P A$ and $P C
1$, let $a_{n}$ be the largest real number such that\n\n$$\n4\\left(a_{n-1}^{2}+a_{n}^{2}\\right)=10 a_{n-1} a_{n}-9\n$$\n\nWhat is the largest positive integer less than $a_{8}$ ?", "solution": "335 Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\\frac{1}{t_{n}}$. Then $t_{1}=\\frac{3+\\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have:\n\n$$\n4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0\n$$\n\nUsing the quadratic formula, we see that $a_{n}=\\frac{5}{4} a_{n-1}+\\frac{3}{4} \\sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\\sqrt{a_{n-1}^{2}-4}=\\sqrt{t_{n-1}^{2}-2+\\frac{1}{t_{n-1}^{2}}}$, so we have:\n\n$$\na_{n}=\\frac{5}{4}\\left(t_{n-1}+\\frac{1}{t_{n-1}}\\right)+\\frac{3}{4} \\sqrt{\\left(t_{n-1}-\\frac{1}{t_{n-1}}\\right)^{2}}=2 t_{n-1}+\\frac{1}{2 t_{n-1}}\n$$\n\nso $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\\frac{128(3+\\sqrt{5})}{2}+\\frac{2}{128(3+\\sqrt{5})}$. The second term is vanishingly small, so $\\left\\lfloor a_{8}\\right\\rfloor=\\lfloor 64(3+\\sqrt{5})\\rfloor$. We approximate $\\sqrt{5}$ to two decimal places as 2.24 , making this expression $\\lfloor 335.36\\rfloor=335$. Since our value of $\\sqrt{5}$ is correct to within 0.005 , the decimal is correct to within 0.32 , which means the final answer is exact.\n$13^{\\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl", "problem_match": "\n33. [21]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "34", "problem_type": null, "exam": "HMMT", "problem": "3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000 . Otherwise, it will be $25-27 \\frac{|A-C|}{\\min (A, C)-1000}$.", "solution": "1019 To get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution ${ }^{6}$ in this case with a mean of 1000 and a variance of $3000 \\cdot \\frac{2}{9}=667$. Since $\\sqrt{667} \\approx 26,1026$ is a good guess, as Gaussians tend to differ from their mean by approximately their variance.\nThe actual answer was computed with the following python program:\n\n```\nfacts = [0]*3001\nfacts[0]=1\nfor a in range(1,3001):\n facts[a]=a*facts[a-1]\ndef binom(n,k):\n return facts[n]/(facts[k]*facts[n-k])\n```\n\n[^3]```\nmaxes = [0]*3001\nM = 1075\nfor a in range(0,3001):\n for b in range(0,3001-a):\n c = 3000-a-b\n m = max (a,max (b,c))\n if m < M:\n maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])\n print [a,b]\nbest = 1000\nfor a in range(1000,1050):\n print maxes[a],a\n if maxes[best] <= maxes[a]:\n best = a\nprint maxes[best]\nprint best\n```\n\nWe can use arguments involving the Chernoff bound ${ }^{7}$ to show that the answer is necessarily less than 1075. Alternately, if we wanted to be really careful, we could just set $M=3001$, but then we'd have to wait a while for the script to finish.