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internet!\nThanks for stopping by. There's nothing much here, but feel free to explore.\n","description":"","id":"https://kibibytium.github.io/","path":"/","title":"Kibibytium"},"https://kibibytium.github.io/about/":{"body":"Wait, kibi-what?..\nIt's Kibibytium, as in a (KiB)ibyte that is also a fake element in the periodic table :)\nThis site is made with Zola, using\na slightly modified version of the apollo theme.\n","description":"","id":"https://kibibytium.github.io/about/","path":"/about/","title":"About"},"https://kibibytium.github.io/posts/":{"body":"","description":"","id":"https://kibibytium.github.io/posts/","path":"/posts/","title":"Posts"},"https://kibibytium.github.io/posts/remainder-theorem/":{"body":"The Puzzle\nA woman goes to the market with a basket full of eggs. On her way, a horse bumps into her, causing her to drop the\nbasket and break all the eggs. The horse's owner feels guilty and asks how many eggs she had so he can compensate her.\nShe replies:\n\n\"I don't remember the exact number of eggs, but I do remember this:\n\nWhen I tried to group the eggs two at a time, there was one egg left over.\nThe same happened when I grouped them three at a time, four at a time, five at a time, and six at a time—there was\nalways one egg left over.\nHowever, when I grouped them seven at a time, there were no eggs left over.\"\n\n\nWhat is the smallest number of eggs she could have had?\nThe explanation\nWe have to search for a number that is divisible by seven but always leaves a remainder of one when divided by $2, 3, 4,\n5$, or $6$. So your first question should be: why should such a number exist? Can I define arbitrary conditions of\nremainders, as above in the puzzle and find solutions for them, or is this puzzle unique in some way?\nIt is fairly easy to find a number not divisible by any of the integers $2$ to $6$ with a reminder of $1$. Whether it\nis divisible by $7$, is a different story. An answer to this question is provided by the\n\"Chinese remainder theorem\" and I will discuss it a bit further down below.\nKeep in mind that solving the puzzle is straightforward, but I’m more interested in a formal approach.\nSpecifically, I’m looking for a method to determine whether a general solution exists, along with a\nformula or algorithmic approach to compute it.\nNaturally, I realize that the abstract mathematical questions I posed above are not the first questions which come to\nmind when one encounters this puzzle, so I will start by presenting the direct easy solution first.\nA naive approach\nConsider $6! =720$ which by definition is divisible by the integers $2$ to $6$, and add one to it.\nNow you have constructed an integer which leaves a remainder of $1$ when divided by $2,3,4,5,$ or $6$ and therefore\nsatisfies the first of the puzzle's conditions. Coincidentally, $(721)$ is also divisible by $7$, but it does not\nsolve the puzzle, because it is not the smallest possible solution.\nAn educated approach\nThe idea is simple: compute the least common multiple\nof the integers $2,3,4,5,6$ and add $1$ to it. Then check whether the result is divisible by 7.\nIf not, find the next smallest common multiple of $2,3,4,5,6$ add 1 to it, then divide by seven and so on.\nThe first number divisible by $7$ is the solution to the puzzle, and it's $301$!\nHere be Mathematics\nLet's start by writing the idea above formally. The least common multiple of these integers is:\n$$\n\\text{lcm}(2,3,4,5,6) = 2 \\cdot 3 \\cdot 2 \\cdot 5 = 60\n$$\nUsing a congruence relation, we can write a general form for the solution: The solution - let's call it $x$ - has the\nform $x = 60n +1$ for the smallest integer $n$ satisfying:\n$$\n60n + 1 \\equiv 0 \\quad (\\text{mod} 7)\n$$\nThe first multiple of $7$ that can be written in this form is $301$ which we get for $n=5$.