language:en license:mit size_categories:n<1K format:parquet modality:text library:datasets library:pandas library:mlcroissant library:polars region:us
---
language: en
license: mit
tag: curator
configs:
- config_name: default
  data_files:
  - split: train
    path: data/train-*
dataset_info:
  features:
  - name: name
    dtype: string
  - name: lean4_statement
    dtype: string
  - name: informal_statement
    dtype: string
  - name: informal_solution
    dtype: string
  - name: tags
    dtype: string
  - name: coq_statement
    dtype: string
  - name: isabelle_statement
    dtype: string
  - name: r1_reasoning
    dtype: string
  - name: r1_solution
    dtype: string
  - name: judge_correct
    dtype: bool
  - name: judge_reasoning
    dtype: string
  - name: math_verify_extracted_r1_answer
    dtype: string
  - name: math_verify_extracted_ground_truth_answer
    dtype: string
  - name: math_verify_correct
    dtype: bool
  - name: math_verify_error
    dtype: string
  splits:
  - name: train
    num_bytes: 17861397
    num_examples: 492
  download_size: 8773725
  dataset_size: 17861397
---

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 <img src="https://huggingface.co/datasets/bespokelabs/Bespoke-Stratos-17k/resolve/main/made_with_curator.png" alt="Made with Curator" width=200px>
</a>

## Dataset card for putnambench_r1

This dataset was made with [Curator](https://github.com/bespokelabsai/curator/).

## Dataset details

A sample from the dataset:

```python
{
    "name": "putnam_1981_a1",
    "lean4_statement": "abbrev putnam_1981_a1_solution : \u211d := sorry\n-- 1/8\ntheorem putnam_1981_a1\n(P : \u2115 \u2192 \u2115 \u2192 Prop := fun n k : \u2115 => 5^k \u2223 \u220f m in Finset.Icc 1 n, (m^m : \u2124))\n(E : \u2115 \u2192 \u2115)\n(hE : \u2200 n \u2208 Ici 1, P n (E n) \u2227 \u2200 k : \u2115, P n k \u2192 k \u2264 E n)\n: Tendsto (fun n : \u2115 => ((E n) : \u211d)/n^2) atTop (\ud835\udcdd putnam_1981_a1_solution) :=\nsorry",
    "informal_statement": "Let $E(n)$ be the greatest integer $k$ such that $5^k$ divides $1^1 2^2 3^3 \\cdots n^n$. Find $\\lim_{n \\rightarrow \\infty} \\frac{E(n)}{n^2}$.",
    "informal_solution": "The limit equals $\\frac{1}{8}$.",
    "tags": "['analysis', 'number_theory']",
    "coq_statement": "Section putnam_1981_a1.\nRequire Import Nat Reals Coquelicot.Coquelicot. From mathcomp Require Import div.\nDefinition putnam_1981_a1_solution := 1 / 8.\nTheorem putnam_1981_a1\n    (prod_n : (nat -> nat) -> nat -> nat := fix prod_n (m: nat -> nat) (n : nat) :=\n        match n with\n        | O => m 0%nat\n        | S n' => mul (m n') (prod_n m n')\n    end)\n    (P : nat -> nat -> Prop := fun n k => 5 ^ k %| prod_n (fun m => Nat.pow m m) (S n) = true)\n    (f : nat -> nat)\n    (hf : forall (n: nat), gt n 1 -> P n (f n) /\\ forall (k: nat), P n k -> le k (f n))\n    : Lim_seq (fun n => INR (f n) / INR n ^ 2) = putnam_1981_a1_solution.\nProof. Admitted.\nEnd putnam_1981_a1.",
    "isabelle_statement": "theory putnam_1981_a1 imports Complex_Main\nbegin\n\ndefinition putnam_1981_a1_solution::real where \"putnam_1981_a1_solution \\<equiv> undefined\"\n(* 1/8 *)\ntheorem putnam_1981_a1:\n  fixes P::\"nat\\<Rightarrow>nat\\<Rightarrow>bool\" and E::\"nat\\<Rightarrow>nat\"\n  defines \"P \\<equiv> \\<lambda>n. \\<lambda>k. 5^k dvd (\\<Prod>m=1..n. m^m)\"\n    and \"E \\<equiv> \\<lambda>n. (GREATEST k. P n k)\"\n  shows \"(\\<lambda>n. (E n) / n^2) \\<longlonglongrightarrow> putnam_1981_a1_solution\"\n  sorry\n\nend",
