1. Chinese Remainder Theorem to solve `2x=5 (mod 7),3x=4 (mod 8)`

Solution:
Method-2: Successive substitution method
`2x-=5\ ("mod "7)->(1)`

`3x-=4\ ("mod "8)->(2)`

from (1)
`2x-=5\ ("mod "7)`

Here `2^(-1)-=4\ ("mod "7)`

`:. 4xx2x-=4xx5\ ("mod "7)`

`:. 8x-=20\ ("mod "7)`

`:. x-=6\ ("mod "7)`

Hence `x=6+7a`

Now from (2), `3x-=4\ ("mod "8)`

`:. 3(6+7a)-=4\ ("mod "8)`

`:. 18+21a-=4\ ("mod "8)`

`:. 2+5a-=4\ ("mod "8)`

`:. 5a-=2\ ("mod "8)`

Here `5^(-1)-=5\ ("mod "8)`

`:. 5xx5a-=5xx2\ ("mod "8)`

`:. 25a-=10\ ("mod "8)`

`:. a-=2\ ("mod "8)`

Hence `a=2+8b`

But `x=6+7a`

`:.x=6+7(2+8b)`

`:.x=6+14+56b`

`:.x=20+56b`

`:.x-=20\ ("mod "56)`

---

2. Chinese Remainder Theorem to solve `2x=6 (mod 14),3x=9 (mod 15),5x=20 (mod 60)`

Solution:
Method-2: Successive substitution method
`2x-=6\ ("mod "14)`

divide by 2, we get
`x-=3\ ("mod "7)->(1)`


`3x-=9\ ("mod "15)`

divide by 3, we get
`x-=3\ ("mod "5)->(2)`


`5x-=20\ ("mod "60)`

divide by 5, we get
`x-=4\ ("mod "12)->(3)`


from (1)
`x-=3\ ("mod "7)`

Hence `x=3+7a`

Now from (2), `x-=3\ ("mod "5)`

`:. (3+7a)-=3\ ("mod "5)`

`:. 3+2a-=3\ ("mod "5)`

`:. 2a-=0\ ("mod "5)`

Here `2^(-1)-=3\ ("mod "5)`

`:. 3xx2a-=3xx0\ ("mod "5)`

`:. 6a-=0\ ("mod "5)`

`:. a-=0\ ("mod "5)`

Hence `a=0+5b`

But `x=3+7a`

`:.x=3+7(0+5b)`

`:.x=3+0+35b`

`:.x=3+35b`

Now from (3), `x-=4\ ("mod "12)`

`:. (3+35b)-=4\ ("mod "12)`

`:. 3+11b-=4\ ("mod "12)`

`:. 11b-=1\ ("mod "12)`

Here `11^(-1)-=11\ ("mod "12)`

`:. 11xx11b-=11xx1\ ("mod "12)`

`:. 121b-=11\ ("mod "12)`

`:. b-=11\ ("mod "12)`

Hence `b=11+12c`

But `x=3+35b`

`:.x=3+35(11+12c)`

`:.x=3+385+420c`

`:.x=388+420c`

`:.x-=388\ ("mod "420)`

---

3. Chinese Remainder Theorem to solve `x=1 (mod 4),x=0 (mod 6)`

Solution:
`x-=1\ ("mod "4)->(1)`

`x-=0\ ("mod "6)->(2)`

`a_1=1,a_2=0`

`n_1=4,n_2=6`

Check if each `n_i` is pairwise coprime

`GCD(4,6)=2`

Here gcd is not 1, so the given equations has no solution.

---

Method-2: Successive substitution method
`x-=1\ ("mod "4)->(1)`

`x-=0\ ("mod "6)->(2)`

from (1)
`x-=1\ ("mod "4)`

Hence `x=1+4a`

Now from (2), `x-=0\ ("mod "6)`

`:. (1+4a)-=0\ ("mod "6)`

`:. 4a-=-1\ ("mod "6)`

`:. 4a-=5\ ("mod "6)`

Here, `gcd(4,6)=2` and 2 does not divide 5

So this is impossible

---

---