Chinese remainder theorem example

Chinese Remainder Theorem: If a number N = a×b, where a and b are prime to each other, and M is a number such that the remainders obtained when M is divided by a and b are \( \to R\left( 1 \right)\) Now, according to Chinese Remainder Theorem, the final remainder is in the form of 7x +2 or 11y +1. Equating both to get the smallest solution we get, 7x +2 = 11y + 1 Or 7x + 1 = 11y With some hit and trial, we get x = 3 and y = 2. And the Final remainder is 7x+2 = 7*3+2 = 23. you may also like:

\( \newcommand\) No headers \(\newcommand\) In this section, we discuss solutions of systems of congruences having different moduli. An example of this kind of systems is the following: find a number that leaves a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by three, and a remainder of \(3\) when divided by \(5\). We shall see that there is a systematic way of solving this kind of system. Theorem \(\PageIndex\), or \(n_j\mid x-y\), for all \(1\leq j\leq k\). But then, since the moduli are pairwise relatively prime, using \(k\) times (formally, this needs to be done by induction!), we can conclude that \(N=n_1\dots n_k\mid x-y\) or \(x\equiv y\pmod N\). Example \(\PageIndex.\] Exercise \(\PageIndex\) • Find an integer that leaves a remainder of \(2\) when divided by either \(3\) or \(5\), but that is divisible by \(4\). • Find all integers that leave a remainder of \(4\) when divided by \(11\) and leaves a remainder of \(3\) when divided by \(17\). • Find all integers that leave a remainder of \(1\) when divided by \(2\), a remainder of \(2\) when divided by \(3\), and a remainder of \(3\) when divided by \(5\). • A band of \(17\) pirates steal some gold bars. When they try to divide the spoils equally, \(3\) bars are left over – so a fight breaks out, killing one. This immediately brings calm as they see if the gold can now be evenly shared. Unfortunately, there are now \(10\) bars left out, so they fight again. After the inevitable (single)...

\( \newcommand\) • • • • • • • • • • • • • • The Chinese Remainder Theorem is an important theorem appearing for perhaps the first time in Sunzi Suanjing, a Chinese mathematical text written sometime during the 3rd to 5th centuries AD. We will illustrate its usefulness with an anecdote. The child of a number theorist is sorting a large pile of pennies (worth less than a dollar) into groups of 3 pennies each. At the end, the child reports that 2 pennies are left over. The child starts over, instead sorting the pennies into groups of 4 and reports that 1 penny is left over. The child starts over again, sorting the pennies into groups of 11 and reports that 7 pennies are left over. The number theorist didn’t originally know how many pennies were in the pile, but at this point she speaks up. What does she say? Did the child make a mistake in sorting the pennies? Or does the number theorist have enough information to tell how many pennies are in the pile? We will answer this question with the Chinese Remainder Theorem. Here it is: Theorem Let \(m_1,m_2,\dots,m_k\) be natural numbers such that each is greater than 1, and every pair of them is relatively prime. Let \(M = m_1m_2\cdots m_k\), and let \(b_1,b_2,\dots,b_k\) be integers. The system of congruences \[\begin\). Before we prove the theorem, let’s see what the number theorist from our story may have said. Example Suppose that \(x\) is the number of pennies in the child’s pile. If we assume for a moment that the child didn’...

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