    "r1_reasoning": "Alright, let's see... I need to find the limit as n approaches infinity of E(n)/n2, where E(n) is the greatest integer k such that 5^k divides the product 11\u00b722\u00b733\u00b7...\u00b7nn. Hmm, okay, so the problem is about determining how many times 5 divides into this big product, right? And then we need the asymptotic behavior of this count divided by n2 as n becomes large.\n\nFirst, let me recall that for any prime p, the exponent of p in the prime factorization of a number is given by the sum of the floor divisions of that number by p, p2, p3, etc., until p^k exceeds the number. This is Legendre's formula. But in this case, we're dealing with a product of terms where each term is m^m, so the exponent of 5 in the entire product would be the sum over m from 1 to n of m times the exponent of 5 in m.\n\nWait, let me confirm that. If I have a term like m^m, the exponent of 5 in m^m would be m multiplied by the exponent of 5 in m itself. For example, if m is divisible by 5 once, then m^m would contribute m exponents of 5. So, in general, for each m, if v_5(m) is the exponent of 5 in m, then the total exponent contributed by m^m is m*v_5(m). So then, E(n) should be the sum from m=1 to n of m*v_5(m). Is that correct?\n\nYes, that makes sense. So E(n) = \u03a3m=1n m * v5(m). Then we need to evaluate the limit of E(n)/n2. The challenge then is to compute this sum and find its asymptotic growth rate.\n\nSo, let me write that sum down again: E(n) = \u03a3_{m=1}^n m * v5(m). How can I compute this? For each m, v5(m) is the number of times 5 divides into m. For example, if m is not divisible by 5, then v5(m) = 0. If m is divisible by 5 once, v5(m) = 1; if divisible by 25, then v5(m) = 2; if divisible by 125, then v5(m)=3, etc.\n\nSo, this summation can be handled by grouping numbers according to their divisibility by powers of 5. Let's think: for each integer k \u2265 1, the numbers divisible by 5^k but not by 5^{k+1} will contribute k to v5(m). So to compute the total \u03a3 m*v5(m), we can consider the layers where each layer corresponds to numbers m divisible by 5^j for each j.\n\nWait, but v5(m) is additive over the exponents. For example, if m is divisible by 5^j but not 5^{j+1}, then v5(m) = j.\n\nTherefore, we can express E(n) as \u03a3_{j=1}^\u221e \u03a3_{m=1}^n [m contributes j to E(n) if 5^j divides m but 5^{j+1} does not]. However, m contributes j to the total sum only if its exponent v5(m) is j, so each m is counted in the sum for each j from 1 up to v5(m). Wait, no. The total contribution from m is m * v5(m) = m * (k1 + k2 + ...) where k1 etc. are the exponents in each prime factor. Wait, but 5 is a prime, so v5(m)=max{k | 5^k divides m}. So actually, for m, it's contributing m*v5(m) = m*k where k is the highest power of 5 dividing m. Which is why when you consider m divisible by 5^j but not 5^{j+1}, then v5(m)=j.\n\nTherefore, the entire sum E(n) can be written as \u03a3_{j=1}^\u221e \u03a3_{m=1}^n [m \u2261 0 mod 5^j and m \u2261\u0338 0 mod 5^{j+1}}] * m *j. Alternatively, we might rephrase. Alternatively, we might instead write each m's contribution as contributing 1 to each j from 1 up to v5(m). But here the contribution is m*j. Wait, is there a way to split up the sum over m?\n\nAlternatively, since for each m, v5(m) = the number of times 5 divides into m. So, if m is divisible by 5, that adds 1, if it is divisible by 25, adds another 1, so the total v5(m) is the sum over j=1}^\u221e [if 5^j divides m, then 1 else 0].\n\nWait, yes, exactly. Which is Legendre's formula: v5(m!) = \u03a3_{j=1}^\u221e floor(m /5^j), but here we're dealing with v5(m). Wait, v5(m) is not a sum here. But for each m, v5(m) is the maximum k such that 5^k divides m. But we can express v5(m) as the sum over 1 if 5 divides m, plus 1 if 25 divides m, etc., up to the highest power. So in that linear algebra way, it's similar to the exponent in the factorial.\n\nHmm, perhaps I can use that for the generating function or summation.