8 results.
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving an equation of the form ax≡bmodn where a and n are coprime.
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Given two numbers, find the gcd, then use Bézout's algorithm to find s and t such that as+bt=gcd(a,b).
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Given two numbers, find the gcd, then use Bézout's algorithm to find s and t such that as+bt=gcd(a,b). Finally, find all solutions of an equation \mod b.
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving a pair of congruences of the form \begin{align}x &\equiv b_1\;\textrm{mod} \;n_1\\x &\equiv b_2\;\textrm{mod}\;n_2 \end{align} where n_1,\;n_2 are coprime.
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving two simultaneous congruences:
\begin{eqnarray*} c_1x\;&\equiv&\;b_1\;&\mod&\;n_1\\ c_2x\;&\equiv&\;b_2\;&\mod&\;n_2\\ \end{eqnarray*} where \operatorname{gcd}(c_1,n_1)=1,\;\operatorname{gcd}(c_2,n_2)=1,\;\operatorname{gcd}(n_1,n_2)=1
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving an equation of the form ax \equiv\;b\;\textrm{mod}\;n where \operatorname{gcd}(a,n)|r. In this case we can find all solutions. The user is asked for the two greatest.
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving three simultaneous congruences using the Chinese Remainder Theorem:
\begin{eqnarray*} x\;&\equiv&\;b_1\;&\mod&\;n_1\\ x\;&\equiv&\;b_2\;&\mod&\;n_2\\x\;&\equiv&\;b_3\;&\mod&\;n_3 \end{eqnarray*} where \operatorname{gcd}(n_1,n_2,n_3)=1
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Question in Content created by Newcastle University by
Newcastle University Mathematics and Statistics
Solving three simultaneous congruences using the Chinese Remainder Theorem:
\begin{eqnarray*} x\;&\equiv&\;b_1\;&\mod&\;n_1\\ x\;&\equiv&\;b_2\;&\mod&\;n_2\\x\;&\equiv&\;b_3\;&\mod&\;n_3 \end{eqnarray*} where \operatorname{gcd}(n_1,n_2,n_3)=1