\nA trick using congruence relations\nThere is a simple trick that makes use of congruence relations.\nThis trick makes the arithmetic - and therefore the process of trial and error - for finding $n$ a bit easier.\nConsider the following equivalent statements:\n\n$$\n\\begin{aligned}\n60 &amp;\\equiv 4 \\quad &amp;(\\text{mod } 7)\\\\\n60n &amp;\\equiv 4n \\quad &amp;(\\text{mod } 7)\\\\\n60n + 1 &amp;\\equiv 4n +1 \\quad &amp;(\\text{mod } 7)\n\\end{aligned}\n$$\n\nThe first n which makes $( 4n +1$) divisible by 7 solves the puzzle, which is of course $5$.\n\n  \n    \n      \n        \n        Note\n\n        \n      \n    \n    \n      These statements are equivalent because congruence relations are compatible with algebraic\nstructures. In this case the algebraic structure is the integer ring $\\mathbb{Z}$, with the conventional addition and\nmultiplication.\n\n    \n  \n\nThe Chinese Remainder Theorem\nTheorem\nGiven a system of $k$ congruence relations with pairwise\ncoprime moduli $m_1,\\cdots, m_k$ then there exists a unique solution\nup to $ \\text{mod } M$, where $M= m_1m_2\\cdots m_k$.\n\n$$\n\\begin{aligned}\nx &amp;\\equiv a_1 \\quad &amp;(\\text{mod } m_1)\\\\\nx &amp;\\equiv a_2 \\quad &amp;(\\text{mod } m_2)\\\\\n\\vdots\\\\\nx &amp;\\equiv a_k \\quad &amp;(\\text{mod } m_k)\n\\end{aligned}\n$$\n\nFurthermore, if $0 \\le a_i \\lt m_i: \\forall i$, then there exists one and only one $0 \\le x \\lt M$ and $a_i$ is the\nremainder of the Euclidean division of $x$ by $m_i$\n\nThis is one of multiple possible statements of the theorem. It is of course a modern formulation\nbut the theorem originates from 5th century china, in a book on mathematics.\nIt appeared as conjecture without proof or algorithm.\nGreat… how does this help us in actually finding the solution? Well, the theorem has a proof by construction.\nThis means that we can prove the theorem by constructing a solution for a general form of the puzzle above.\nProof of existence by construction\nLet $M_i = \\frac{M}{m_i} = m_1 \\cdot m_2 \\cdots m_{i-1} \\cdot m_{i+1} \\cdots m_k$.\nSince all the moduli $(m_i$) are coprime, then it is easy to see\nthat the greatest common divisor of $(M_i$) and $(m_i$) is $(1$). Write:\n$$\n\\text{gcd}(M_i,m_i) = 1\n$$\nA well known result in number theory says that if two integers are coprime - as is the case with $(M_i, m_i$), then\nthere exists an integer1 - call it $x_i$ - such that:\n$$\nx_iM_i \\equiv 1 \\quad (\\text{mod } m_i)\n$$\nFor any other index $j \\neq i$, $m_j$ will divide $M_i$ by definition. In other words:\n$$\nx_iM_i \\equiv 0 \\quad (\\text{mod } m_j)\\quad \\forall j\\neq i\n$$\nThis is all we need to construct an integer $x$ which solves the system of congruence relations and proves the\ntheorem. Consider:\n$$\nx = \\sum_{i=1}^{k} x_i M_i a_i\n$$\nand check for any $i$:\n\n$$\n\\begin{aligned}\nx \\quad \\text{mod} \\quad m_i &amp;= \\sum_{j=1}^{k} x_j M_j a_j \\quad &amp;(\\text{mod } m_i)\\\\\n&amp;= 0 + 0 + \\cdots + x_iM_ia_i + \\cdots + 0 \\quad &amp;(\\text{mod } m_i)\\\\\n&amp;= x_iM_ia_i \\quad &amp;(\\text{mod } m_i)\n\\end{aligned}\n$$\n\nso in short:\n$$\nx \\equiv x_iM_ia_i \\quad (\\text{mod } m_i)\n$$\nSince we already know that $x_iM_i \\equiv 1 \\quad (\\text{mod } m_i)$, the equivalence above simplifies to:\n$$\nx \\equiv a_i \\quad (\\text{mod } m_i)\n$$\nwhich proves the statement of the theorem.