\n\nBut how does that help here? Wait, perhaps through double summation. For each m, v5(m) = \u03a3_{j=1}^\u221e [if 5^j divides m, then 1 else 0]. So interchange the order: \u03a3_{m=1}^n m*v5(m) = \u03a3_{m=1}^n m * \u03a3_{j=1}^{\u221e}[m divisible by 5^j] = \u03a3_{j=1}^\u221e \u03a3_{m=1}^n m*[m divisible by 5^j]\n\nThat is, exchanging the order of summation gives me \u03a3_{j=1}^\u221e \u03a3_{m=1}^n m*[m divisible by 5^j] = \u03a3_{j=1}^\u221e \u03a3_{m divisible by5^j, 1 \u2264 m \u2264n} m. But wait, v5(m) counts how many j satisfy 5^j divides m, right? Wait, no, actually, Legendre's formula for v5(m!) was counting with multiplicities. Here, v5(m) is similar but not exactly the same. Wait, rather for the factorial it's the sum over floor(m/5^j). Here, if we consider v5(m) for each m, it's equal to the max j where 5^j divides m. Therefore, v5(m) can also be written as the sum_{j=1}^{\u221e} [m divisible by 5^j]. But that would overcount. For example, if 5 divides m with exponent k, v5(m) is k, not 1. So this cannot hold. Wait, no, overcounting since each j in 1 to k adds 1, leading to summation over [divisibility] gives k.\n\nWait, let me check: Let me take m = 50, which factors as 2 * 5^2 * 1. So, v5(50) = 2. Using this formula, sum_{j=1}^\u221e [5^j divides 50] would be j=1: 5 divides 50? yes, so 1. j=2: 25 divides 50? yes, so another 1. j=3: 125 divides 50? no. So sum is 2, which is equal to v5(m). So in fact, v5(m) = sum_{j=1}^\u221e [5^j divides m]. Since the number of yeses is which is just floor(log_5 m), but written as multiplicative.\n\nTherefore, v5(m) = \u03a3_{j=1}^\u221e [5^j | m]. Therefore, E(n) = \u03a3_{m=1}^n m * \u03a3_{j=1}^\u221e [5^j | m] = \u03a3_{j=1}^\u221e \u03a3_{m=1}^n [5^j | m] * m.\n\nSo, swapping the order of summation, we can write E(n) as the sum over j of the sum over multiples of 5^j of m. So, for each j, we need to compute the sum of m from 1 to n for m divisible by 5^j, multiplied by 1 (since we just need to sum m, such sums for each j). Therefore:\n\nE(n) = \u03a3_{j=1}^\u221e \u03a3_{m=1}^n [5^j | m] * m\n\nWhich is the same as sum_{j=1}^\u221e (sum_{k=1}^{floor(n/5^j)} k*5^j )\n\nWait, m must be a multiple of 5^j: m = 5^j*k, where k ranges from 1 up to floor(n /5^j). Then:\n\nsum_{m=1}^{n} [5^j divides m] * m = sum_{k=1}^{floor(n/5^j)} (5^j * k) =5^j *sum_{k=1}^{floor(n/5^j)} k\n\nSince that is an arithmetic series, sum_{k=1}^K k = K(K+1)/2. So:\n\nsum_{k=1}^{K}5^j k =5^j K(K+1)/2, where K=floor(n/5^j)\n\nTherefore,\n\nE(n) = \u03a3_{j=1}^\u221e 5^j \\* floor(n/5^j)(floor(n/5^j)+1)/2\n\nOkay, that's an exact expression. But we need to find the asymptotic limit as n approaches infinity. So, when n is large, floor(n/5^j) \u2248 n/5^j. And floor(n/5^j) +1 \u2248 n/5^j +1\u2248n/5^j (since the +1 is negligible compared to the n term for large n).\n\nWait, but floor(n/5^j) +1 is equal to floor(n/5^j +1), right? However, given that n is large, maybe we can approximate floor(n/5^j) \u2248 n/5^j - \u03b4 where 0 \u2264 \u03b4 <1. Then floor(n/5^j)+1 \u2248 n/5^j - \u03b4 +1, not sure. Maybe it's better just to approximate as \u2248 (n/5^j)^2 /2 for each term.\n\nBut perhaps, given that as n becomes larger, the floor operations can be approximated by their real values, so:\n\nE(n) \u2248 \u03a3_{j=1}^\u221e 5^j \\* (n/5^j)(n/5^j +1)/2 \u2248 \u03a3_{j=1}^\u221e 5^j \\* (n2 /5^{2j} + n/5^j)/2 \u2248 sum_{j=1}^\\infty (n2 / (2*5^j)) + \u03a3_{j=1}^\\infty (n / 2).\n\nWait, but that second term \u03a3_{j=1}^\\infty n / 2 would diverge. Clearly something is wrong here.\n\nWait, probably we need to not expand it that way. Let me compute each term:\n\nsum_{j=1}^\\infty 5^j \\* floor(n/5^j)\\*(floor(n/5^j) +1)/2\n\nwhen floor(n/5^j) is roughly n/5^j, so approximately n/5^j * (n/5^j) for the quadratic term?\n\nSo 5^j * [ (n/5^j)2 + n/5^j ] /2 = 5^j [ n2/5^{2j} + n/5^j ] /2 = [ n2 /5^j + n ] /2.\n\nTherefore, the j-th term is approximately (n2 /5^j + n)/2.