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x = 1(mod 3)

x = 1(mod 4)

x = 1(mod 5)

x = 0(mod 7)

We know, linear congurency is of form x = a_{n} (mod m_{n}) where n is an integer. So, Let us suppose

a_{1} = 1 |
m_{1} = 3 |

a_{2} = 1 |
m_{2} = 4 |

a_{3} = 1 |
m_{3} = 5 |

a_{4} = 0 |
m_{4} = 7 |

Now, m = m_{1} × m_{2} × m_{3} × m_{4}

= 3 × 4 × 5 × 7

= 420

Also,

M_{1} = \(\frac{m}{m_1} = \frac{420}{3}\) = 140

M_{2} = \(\frac{m}{m_2} = \frac{420}{4}\) = 105

M_{3} = \(\frac{m}{m_3} = \frac{420}{5}\) = 84

M_{4} = \(\frac{m}{m_4} = \frac{420}{7}\) = 60

Now, we need to find the inverse of M_{1} mod m_{1}, M_{2} mod m_{2}, M_{3} mod m_{3} and M_{4} mod m_{4}.

Now, inverse of M_{1} mod m_{1} i.e 140 mod 3 is

140 × 0 = 0 (mod 3)

140 × 1 = 2 (mod 3)

140 × 2 = 1 (mod 3)

∴ 2 is the invers of 140 mod 3

Let y_{1} = 2

Also, inverse of 105 mod 4 is

105 × 0 = 0 (mod 4)

105 × 1 = 1 (mod 4)

∴ 1 is the inverse of 105 mod 4.

Let y_{2} = 1

Also, inverse of 84 mod 5

84 × 0 = 0 (mod 5)

84 × 1 = 4 (mod 5)

84 × 2 = 3 (mod 5)

84 × 3 = 2 (mod 5)

84 × 4 = 1 (mod 5)

∴ 4 is the inverse of 84 mod 5

Let y_{3} = 4

Similarly, inverse of 60 mod 7

60 × 0 = 0 (mod 7)

60 × 1 = 4 (mod 7)

60 × 2 = 1 (mod 7)

∴ 2 is the inverse of 60 mod 7

Let y_{4} = 2

Now, we know,

a_{1} = 1, a_{2} = 1, a_{3} = 1, a_{4} = 0

M_{1} = 140, M_{2} = 105, M_{3} = 84, M_{4} = 60

y_{1} = 2, y_{2} = 1, y_{3} = 4, y_{4} = 2

The solutions to these systems are those

x such that

x = a_{1}M_{1}y_{1} + a_{2}M_{2}y_{2} + a_{3}M_{3}y_{3} + a_{4}M_{4}y_{4} mod m

= 1 × 140 × 2 + 1× 105 × 1 + 1 × 84 × 4 + 0 × 60 × 2 mo 420

= 280 + 105 + 446 (mod 420)

= 301

Hence, we conclde that 301 is the smallest +ve integer that leaves remainder 1 when divided by 3,4,5 and 7.

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