $(a^2 - b^2)$ and the imaginary part is $2ab$.
We are given that the imaginary part of $z^2$ is 8 times the real part, so:... | 010934c19fcdbfae0d9b8114702eac379d36f92e9532ce3427de20f5320881fd | 441,074 | train | 799 |
x^3 - x^2 - 2x + 2 | b660c6e2d9fb26fc8a465ce797f7a1d82824974f873e9514fc04bb60dcc7c408 | [
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[asy]
size(8cm);
import graph;
Label f;
f.p=fontsize(6);
real f(real x) {return (x-1)/((x-1)*(x+1)*(... | 28df9e46fc64ab53a503602a1ba69e317f59a34325e0af92b1b16af725bd3b6e | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: The graph of the rational function $\frac{p(x)}{q(x)}$ is shown below, with a horizontal asymptote of $y = 1$ and a vertical asymptote of $x = 2$. If $q(x)$ is cubic, $p(0) = 0$, and $q(0) = 2$, find $p(x) + q(x)$.
[asy]
s... | 465 | 0020b4a8cbda571d6ad265a24f47d6540642201bacd83dc7995ab11d2a8503ad | Since the graph has a horizontal asymptote at $y = 1$, the degrees of the polynomials $p(x)$ and $q(x)$ must be the same.
Given that $q(x)$ is cubic, we know $p(x)$ is also cubic.
The presence of a hole in the graph at $x = 1$ indicates a factor of $(x-1)$ in both $p(x)$ and $q(x)$.
The vertical asymptote at $x = 2$... | b9770ee78acebcdf5f91bfe7c53eb4bf2dbaebfb7c54a51a48571b2d0e3fdac1 | 134,160 | train | 460 |
0 | 0d4328cd9fbd7dd237cda0ea8f0415637e36cffef99c7f3e7cdbfb577cc917a6 | [
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Problem: Let $f(x)$ be a rational function such that its graph has a horizontal asymptote $y = 3$ and two vertical asymptotes $x = -1$ and $x = 2.$ Also, the graph of $y = f(x)$ intersects the line $y = x$ at exactly two points. Fin... | 99 | 0020bc4684acd35dd6879232e1f38ca2b01c93d7c8f00ae3bd6b646d970031cb | Given the conditions, $f(x)$ can be written in the form:
\[ f(x) = \frac{3(x - r_1)(x - r_2)}{(x + 1)(x - 2)} \]
where $r_1$ and $r_2$ are the roots of $f(x) = x$.
Since the graph of $y = f(x)$ intersects the line $y = x$ at exactly two points, $f(x) = x$ has exactly two roots. This means that $f(x) - x$ has exactly ... | 06d584cd7fea26b9abbe9c9ebb83497dbdbd5b68e0d12298c8bb358a14d23f18 | 162,412 | train | 963 |
15 | d9174bb5dd30eb3a03102fded83b52489d79dc1b3b8525e4f562e808c74458a4 | [
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-100... | 673 | The sides of a triangle are three consecutive integers, and its area is two-thirds the perimeter. Find the sum of all possible values of the perimeter. | e01c597504acdefa756d5043911148cea54c5c229ce92e92a6c5631db3a529cb | augmented_math | Solve the following problem. Show your reasoning and put the final answer in \boxed{}.
Problem: The sides of a triangle are three consecutive integers, and its area is two-thirds the perimeter. Find the sum of all possible values of the perimeter.
Solution:
| 55 | 0020f08bc722fd968a80ff09f98715a2abf085b2268c73c40dda236b476a5952 | Let the sides of the triangle be $x$, $x+1$, and $x+2$. The perimeter is $P = 3x + 3$.
The semi-perimeter is $s = \frac{P}{2} = \frac{3x + 3}{2}$.
Using Heron's formula, the area of the triangle is:
\[ A = \sqrt{s(s - x)(s - (x + 1))(s - (x + 2))} \]
We know that the area is two-thirds the perimeter:
\[ A = \frac{2}... | 764bbb82dbabced0d65e344dcd56e3ddfab68f64674bece24e5a7be54027a5a3 | 979,738 | train | 618 |
TailSFT with LFM2.5-350M and Trackio
Status, 2026-09-16: GPU preflight completed all three training arms and four evaluations. The strict final-answer grader and namespace/resume checks pass (50 tests). A full-length batch of 256 samples completed on the A100; the full comparison has been submitted. Full experiment results are not yet available.
Small, hackable reproduction of the TailSFT filtering method. The objective is higher pass@16 after supervised training. This experiment does not test downstream RL or claim to match the paper's reported scores.
- Model: LiquidAI/LFM2.5-350M-Base, revision
9960764e30892e01f29a6dc23df2533fcd8bd5ae. - Training: 8,192 distinct math problems from OpenMathInstruct-2; 256 held-out validation problems; two epochs.
- Arms: ordinary SFT, TailSFT dropping the most-improved 25%, random dropping 25%. Same initialization, data order, learning rate, steps, and training seed 42.
- Evaluation: base and all three trained models, MATH-500, 16 samples/problem, batch 256, 3,072-token generation cap, temperature/top-p 1, no top-k. Math-Verify grading of the last explicit boxed answer in the completion; absent, unfinished, or unparseable answers count as incorrect.