\n\nWait, summing the first term over j would give geometric series with ratio 1/5: sum_{j=1}^\\infty n2/(2 *5^j) = (n2/2) * (1/5 +1/25 +1/125 + ...) ) = (n2/2) * (1/5 / (1 -1/5)) )= (n2/2)(1/5 /4/5)= n2/2 *1/4 = n2/8.\n\nThe second term is sum_{j=1}^\\infty n/2, which diverges. But that is a problem, which suggests that the approximation floor(n/5^j)(floor(n/5^j)+1)/2 \u2248 n2 / (2*5^{2j}) is invalid.\n\nWait, let me re-examine:\n\nThe original sum for E(n) was sum_{j=1}^\\infty 5^j [floor(n/5^j)(floor(n/5^j) + 1)/2].\n\nIf I let for large n, floor(n /5^j) \u2248 n /5^j - fractional part.\n\nBut then floor(n /5^j) +1 \u2248 n/5^j - fractional part +1 \u2248 n /5^j + 1 - fractional part. But unless n is an exact multiple of 5^j, they differ. However, for large n, unless j is exponentially large, which it isn't, 5^j << n, right? The number of significant terms j is where 5^j \u2264n, so j \u2264 log_5 n, so even for the largest j, 5^j ~ n. But when j is at the upper end, floor(n /5^j) is about 1. So we can split the sum into two parts:\n\nFor j \u2264 log_5 n - \u03b5, so 5^j <<n. In such cases, floor(n /5^j ) \u2248n /5^j and floor(n/5^j)+1\u2248n/5^j +o(1). Therefore, for j << log_5 n, we can approximate, floor(n /5^j) \u2248n /5^j, and floor(n//5^j) +1 \u2248n /5^j +1\u2248n/5^j.\n\nBut floor(n/5^j)(floor(n/5^j) +1)/2 \u2248 (n /5^j)(n /5^j+1)/2\u2248 (n2 /5^{2j} +n /5^j ) /2.\n\nSo then, the term with 5^j multiplied on the outside becomes:\n\n5^j * [ (n2 /5^{2j} +n /5^j ) /2 ]= [n2/(5^{j}) +n ] /2.\n\nSum that over j.\n\nSum_{j=1}^\\infty [n2/(2*5^j) + n/2]. So same as before, first term converges to n2/(2*(5-1)) )= n2/8? Let me check: sum1/(5^j) from j=1 is 1/5 +1/25 +...= (1/5)/(1 -1/5)))=1/4. Therefore sum n2/(2*5^j)=n2/8. The second term is sum_{j=1}^\\infty n/2 which is infinite, which can't be.\n\nSo this approach is leading to a problem. Therefore, my earlier approximation is insufficient. So perhaps not every term floor(n/5^j) can be approximated as n /5^j.\n\nAlternatively, perhaps floor(n/5^j)(floor(n/5^j)+1)/2 is approximately (n /5^j)^2 /2 when j is not near log_5 n and smaller terms when j is near log_5 n. But perhaps the sum can be approximated as \u03a3 (n^2 / (2*5^{2j})) *5^j = \u03a3 n2/(2*5^j). So then that gives \u03a3 n2/(2*5^j). So same as sum j\u22651 of n2/(2*5^j )= n2/8, as sum 1/5^j = 1/4.\n\nBut if so, then the leading term might be n2/8, but we have higher order terms. But then what about the rest?\n\nWait, why did that approach neglect the floor(n/5^j)^2 vs (n/5^j)^2?\n\nPerhaps note that (floor(n/5^j) +1)(floor(n/5^j)))/2 \u2248 (n/5^j)(n/5^j)/2 for n/5^j large, but when 5^j is comparable to n, then it's more variable.\n\nAlternatively, perhaps a better approach is to use the formula (k)(k+1)/2= (k^2 +k)/2. So expand the expression within the sum:\n\nE(n)= \u03a3_{j=1}^{\u221e} 5^j*( (floor(n/5^{j})^2 + floor(n/5^{j}) ) /2 )\n\n= (1/2) \u03a3_{j=1}^\u221e 5^j*floor(n/5^{j})^2 + (1/2) \u03a3_{j=1}^\u221e5^j*floor(n/5^{j})\n\nSo E(n)= [\u03a31 + \u03a32]/2, where \u03a31=\u03a35^j floor(n/5^{j})2, \u03a32=\u03a35^j floor(n/5^{j})\n\nHmm. Maybe compute each of these sums. Let's see:\n\nFirst, let's compute \u03a32 as sum_{j=1}^\u221e5^j floor(n/5^j). That's like multiplying each term of the original thing.\n\nBut to handle these sums, perhaps consider how floor(n /5^j) behaves.\n\nOur goal is to compute \u03a3_{j=1}^\\infty floor(n /5^j) *5^j. Hmm. Let's think for finite n.\n\nBut \u03a3_{j=1}^\\infty floor(n /5^j) *5^j. For j with 5^{j} > n, floor(n/5^j) =0, so the terms become zero. So the number of terms is up to log_5 n.\n\nWait, so \u03a3_{j=1}^\\infty floor(n /5^j) *5^j = floor(n/5)*5 + floor(n/25)*25 + floor(n/125)*125 + ... + floor(n/5^J)*5^J where J is the largest integer such that 5^J <= n.\n\nSimilarly for \u03a31.\n\nBut how can I estimate these sums? Let's try approximating floor(n /5^j) as n /5^j - {n /5^j }, where {x} = x -floor(x) is the fractional part.