{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "(a) Decompose 1 into unit fractions.", "solution": "$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}$\n(b) Decompose $\\frac{1}{4}$ into unit fractions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "(a) Decompose 1 into unit fractions.", "solution": "$\\frac{1}{8}+\\frac{1}{12}+\\frac{1}{24}$\n(c) Decompose $\\frac{2}{5}$ into unit fractions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "(a) Decompose 1 into unit fractions.", "solution": "$\\frac{1}{5}+\\frac{1}{10}+\\frac{1}{15}+\\frac{1}{30}$", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Explain how any unit fraction $\\frac{1}{n}$ can be decomposed into other unit fractions.", "solution": "$\\frac{1}{2 n}+\\frac{1}{3 n}+\\frac{1}{6 n}$", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "(a) Write 1 as a sum of 4 distinct unit fractions.", "solution": "$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{7}+\\frac{1}{42}$\n(b) Write 1 as a sum of 5 distinct unit fractions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "(a) Write 1 as a sum of 4 distinct unit fractions.", "solution": "$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{7}+\\frac{1}{43}+\\frac{1}{43 \\cdot 42}$\n(c) Show that, for any integer $k>3,1$ can be decomposed into $k$ unit fractions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "(a) Write 1 as a sum of 4 distinct unit fractions.", "solution": "If we can do it for $k$ fractions, simply replace the last one (say $\\frac{1}{n}$ ) with $\\frac{1}{n+1}+\\frac{1}{n(n+1)}$. Then we can do it for $k+1$ fractions. So, since we can do it for $k=3$, we can do it for any $k>3$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Say that $\\frac{a}{b}$ is a positive rational number in simplest form, with $a \\neq 1$. Further, say that $n$ is an integer such that:\n\n$$\n\\frac{1}{n}>\\frac{a}{b}>\\frac{1}{n+1}\n$$\n\nShow that when $\\frac{a}{b}-\\frac{1}{n+1}$ is written in simplest form, its numerator is smaller than $a$.", "solution": "$\\quad \\frac{a}{b}-\\frac{1}{n+1}=\\frac{a(n+1)-b}{b(n+1)}$. Therefore, when we write it in simplest form, its numerator will be at most $a(n+1)-b$. We claim that $a(n+1)-b<a$. Indeed, this is the same as $a n-b<0 \\Longleftrightarrow a n<b \\Longleftrightarrow \\frac{b}{a}>n$, which is given.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "An aside: the sum of all the unit fractions\n\nIt is possible to show that, given any real M , there exists a positive integer $k$ large enough that:\n\n$$\n\\sum_{n=1}^{k} \\frac{1}{n}=\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{3} \\ldots>M\n$$\n\nNote that this statement means that the infinite harmonic series, $\\sum_{n=1}^{\\infty} \\frac{1}{n}$, grows without bound, or diverges. For the specific example $\\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.", "solution": "Note that $\\frac{1}{n+1}+\\frac{1}{n+2}+\\ldots+\\frac{1}{2 n}>\\frac{1}{2 n}+\\ldots+\\frac{1}{2 n}=\\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get\n\n$$\n\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{3}+\\frac{1}{4}\\right)+\\left(\\frac{1}{5}+\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}\\right)+\\ldots+\\left(\\frac{1}{129}+\\ldots+\\frac{1}{256}\\right)>\\frac{1}{2}+\\ldots+\\frac{1}{2}=4\n$$\n\nso, adding in $\\frac{1}{1}$, we get\n\n$$\n\\sum_{n=1}^{256} \\frac{1}{n}>5\n$$\n\nso $k=256$ will suffice.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n## 5. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Now, using information from problems 4 and 5 , prove that the following method to decompose any positive rational number will always terminate:\nStep 1. Start with the fraction $\\frac{a}{b}$. Let $t_{1}$ be the largest unit fraction $\\frac{1}{n}$ which is less than or equal to $\\frac{a}{b}$.\nStep 2. If we have already chosen $t_{1}$ through $t_{k}$, and if $t_{1}+t_{2}+\\ldots+t_{k}$ is still less than $\\frac{a}{b}$, then let $t_{k+1}$ be the largest unit fraction less than both $t_{k}$ and $\\frac{a}{b}$.\nStep 3. If $t_{1}+\\ldots+t_{k+1}$ equals $\\frac{a}{b}$, the decomposition is found. Otherwise, repeat step 2 .\nWhy does this method never result in an infinite sequence of $t_{i}$ ?", "solution": "Let $\\frac{a_{k}}{b_{k}}=\\frac{a}{b}-t_{1}-\\ldots-t_{k}$, where $\\frac{a_{k}}{b_{k}}$ is a fraction in simplest terms. Initially, this algorithm will have $t_{1}=1, t_{2}=\\frac{1}{2}, t_{3}=\\frac{1}{3}$, etc. until $\\frac{a_{k}}{b_{k}}<\\frac{1}{k+1}$. This will eventually happen by problem 5 , since there exists a $k$ such that $\\frac{1}{1}+\\ldots+\\frac{1}{k+1}>\\frac{a_{k}}{b_{k}}$. At that point, there is some $n$ with $\\frac{1}{n}<t_{k}$ such that $\\frac{1}{n}>\\frac{a_{k}}{b_{k}}>\\frac{1}{n+1}$. In this case, $t_{k+1}=\\frac{1}{n+1}$.