Solve three congruences with the Chinese Remainder Theorem

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			Name	Type	Generated Value

pans	integer	67
co	list	Nested 9×3 list
na	integer	45
age	integer	67
nc	integer	20
ea	integer	45
ec	integer	-80
n	integer	180
t	integer	8
rb	integer	2
rc	integer	4
sc	integer	9
sb	integer	5
sa	integer	4
eb	integer	36
nb	integer	36
ra	integer	3

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xxxxxxxxxx
 
mod(ra*ea+rb*eb+rc*ec,n)

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Solve the following simultaneous congruences.
On division by $\var{sa}$ my age gives remainder $\var{ra}$ .
On division by $\var{sb}$ my age gives remainder $\var{rb}$ .
On division by $\var{sc}$ my age gives remainder $\var{rc}$ .

Hold old am I?

age $\;=$ Unnamed gap

Your value of the age should satisfy $29 \leq x \leq 73$

Steps gives you more information on solving the Chinese Remainder Theorem.

Advice

After converting the statements about age into congruences we obtain the following congruences:

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

Following the notation as set out 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 solution is:

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

$\var{age}$ .

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Solve three congruences with the Chinese Remainder Theorem

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Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Solve three congruences with the Chinese Remainder Theorem

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

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Generated value: `integer`