\n\nTherefore, floor(n/5^j) = n/5^j - {n/5^j}\n\nTherefore:\n\n\u03a32=\u03a3_{j=1}^\\infty 5^j (n/5^j - {n/5^j})= \u03a3_{j=1}^\\infty n -5^j {n/5^j}= \u03a3_{j=1}^\\infty n -5^j {n/5^j}\n\nThe sum \u03a3_{j=1}^\\infty n seems divergent as infinite, but of course j really goes only up to log_5 n, since for j > log_5 n, 5^j >n thus floor(n / 5^j)=0. Let's denote J = floor(log_5 n). So actually:\n\n\u03a32=\u03a3_{j=1}^J floor(n/5^j )5^j= n*J - \u03a3_{j=1}^J5^j {n/5^j}\n\nBut probably, the fractional parts {n /5^j} are evenly distributed and don't contribute significantly to the sum when n is considered asymptotically. Alternatively, the term \u03a35^{j}{n/5^j } is bounded to within a multiple of n, since each {n/5^j } \u22641, so \u03a3_{j=1}^J5^{j } {n/5^j } <= \u03a3_{j=1}^J5^j= (5^{J+1 } -5)/4 which is \u22485^{log_5 n } *5=5n? But because 5^{J } <=n <5^{J+1} therefore 5^{J} \u2208[n/5, n). Therefore, sum_{j=1}^J5^j <5^{J+1} -5 ~5n . Therefore, the term \u03a35^{j}{n/5^j } is O(n). Then total \u03a32= n*J -O(n)\n\nGiven that n*J corresponds to n*log_5 n because J= floor(log_5 n). Therefore, asymptotically \u03a32= n log_5 n -O(n). Therefore, leading term is n log n. But in our original expression, we need \u03a32. But even more, when we think about E(n), we have:\n\nE(n)\u2248[\u03a31 + \u03a32 ]/2. But for E(n)/n2.\n\nWait, but our goal is the limit of E(n)/n2. So even if \u03a32\u2248n log n, then \u03a32/(2n2)\u2248log n / (2n) which goes to zero as n approaches infinity. Therefore, \u03a32 is lower order compared to n squared, hence E(n) would be dominated by \u03a31. So leading term is \u03a31/2. Then \u03a31\u2248 sum_{j=1}^infty 5^j*(n/5^j)^2=\u03a3 n2/5^j. Therefore, sum n2(1/5 +1/25 +1/125 +...)= n2*(1/4). Then \u03a31\u2248n2/4. So E(n)\u2248\u03a31/2 \u2248n2/8?\n\nBut then shouldn't the answer be 1/8? Wait but perhaps there's an error?\n\nAlternatively, note that E(n)= \u03a3_{m=1}^n m v_5(m). Let me try handling this sum differently, via generating functions or directly by using properties of v_5(m).\n\nv5(m) is the exponent of 5 in prime decomposition of m, so for each m, we can write m as 5^{v5(m)}*t where t not divisible by 5. Therefore, m=5^{v_5(m)}*t. Then v5(m) = logarithm base 5 of m/t, which is integer, but perhaps not useful here.\n\nAlternatively, we might note that:\n\nThe sum E(n) = sum_{m=1}^n m v_5(m) can be rewritten by splitting the m's into their different exponents. Let\u2019s for integers k\u22651, define the set S_k = { m \u2208 N : 1\u2264m \u2264n, v5(m)=k }, so numbers divisible by 5^k but not 5^{k+1}. Then, for each k, the numbers in S_k are: m = 5^k * t where t is not divisible by 5 and t \u2264 n/5^k. Then maybe the sum becomes:\n\nsum_{k=1}^\u221esum_{t=1}^{floor(n/5^k)} (5^k t) *k, where t is such that t not divisible by 5. Because m =5^k t with t not divisible by 5.\n\nSo the total sum is sum_{k=1}^\u221e5^k k sum_{t=1}^{floor(n/5^k)} t - sum_{k=1}^\u221e5^k k sum_{t=1}^{floor(n/5^{k+1})}t*5\nWait, no. Wait, because each t in S_k is not divisible by 5. So to count t, we can split sum over t: t \u2264 floor(n /5^k)} and t coprime to 5. Then:\n\nsum_{t=1}^{floor(n/5^k)} t * [5 does not divide t].\n\nThat is sum_{t=1}^{M} t - sum_{t=1}^{ \\lfloor M /5 \\rfloor }5t where M= floor(n /5^k).\n\nTherefore,\n\nsum_{t=1}^{M} t - 5 sum_{t=1}^{floor(M /5)} t= M(M+1)/2 -5*(floor(M/5))(floor(M/5)+1)/2.