\nSuppose that there exists $n_{k}$ such that $\\frac{1}{n_{k}}>\\frac{a_{k}}{b_{k}}>\\frac{1}{n_{k}+1}$ for some $k$. Then we have $t_{k+1}=\\frac{1}{n_{k}+1}$ and $\\frac{a_{k+1}}{b_{k+1}}<\\frac{1}{n_{k}\\left(n_{k}+1\\right)}$. This shows that once we have found $n_{k}$ such that $\\frac{1}{n_{k}}>\\frac{a_{k}}{b_{k}}>\\frac{1}{n_{k}+1}$ and $\\frac{1}{n_{k}} \\leq t_{k}$, we no longer have to worry about $t_{k+1}$ being less than $t_{k}$, since $t_{k+1}=\\frac{1}{n_{k}+1}<\\frac{1}{n_{k}}<$ $t_{k}$, and also $n_{k+1} \\geq n_{k}\\left(n_{k}+1\\right)$ while $\\frac{1}{n_{k}\\left(n_{k}+1\\right)} \\leq \\frac{1}{n_{k}+1}=t_{k+1}$.\nOn the other hand, once we have found such an $n_{k}$, the sequence $\\left\\{a_{k}\\right\\}$ must be decreasing by problem 4. Since the $a_{k}$ are all integers, we eventually have to get to 0 (as there is no infinite decreasing sequence of positive integers). Therefore, after some finite number of steps the algorithm terminates with $a_{k+1}=0$, so $0=\\frac{a_{k}}{b_{k}}=\\frac{a}{b}-t_{1}-\\ldots-t_{k}$, so $\\frac{a}{b}=t_{1}+\\ldots+t_{k}$, which is what we wanted.\n\n## Juicy Numbers [100]\n\nA juicy number is an integer $j>1$ for which there is a sequence $a_{1}<a_{2}<\\ldots<a_{k}$ of positive integers such that $a_{k}=j$ and such that the sum of the reciprocals of all the $a_{i}$ is 1 . For example, 6 is a juicy number because $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1$, but 2 is not juicy.\n\nIn this part, you will investigate some of the properties of juicy numbers. Remember that if you do not solve a question, you can still use its result on later questions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Explain why 4 is not a juicy number.", "solution": "If 4 were juicy, then we would have $1=\\ldots+\\frac{1}{4}$. The $\\ldots$ can only possible contain $\\frac{1}{2}$ and $\\frac{1}{3}$, but it is clear that $\\frac{1}{2}+\\frac{1}{4}, \\frac{1}{3}+\\frac{1}{4}$, and $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}$ are all not equal to 1 .", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers, and show a decomposition of 1 into unit fractions for each of these numbers. You do not need to prove that no smaller numbers are juicy.", "solution": "12 and $151=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}, 1=\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{10}+\\frac{1}{15}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let $p$ be a prime. Given a sequence of positive integers $b_{1}$ through $b_{n}$, exactly one of which is divisible by $p$, show that when\n\n$$\n\\frac{1}{b_{1}}+\\frac{1}{b_{2}}+\\ldots+\\frac{1}{b_{n}}\n$$\n\nis written as a fraction in lowest terms, then its denominator is divisible by $p$. Use this fact to explain why no prime $p$ is ever juicy.", "solution": "We can assume that $b_{n}$ is the term divisible by $p$ (i.e. $b_{n}=k p$ ) since the order of addition doesn't matter. We can then write\n\n$$\n\\frac{1}{b_{1}}+\\frac{1}{b_{2}}+\\ldots+\\frac{1}{b_{n-1}}=\\frac{a}{b}\n$$\n\nwhere $b$ is not divisible by $p$ (since none of the $b_{i}$ are). But then $\\frac{a}{b}+\\frac{1}{k p}=\\frac{k p a+b}{k p b}$. Since $b$ is not divisible by $p, k p a+b$ is not divisible by $p$, so we cannot remove the factor of $p$ from the denominator. In particular, $p$ cannot be juicy as 1 can be written as $\\frac{1}{1}$, which has a denominator not divisible by $p$, whereas being juicy means we have a sum $\\frac{1}{b_{1}}+\\ldots+\\frac{1}{b_{n}}=1$, where $b_{1}<b_{2}<\\ldots<b_{n}=p$, and so in particular none of the $b_{i}$ with $i<n$ are divisible by p.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Show that if $j$ is a juicy integer, then $2 j$ is juicy as well.", "solution": "Just replace $\\frac{1}{b_{1}}+\\ldots+\\frac{1}{b_{n}}$ with $\\frac{1}{2}+\\frac{1}{2 b_{1}}+\\frac{1}{2 b_{2}}+\\ldots+\\frac{1}{2 b_{n}}$. Since $n>1,2 b_{1}>2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Prove that the product of two juicy numbers (not necessarily distinct) is always a juicy number. Hint: if $j_{1}$ and $j_{2}$ are the two numbers, how can you change the decompositions of 1 ending in $\\frac{1}{j_{1}}$ or $\\frac{1}{j_{2}}$ to make them end in $\\frac{1}{j_{1} j_{2}}$ ?", "solution": "Let $1=\\frac{1}{b_{1}}+\\ldots+\\frac{1}{b_{n}}=\\frac{1}{c_{1}}+\\ldots+\\frac{1}{c_{m}}$, where $b_{n}=j_{1}$ and $c_{m}=j_{2}$. Then\n\n$$\n1=\\frac{1}{b_{1}}+\\ldots+\\frac{1}{b_{n-1}}+\\left(\\frac{1}{b_{n} c_{1}}+\\frac{1}{b_{n} c_{2}}+\\ldots+\\frac{1}{b_{n} c_{m}}\\right)\n$$\n\nand so $j_{1} j_{2}$ is juicy.", "metadata": {"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution: "}}