Using the Chinese Remainder Theorem, solve:

x ≡ 1 mod 2

x ≡ 2 mod 3

x ≡ 3 mod 5

x ≡ 4 mod 11

GCF(2,3) = 1

GCF(2,5) = 1

GCF(2,11) = 1

GCF(3,5) = 1

GCF(3,11) = 1

GCF(5,11) = 1

Since all 6 GCF calculations equal 1

the n_{i}'s are pairwise coprime

We can use the regular CRT Formula

Take the product of each n_{i}

N = n_{1} x n_{2} x n_{3} x n_{4}

N = 2 x 3 x 5 x 11

N = 330

c_{i} = | N |

n_{i} |

c_{1} = | 330 |

2 |

c_{1} = 165

c_{2} = | 330 |

3 |

c_{2} = 110

c_{3} = | 330 |

5 |

c_{3} = 66

c_{4} = | 330 |

11 |

c_{4} = 30

x = a_{1}(c_{1}y_{1}) + a_{2}(c_{2}y_{2}) + a_{3}(c_{3}y_{3}) + a_{4}(c_{4}y_{4})

x = a_{1}(165y_{1}) + a_{2}(110y_{2}) + a_{3}(66y_{3}) + a_{4}(30y_{4})

Note: The a_{i} piece is factored out

We will use this below

Using 1 modulus of 2 and c_{1} = 165

calculate y_{1} in the equation below:

Using 2 modulus of 3 and c_{2} = 110

calculate y_{2} in the equation below:

Using 3 modulus of 5 and c_{3} = 66

calculate y_{3} in the equation below:

Using 4 modulus of 11 and c_{4} = 30

calculate y_{4} in the equation below:

x = a_{1}(165y_{1}) + a_{2}(110y_{2}) + a_{3}(66y_{3}) + a_{4}(30y_{4})

x = 1 x 165 x 1 + 2 x 110 x -1 + 3 x 66 x 1 + 4 x 30 x -4

x = 165 - 220 + 198 - 480

x = **-337**

-337 ≡ 1 mod 2

Add remainder of 1 to -338 = -337

-337 ≡ 2 mod 3

Add remainder of 2 to -339 = -337

-337 ≡ 3 mod 5

Add remainder of 3 to -340 = -337

-337 ≡ 4 mod 11

Add remainder of 4 to -341 = -337

-337

-337

Free Chinese Remainder Theorem Calculator - Given a set of modulo equations in the form:

x ≡ a mod b

x ≡ c mod d

x ≡ e mod f

the calculator will use the Chinese Remainder Theorem to find the lowest possible solution for x in each modulus equation.

Given that the n_{i} portions are not pairwise coprime and you entered two modulo equations, then the calculator will attempt to solve using the Method of Successive Subsitution

This calculator has 1 input.

x ≡ a mod b

x ≡ c mod d

x ≡ e mod f

the calculator will use the Chinese Remainder Theorem to find the lowest possible solution for x in each modulus equation.

Given that the n

This calculator has 1 input.

- algorithm
- A process to solve a problem in a set amount of time
- chinese remainder theorem
- ancient theorem that gives the conditions necessary for multiple equations to have a simultaneous integer solution
- coefficient
- a numerical or constant quantity placed before and multiplying the variable in an algebraic expression
- equation
- a statement declaring two mathematical expressions are equal
- gcf
- greatest common factor - largest positive integer dividing a set of integers
- modulus
- the remainder of a division, after one number is divided by another.

a mod b - product
- The answer when two or more values are multiplied together
- remainder
- The portion of a division operation leftover after dividing two integers
- substitution
- a simple way to solve linear equations algebraically and find the solutions of the variables.
- theorem
- A statement provable using logic

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