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl", "problem_match": "\n34. [25]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "35", "problem_type": null, "exam": "HMMT", "problem": "Call an positive integer almost-square if it can be written as $a \\cdot b$, where $a$ and $b$ are integers and $a \\leq b \\leq \\frac{4}{3} a$. How many almost-square positive integers are less than or equal to 1000000 ? Your score will be equal to $25-65 \\frac{|A-C|}{\\min (A, C)}$.", "solution": "130348 To get a good estimate for the number of almost-square integers, note that any number of the form $a \\cdot b$, with $b \\leq \\frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then the number of almostsquare numbers less than $n$ will be approximately\n\n$$\n\\sum_{a=1}^{\\sqrt{n}} \\sum_{b=a}^{\\frac{4}{3} a} 1=\\frac{1}{3} \\sum_{a=1}^{\\sqrt{n}} a=\\frac{1}{6} \\sqrt{n}(\\sqrt{n}+1)\n$$\n\nwhich is about $\\frac{n}{6}$. So, $\\frac{n}{6}$ will be a fairly good estimate for the number of almost-square numbers less than $n$, making 160000 a reasonable guess.\n\nWe can do better, though. For example, we summed $\\frac{a}{3}$ all the way up to $\\sqrt{n}$, but we are really overcounting here because when $a$ is close to $\\sqrt{n}, a \\cdot b$ will be less than $n$ only when $b \\leq \\frac{n}{a}$, as opposed to $b \\leq \\frac{4 a}{3}$. So we should really be taking the sum\n\n[^4]\\[\n\n$$\n\\begin{aligned}\n& \\sum_{a=1}^{\\sqrt{\\frac{3 n}{4}}} \\sum_{b=a}^{\\frac{4 a}{3}} 1+\\sum_{a=\\sqrt{\\frac{3 n}{4}}}^{\\sqrt{n}} \\sum_{b=a}^{\\frac{n}{a}} 1 \\\\\n& \\quad=\\sum_{a=1}^{\\sqrt{\\frac{3 n}{4}}} \\frac{a}{3}+\\sum_{a=\\sqrt{\\frac{3 n}{4}}}^{\\sqrt{n}}\\left(\\frac{n}{a}-a\\right) \\\\\n& \\quad \\approx \\frac{1}{6} \\frac{3 n}{4}+n\\left(\\log (\\sqrt{n})-\\log \\left(\\sqrt{\\frac{3 n}{4}}\\right)\\right)-\\left(\\frac{n}{2}-\\frac{3 n}{8}\\right) \\\\\n& \\quad=\\frac{n}{8}+n \\frac{\\log (4)-\\log (3)}{2}-\\frac{n}{8} \\\\\n& \\quad=n \\frac{\\log (4)-\\log (3)}{2}\n\\end{aligned}\n$$\n\\]\n\nIn the process of taking the sum, we saw that we had something between $\\frac{n}{8}$ and $\\frac{n}{6}$, so we could also guess something between 166000 and 125000 , which would give us about 145000 , an even better answer. If we actually calculate $\\frac{\\log (4)-\\log (3)}{2}$, we see that it's about 0.14384 , so 143840 would be the best guess if we were to use this strategy. In reality, we would want to round down a bit in both cases, since we are overcounting (because numbers could be square-free in multiple ways), so we should probably answer something like 140000 .\n\nA final refinement to our calculation (and perhaps easier than the previous one), is to assume that the products $a \\cdot b$ that we consider are randomly distributed between 1 and $n$, and to compute the expected number of distinct numbers we end up with. This is the same type of problem as number 31 on this contest, and we compute that if we randomly distribute $k$ numbers between 1 and $n$ then we expect to end up with $n\\left(1-\\left(1-\\frac{1}{n}\\right)^{k}\\right)$ distinct numbers. When $k=n \\frac{\\log (4)-\\log (3)}{2}$, we get that this equals\n\n$$\n\\begin{aligned}\nn\\left(1-\\left(\\left(1-\\frac{1}{n}\\right)^{n}\\right)^{\\frac{\\log (4)-\\log (3)}{2}}\\right) & =n\\left(1-\\sqrt{e^{\\log (3)-\\log (4)}}\\right) \\\\\n& =n\\left(1-\\sqrt{\\frac{3}{4}}\\right) \\\\\n& =n\\left(1-\\frac{\\sqrt{3}}{2}\\right) \\\\\n& \\approx 0.134 n\n\\end{aligned}\n$$\n\nGiving us an answer of 134000 , which is very close to the correct answer.\nThe actual answer was found by computer, using the following $\\mathrm{C}++$ program:\n\n```\n#include