\nremarks\nIn the proof above we have constructed a solution $x$ for a system of congruences. However, our solution still relies\non an existence theorem for all the $x_i$. Unless we find a systematic way to compute these integers, we would not\nbe able to find a solution to the puzzle easily.\nWith that in mind, consider again the statement:\n$$\nx_iM_i \\equiv 1 \\quad (\\text{mod } m_i)\n$$\n$x_iM_i$ divided by $m_i$ will have a remainder of 1. Meaning that $x_iM_i \\pm 1 $ is a multiple of $m_i$.\nWrite:\n\n$$\n\\begin{aligned}\nx_iM_i - 1 &amp;= m_i\\\\\n\\end{aligned}\n$$\n\nWithout loss of generality we can write this as:\n\n$$\n\\begin{equation}\n1 = x_iM_i + sm_i\n\\end{equation}\n$$\n\n\n  \n    \n      \n        \n        Note\n\n        \n      \n    \n    \n      Note that 1 is the greatest common divisor of $x_iM_i$ and $m_i$ and the equation above is called Bézout's\nidentity.\nThe factors $x_i, s$ are called Bézout's coefficients, and they are computed using\nthe extended Euclidean algorithm of division at the step before last.\nBézout's identity and the Euclidean algorithm warrant an article by themselves and are therefore beyond the scope of\nthis post.\n\n    \n  \n\nApplication of the theorem\nFirst we need to formulate the puzzle in a way that is compatible with the theorem. We could write:\n\n$$\n\\begin{aligned}\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 2)\\\\\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 3)\\\\\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 4)\\\\\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 5)\\\\\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 6)\\\\\nx &amp;\\equiv 0 \\quad &amp;(\\text{mod } 7)\\\\\n\\end{aligned}\n$$\n\nWe can improve on this by using the least common multiple of these integers,$60$, since it encodes the first five\ncongruences:\n\n$$\n\\begin{aligned}\nx &amp;\\equiv 1 \\quad &amp;(\\text{mod } 60)\\\\\nx &amp;\\equiv 0 \\quad &amp;(\\text{mod } 7)\\\\\n\\end{aligned}\n$$\n\n$60$ and $7$ are coprime, and the divisors $1$ and $0$ are less than $60$ and $7$ respectively.\nApplying the theorem, the solution becomes:\n$$\nx \\equiv x_1 M_1 a_1 + x_2 M_2 a_2 \\quad (\\text{mod } 7\\cdot60)\n$$\nwhere\n\n$$\n\\begin{aligned}\nM_1 &amp;= 7\\\\\nM_2 &amp;= 60\\\\\na_1 &amp;= 1\\\\\na_2 &amp;= 0\\\\\n\\end{aligned}\n$$\n\n$$\n\\Rightarrow x \\equiv 7x_1 + 0 \\quad (\\text{mod } 420)\n$$\nNow it only remains to compute Bézout's coefficients for the identity:\n$$\n1 = 7x_1 + s60\n$$\nAs mentioned before, those coefficients are computed from the Euclidean algorithm of division. In this case we have:\n$$\n1 = 7(-17)  + (2)60\n$$\n$$\n\\Rightarrow x \\equiv -119 \\quad (\\text{mod } 420)\n$$\nThe solution $x$ is unique up to modulo $420$, meaning that there are multiple solutions, which are $420$ apart.\nWe know from the theorem that there is one and only one solution between $0$ and $420$:\n$$\n\\Rightarrow \\boxed{x = -119 + 420 = 301}\n$$\n1\n$(x_i)$ is called a modular multiplicative inverse of $M_i$, and it exists if and only iff both integers are\ncoprime.\n\n","description":"","id":"https://kibibytium.github.io/posts/remainder-theorem/","path":"/posts/remainder-theorem/","title":"Puzzle of the Egg Basket"}},"docInfo":{"https://kibibytium.github.io/":{"body":11,"description":0,"path":0,"title":1},"https://kibibytium.github.io/about/":{"body":18,"description":0,"path":0,"title":0},"https://kibibytium.github.io/posts/":{"body":0,"description":0,"path":1,"title":1},"https://kibibytium.github.io/posts/remainder-theorem/":{"body":889,"description":0,"path":3,"title":3}},"length":4},"lang":"English"}