\n\nTherefore,\n\nsum_{t=1}^{M} t [5\u2224t] = M(M+1)/2 - 5* [floor(M/5) (floor(M/5)+1)/2 ]\n\nBut this seems quite complicated. Let's substitute back:\n\nSo,\n\nE(n)=sum_{k=1}^\u221e5^k *k *[ sum_{t=1}^{floor(n/5^k)} t -5 sum_{t=1}^{floor(n/5^{k+1}})t ]\n\n= sum_{k=1}^\u221e5^k *k * [ floor(n/5^k)(floor(n/5^k)+1)/2 -5 * floor(n/5^{k+1})(floor(n/5^{k+1})+1)/2 ]\n\nSo,\n\nE(n)= sum_{k=1}^\\infty k5^k [ ( [n/5^k]^2 + [n/5^k] )/2 -5 ( [n/5^{k+1}]^2 + [n/5^{k+1}])/2 )\n\nBut maybe again approximate when n is large, the floor terms can be replaced by n /5^{k} and n /5^{k+1}:\n\nE(n)\u2248 sum_{k=1}^\\infty k5^k [ ( (n/5^k)^2 +n/5^k)/2 - 5( (n/5^{k+1})^2 +n/5^{k+1} )/2 ]\n\n= sum_{k=1}^\\infty k5^k [ (n2 /5^{2k} +n /5^k )/2 -5(n2 /5^{2k +2} +n /5^{k +1}) /2 ]\n\n= sum_{k=1}^\\infty k5^k /2 [ n2 /5^{2k} +n /5^k -5 n2/5^{2k +2} -5 n /5^{k +1 } ]\n\nSimplify term by term:\n\nInside the brackets:\n\nn2 /5^{2k} +n /5^k - 5n2 /5^{2k +2} -5n /5^{k +1 }\n\n= n2 /5^{2k} +n /5^k - 5n2/(5^{2k }* 25 ) -5n / (5*5^k )\n\n= n2 /5^{2k} +n /5^k - 5n2/(25*5^{2k }) -5n / (5^{k +1 })\n\n= n2/5^{2k} -5n2/(25*5^{2k }) + n /5^k -5n/(5^{k +1})\n\n= n2/5^{2k} (1 -1/5 ) +n /5^k (1 -1 )\n\n= (4/5) n2 /5^{2k ) +0\n\nWait, hold on: n2/5^{2k} -5n2/(25*5^{2k } ) = n2/5^{2k} -n2/(5^{2k} *5 )=n2/5^{2k }(1 -1/5)= (4/5)n2/5^{2k }\n\nSimilarly, the n terms:\n\nn/5^k -5n/(5^{k+1}})=n/5^k -5n /(5^k*5)=n/5^k -n /5^k=0.\n\nTherefore, the entire expression reduces to sum_{k=1}^\\infty k5^k /2 * (4 /5 )n2 /5^{2k }\n\n= sum_{k=1}^\\infty k*(4 n2)/10 *1 /5^k\n\n= (4 n2)/10 * sum_{k=1}^\\infty k/5^{k }\n\nSum_{k=1}^\\infty k x^k= x/(1 -x )2 when |x|<1. Hence sum k/5^k= (1/5 ) / (1 -1/5 )2= (1/5)/(16/25)= (1/5 )*(25/16 )=5/16.\n\nTherefore, the sum above is:\n\n4n2/10 *5/16= (2n2/5 )*(5/16 )=2n2/16= n2/8\n\nTherefore, E(n)~n2/8 as n\u2192\u221e. Hence, the lim E(n)/n2=1/8\u22480.125.\n\nTherefore, after going through that elaborate approximation, the leading term of E(n) is n2/8, so the desired limit is 1/8.\n\nBut let me verify this asymptotic with explicit examples. Take n being a power of 5, say n=5^m, how is E(n)? Let's choose n=5.\n\nFor n=5:\n\nThe product is 11 22334455.\n\nCalculating exponents of 5: for each m from 1 to5,\n\nv5(1) =0, so contribution from 11 is 0,\n\nv5(2)=0,\n\nv5(3)=0,\n\nv5(4)=0,\n\nv5(5)=1.\n\nThus, E(5)= 1*1=1. Also, compute E(5)/52=1/25=0.04, but 1/8 is~0.125. Hmm, it maybe small n.\n\nn=25:\n\nE(25)= \u03a3_{m=1}^{25} m*v5(m). For numbers m divisible by 5 but not 25, m=5,10,15,20: their v5(m)=1, and m=25: v5(m)=2. Thus,\n\nContribution from m=5:5*1,\n\nm=10:10*1,\n\nm=15:15*1,\n\nm=20:20*1,\n\nm=25:25*2.\n\nThe rest (total 20 numbers m=1,...5,...25 exclude multiples of5 :5 terms.\n\nTherefore:\n\nE(25)= (5+10+15+20)*1 +25*2=50 +50=100. Then E(25)/252=100/625=0.16, but 1/8=0.125.\n\nBut for n=25 approximation suggests 252 /8= \u224878.125. So here, actual E(25) 100> approximate 78.125. So maybe even for n=25 it's larger?\n\nAlternatively, if I compute for higher n. However, seems that the actual E(n)/n2 approaches certain value.\n\nBut according to our analysis, 1/8.\n\nBut let me see, with the previous derivation:\n\nWe found in approximating, E(n) ~ \u03a3_{j=1}^\\infty5^j*( (n/5^j )2)/2=\u03a3_{j=1}^\\inftyn2/(2*5^j ))=n2/(2*(5-1)))=n2/8.\n\nSo if lim_{n}E(n)/n2=1/8. Then you need to check.\n\nBut for example, let us compute n=125:\n\nE(n)= \u03a3_{m=1}^125 m*v5(m).\n\nGroup numbers as per v5(m):\n\n1. Numbers not divisible by 5: 1-125, number of =125 -25=100. Each contributes 0. Total 0.\n\n2. Numbers divisible by5 but not25: multiples of5 but not25. In 1-125, there's 25 numbers divisible by 5. Among them, those divisible by25:5. So this category has25-5=20 numbers. Each contributesv5(m)=1. So contribution= \u03a3m*1 where m=5(t), t not divisible by5 up to25. The sum =5*\u03a3t for t=1,2,3,4,6,...25) where t skips multiples of5. Sum_t= [sum_{t=1}^{25}t - sum_{t=1}^5 t*5 ]= [25*26/2 -5(5*6)/2 ]= [325 -75]=250. Hence contribution=5*250=1250.\n\n3. Numbers divisible by25 but not125: multiples of25 (i.e., m=25,50,75,100). Check, for m up to125: yes, 25,50,75,100,125. Omit 125. So sum is for m=25,50,75,100 (4 terms). For each m=25j where j=1,2,3,4,5. j=5 is m=125. So 4 terms. Each m=25j: v=2. So contribution= \u03a3_{j=1}^4 25j*2=2*25 \u03a3j=1 to4j=50*(1+2+3+4)=50*10=500.