- Optimizer: FP32 parameters and AdamW state, BF16 autocast forward; batch 8, no accumulation/packing; LR 2e-5, weight decay 0.1, clipping 1, 3% warmup then constant.
- Full GPU job.
- Trackio dashboard.
- Source, frozen data, and run artifacts.
This is a single training-seed comparison. Later confirmation seeds are separate experiments.
The part to hack
tail_loss.py contains the objective: calculate response-token cross-entropy, subtract each example's cached initial mean loss, drop the most negative margins, then divide retained token-loss sum by retained token count. drop_fraction=0 recovers ordinary SFT. Selection occurs within each physical batch.
Change the fraction or replace the ranking score to experiment. Keep the data, optimizer, and evaluation fixed for comparisons. Training uses ordinary PyTorch and Transformers; no custom kernels or distributed setup is required.
Reproduce locally on one CUDA GPU
The Hub repository stores these files under source/ and the prepared data under data/. The exact source revision for the current run is 86591e5c7c856c9c503b94b11dc1d22295cb3962; live status is recorded in the repository-root run-state.json. First install the HF CLI, authenticate with hf auth login, and download that immutable snapshot:
hf download burtenshaw/tailsft-lfm350m-experiment --repo-type dataset \
--revision 86591e5c7c856c9c503b94b11dc1d22295cb3962 --include "source/*" "data/*" --local-dir tailsft-reproduction
cd tailsft-reproduction/source
uv venv --python 3.12 .venv
uv pip install --python .venv/bin/python -r requirements-linux.lock
source .venv/bin/activate
python -m pytest -q
# Edit configs/lfm350m.json for your own namespace before the following runs.
python train.py --config configs/lfm350m.json --data ../data --output ./outputs --mode preflight
python train.py --config configs/lfm350m.json --data ../data --output ./outputs-full --mode full
For your own run, edit trackio_space, trackio_project, artifact_repo, and experiment before starting. Authenticate with hf auth login for Trackio; add --push to persist checkpoints/results to your artifact repository. Use a new experiment ID whenever changing the model, data, or settings.
To regenerate the data:
.venv/bin/python prepare_data.py --config configs/lfm350m.json --output-dir ../data
Reproduce on HF Jobs
bootstrap_job.py embeds all 89 pinned Linux dependencies. It downloads one immutable source/data revision, runs the tests, and executes training. The namespace flags below save models, results, and Trackio logs to your account. Replace YOUR_USERNAME with your Hub username. --resume restores persisted checkpoints and raw evaluation samples from your artifact repository; keep the same experiment ID and settings when resuming.
hf jobs uv run --detach --flavor a100-large --timeout 10h --secrets HF_TOKEN \
bootstrap_job.py --source-revision 86591e5c7c856c9c503b94b11dc1d22295cb3962 --mode full \
--artifact-repo YOUR_USERNAME/tailsft-lfm350m-experiment \
--trackio-space YOUR_USERNAME/tailsft-lfm350m-trackio \
--trackio-project tailsft-lfm350m --experiment tailsft-lfm350m-my-run
The current A100 80GB hourly rate is $2.50; a ten-hour timeout bounds one full job's compute at roughly $25. A measured 256-sample, 3,072-token batch took 175.5 seconds, using 15.43 GB peak allocated GPU memory. This projects roughly 6.2 hours for evaluation alone; plan about 7–9 hours including training, with the ten-hour cutoff retained. This estimate is based on one base-model batch and is not a completion guarantee. Preflight runs are separate and are not experiment results. The original preflight used the permissive grader; its scores must not be interpreted as benchmark results.
Reproducibility and artifacts
data/manifest.json records immutable inputs, exact selection rules, row IDs, hashes, masks, EOS, and removal counts. Selection groups identical problems before splitting and removes normalized exact matches against MATH-500. This check does not exclude paraphrases or unknown pretraining exposure.
Prepared training data: mean length 529.58 tokens, maximum 1,822, and 3,538,839 target tokens. No target truncation. Data uses the explicit prompt in evaluate.build_prompt; it does not depend on a chat template.
Each run preserves configuration, exact installed dependencies, cached initial losses, per-step retained IDs/margins, full checkpoints with optimizer/RNG state, unfiltered validation loss, raw generations, graded outputs, and pass@1/2/4/8/16 by problem and difficulty. Checkpoints upload every 512 steps; evaluation uploads at roughly five-minute intervals. Trackio stores live curves and GPU/CPU metrics in a persistent HF bucket.
The GPU throughput check selects an evaluation batch size using runtime and memory only; its accuracy is not used to tune the recipe. Training and evaluation remain valid if they show no TailSFT gain. Test results will not be used to change the recipe. The original research recommendation is preserved in research-plan.md; this configuration supersedes its model choice.
Sources and attribution
- TailSFT: Malladi et al., arXiv 2608.25756.
- LFM weights: Liquid AI, under the LFM Open License v1.0. Fine-tuned weights are modified derivatives; checkpoint directories retain the license and a modification notice.
- Training data: NVIDIA OpenMathInstruct-2, CC-BY-4.0. The snapshot is a selected, tokenized, answer-checked derivative with source IDs retained.
- Evaluation: HuggingFaceH4/MATH-500; source provenance retained, without assigning it a new license.
- Grading: Math-Verify, pinned 0.9.0.
- Tracking: Trackio, pinned 0.38.0.
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