Blathnaid's copy of Solve three congruences using Chinese Remainder Theorem

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Blathnaid Sheridan

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Blathnaid Sheridan 6 years, 10 months ago

Created this as a copy of Solve three congruences using Chinese Remainder Theorem.

Name	Status	Author	Last Modified
Solve three congruences using Chinese Remainder Theorem	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Blathnaid's copy of Solve three congruences using Chinese Remainder Theorem	draft	Blathnaid Sheridan	11/05/2018 12:11
Blathnaid's copy of Solve three congruences using Chinese Remainder Theorem	draft	Blathnaid Sheridan	18/11/2019 13:17

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

pans	number	305
na	number	30
nb	number	26
nc	number	195
ea	number	-90
rc	integer	1
ec	number	195
n	number	390
ra	integer	6
rb	integer	5
ans	number	305
sc	number	2
sb	integer	15
sa	number	13
eb	number	-104

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Data type

Value

xxxxxxxxxx
 
mod(ra*ea+rb*eb+rc*ec,n)

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Description

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Generated value: `number`

305

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Gap-fill
1. Gap Number entry
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Solve the following simultaneous congruences:

$\begin{eqnarray*} x\;&\equiv&\;\var{ra}\;&\mod&\;\var{sa}\\ \\ x\;&\equiv&\;\var{rb}\;&\mod&\;\var{sb}\\ \\ x\;&\equiv&\;\var{rc}\;&\mod&\;\var{sc} \end{eqnarray*}$

$x=\;\;$ Unnamed gap

Your value of $x$ should satisfy $0 \leq x \lt \var{n}$

Advice

Using the notation in Steps we have:

1) $n_1=\var{sa},\;n_2=\var{sb},\;n_3=\var{sc}$ are coprime as they have no factors in common.
$n=n_1n_2n_3=\var{n}$

2)
$N_1=\var{sb}\times \var{sc}=\var{na},\; N_2=\var{sa}\times \var{sc}=\var{nb},\;N_3=\var{sa}\times \var{sb}=\var{nc}$

Using the Extended Euclidean Algorithm for $N_1=\var{na}$ and $n_1=\var{sa}$ we find:

$\simplify[std]{{extendedgcd2(sa,na)}* {na} +{extendedgcd1(sa,na)}*{sa}=1} \Rightarrow E_1=\var{extendedgcd2(sa,na)}\times \var{na}=\var{ea}$
Using the Extended Euclidean Algorithm for

$N_2=\var{nb}$ and

$n_2=\var{sb}$ we find:

$\simplify[std]{{extendedgcd2(sb,nb)}* {nb} +{extendedgcd1(sb,nb)}*{sb}=1} \Rightarrow E_2=\var{extendedgcd2(sb,nb)}\times \var{nb}=\var{eb}$
Using the Extended Euclidean Algorithm for

$N_3=\var{nc}$ and

$n_3=\var{sc}$ we find:

$\simplify[std]{{extendedgcd2(sc,nc)}* {nc} +{extendedgcd1(sc,nc)}*{sc}=1} \Rightarrow E_3=\var{extendedgcd2(sc,nc)}\times \var{nc}=\var{ec}$
4)
Then the general solution is:

$\begin{eqnarray*} x&=& a_1 E_1+a_2 E_2+a_3 E_3\;\mod\;n\\ &=& \simplify[std]{{ra}*{ea}+{rb}*{eb}+{rc}*{ec}}\;\mod\;\var{n}\\ &=&\var{ra*ea+rb*eb+rc*ec}\;\mod\;\var{n}\\ &=& \var{ans}\;\mod\;\var{n} \end{eqnarray*}$

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This question is used in the following exams:

Paul's copy of CMPU2012-CA by Paul Molloy in Paul's workspace.
CMPU2012-CA by Blathnaid Sheridan in Blathnaid's workspace.

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Blathnaid's copy of Solve three congruences using Chinese Remainder Theorem

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Taxonomy: mathcentre

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Blathnaid Sheridan 6 years, 10 months ago

Error in variable testing condition

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Generated value: `number`

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Gap-fill

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Blathnaid's copy of Solve three congruences using Chinese Remainder Theorem

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Blathnaid Sheridan 6 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`