\n\n4. Numbers divisible by125.\n\n125 contributes v=3, and m=125. So contribution=125*3=375.\n\nTotal E(125)=1250+500+375=2125.\n\nCompute E(125)/1252=2125/(15625 )=0.136. That's bigger than 1/8=.125, but approaching.\n\nWait: we can see maybe a lower rate. Hmm.\n\nWait, for n=5^m=5^3=125, gives E(n)=2125, ratio0.136. For n=625 (5^4), let's compute expected E(n)/6252.\n\nE(n)= sum_{m=1}^{625} m*v_5(m). By previous pattern:\n\nCompute the contributions:\n\n1. Numbers divisible by5 not25: 625/5=125 numbers. Subtract those divisible by25:125-25=100 numbers. Each m=5k contributes sum k*5*1=5 sum k where k numbers not divisible by5 up to125\u2192Wait, no. Wait, m=5k where k not divisible by5, from1 to125. So for each j in k=1 to125, j not divisible by5, then m=5j. Total m's in this group is 125-25=100 (since floor(625/5)=125, minus those also divisible by25: floor(625/25)=25). Then sum=5sum_{j=1}^{100} j' where j' is the indices mapped over. Wait, since the numbers k not divisible by5 from1 to125. So sum_{k=1}^{125} k - sum_{k=1}^{25} 5k= [125*126/2 -5*(25*26/2) ]= (7875) -5*(325)=7875 -1625=6250. Therefore, sum_j multiplied by5:5*6250=31250. Contribution from S1:31,250.\n\n2. Numbers divisible by25 but not125: total numbers=25-5=20 numbers. Each m=25j, j not divisible by5, of sums=25j*2=50j. sum_j=1 to20 j (assuming not divisible by5)? Wait, numbers m=25,50,75,...,625 divisible by25 but not125: from j=1 to125 where25j <=625, then j=1,...,25. Those divisible by5^but not25 in \"j\" terms: j not divisible by5?\n\nHold, for m=25j, m is divisible by25 but not by125 iff25j is divisible by25 but not125 => j divisible by1 but not5. So j ranges from1 to25, 25 terms, j not divisible by5: 20 terms. Thus m=25j, j in1,...,25 not divisible by5. sum25j*2=50 sum j=50 *( sum_{j=1}^{25}j - sum_{j=5,10,15,20,25}j )=50*(325 - (5+10+15+20+25))=50*(325-75)=50*250=12,500.\n\n3. Numbers divisible by125 but not625: m=125,250,375,500. These are j=1,2,3,4, of 125j, since125*5=625. Each m=125j (j=1,2,3,4 ), contributing 125j*3=375j. Sum_j=14 375j=375*(1+2+3+4)=375*10=3,750. If in case n=625, perhaps (there is625/125=5 terms, but m=625 is divisible by625=5^4. Wait, check exact numbers. For m=625: v_5(625)=4. So numbers divisible by125 but not625: j where125j divides m=125j<=625, and j satisfies not divisible by5. Wait, m=125j<=625,j<=5. Those j=1,2,3,4,5. But j=5 gives125*5=625, which is divisible by625 .So numbers m=125,250,375,500 with j=1,...,5 not divisible by5. So j=1,2,3,4, where div not by5. Wait, j=1,2,3,4 aren't divisible by5, so 4 elements). Therefore, sum m*v_5(m)=125*3 +250*3 +375*3 +500*3= (125+250+375+500)*3=1250*3=3,750.\n\n4. Numbers divisible by625: m=625 whose v=4. Contribution=625*4=2500.\n\nTotal E(625)=31250 +12500 +3750 +2500=50,000.\n\nTherefore, E(n)/n2=50,000/390625=0.128. So n=625 gives0.128, approaching 0.125.\n\nSimilarly, next term at n=3125 [5^5], which would presumably approach the limit even closer.\n\nThus, with n=5^m, the ratio E(n)/n2 is approaching 0.125 from above.\n\nSimilarly, if we take higher powers of5, similar approach shows convergence to 1/8.\n\nTherefore, it makes sense that the limit is \\boxed{\\dfrac{1}{8}}.\n\n**Final Answer**\n\\boxed{\\dfrac{1}{8}}\n",
    "r1_solution": "To find the limit as \\( n \\) approaches infinity of \\(\\frac{E(n)}{n^2}\\), where \\( E(n) \\) is the greatest integer \\( k \\) such that \\( 5^k \\) divides the product \\( 1^1 2^2 3^3 \\cdots n^n \\), we need to determine the asymptotic behavior of \\( E(n) \\).\n\nFirst, we recognize that \\( E(n) \\) is the sum of the exponents of 5 in each term of the product. For each \\( m \\), the exponent of 5 in \\( m^m \\) is \\( m \\cdot v_5(m) \\), where \\( v_5(m) \\) is the exponent of 5 in the prime factorization of \\( m \\). Therefore, \\( E(n) = \\sum_{m=1}^n m \\cdot v_5(m) \\).\n\nTo evaluate this sum, we consider the contribution of each term \\( m \\cdot v_5(m) \\) for \\( m \\) from 1 to \\( n \\). The value \\( v_5(m) \\) is the number of times 5 divides \\( m \\). We can express \\( v_5(m) \\) as the sum:\n\\[ v_5(m) = \\sum_{j=1}^{\\infty} [5^j \\mid m] \\]\nwhere the indicator function \\([5^j \\mid m]\\) is 1 if \\( 5^j \\) divides \\( m \\) and 0 otherwise.\n\nInterchanging the order of summation, we get:\n\\[ E(n) = \\sum_{m=1}^n m \\cdot v_5(m) = \\sum_{j=1}^{\\infty} \\sum_{\\substack{m=1 \\\\ 5^j \\mid m}}^n m \\]\n\nFor each \\( j \\), the inner sum is over all \\( m \\) divisible by \\( 5^j \\). Writing \\( m = 5^j k \\), we can rewrite the sum as:\n\\[ \\sum_{m=1}^{\\lfloor n/5^j \\rfloor} 5^j k = 5^j \\sum_{k=1}^{\\lfloor n/5^j \\rfloor} k \\]\n\nThe sum \\( \\sum_{k=1}^K k \\) is a triangular number, which is \\( \\frac{K(K+1)}{2} \\). Substituting this, we get:\n\\[ E(n) = \\sum_{j=1}^{\\infty} 5^j \\cdot \\frac{\\lfloor n/5^j \\rfloor (\\lfloor n/5^j \\rfloor + 1)}{2} \\]\n\nFor large \\( n \\), \\( \\lfloor n/5^j \\rfloor \\approx n/5^j \\), leading to:\n\\[ E(n) \\approx \\sum_{j=1}^{\\infty} 5^j \\cdot \\frac{(n/5^j)(n/5^j + 1)}{2} \\]\n\nApproximating further, the dominant term is:\n\\[ E(n) \\approx \\sum_{j=1}^{\\infty} \\frac{n^2}{2 \\cdot 5^j} \\]\n\nThis series is a geometric series with the first term \\( \\frac{n^2}{2 \\cdot 5} \\) and common ratio \\( \\frac{1}{5} \\), summing to:\n\\[ \\sum_{j=1}^{\\infty} \\frac{n^2}{2 \\cdot 5^j} = \\frac{n^2}{2} \\cdot \\frac{1/5}{1 - 1/5}} = \\frac{n^2}{8} \\]\n\nThus, the asymptotic behavior of \\( E(n) \\) is \\( \\frac{n^2}{8} \\), leading to the limit:\n\\[ \\lim_{n \\to \\infty} \\frac{E(n)}{n^2} = \\frac{1}{8} \\]\n\nThe final answer is:\n\\[\n\\boxed{\\dfrac{1}{8}}\n\\]",
    "judge_correct": true,
    "judge_reasoning": "The submitted solution correctly breaks down the problem step by step. Here is a simplified reasoning of why the solution is correct:\n\n1. **Problem Understanding**:\n   - You need to determine the limit of \\( \\frac{E(n)}{n^2} \\) as \\( n \\to \\infty \\), where \\( E(n) \\) counts the maximum power of 5 that divides the product \\( 1^1 2^2 \\cdots n^n \\).\n\n2. **Determining \\( E(n) \\):**\n   - Realize \\( E(n) = \\sum_{m=1}^n m \\cdot v_5(m) \\), where \\( v_5(m) \\) represents the exponent of 5 in \\( m \\).\n   - Sum contributions across all powers recursively using the expression \\( v_5(m) = \\sum_{j=1}^{\\infty} [5^j \\mid m] \\).\n\n3. **Summation Rearrangement**:\n   - Sum can be rearranged to \\( E(n) = \\sum_{j=1}^{\\infty} \\sum_{m=1, \\, 5^j \\mid m}^n m \\) which computes contributions for each multiple of 5.\n\n4. **Summation Simplification**:\n   - Express as \\( E(n) \\approx \\sum_{j=1}^{\\infty} \\frac{n^2}{2 \\times 5^j} \\). The expression takes advantage of triangular number formula for arithmetic sums.\n\n5. **Geometric Series and Asymptotic Limit**:\n   - The geometric series \\( \\sum_{j=1}^{\\infty} \\frac{1}{5^j} \\) converges and contributes the limit result \\( \\frac{n^2}{8} \\) for large \\( n \\).\n\n6. **Correct Limit**:\n   - Leads to the conclusion that \\( \\lim_{n \\to \\infty} \\frac{E(n)}{n^2} = \\frac{1}{8} \\) which matches both the presented solution and the ground truth.\n\nThe neat encapsulation of each mathematical component reflects a coherent understanding of the asymptotic behavior and a systemic approach to convergence. The matching between the presented solution's resulting limit \\( \\boxed{\\frac{1}{8}} \\) and the ground truth confirms correctness."
}
```

## Loading the dataset

You can load this dataset using the following code:

```python
from datasets import load_dataset

dataset = load_dataset("mlfoundations-dev/putnambench_r1", split="default")
```

