Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively.

Here is a general way to solve. It is equivalent to solving the system: {x≡8x≡12mod28,mod32.{x≡8mod28,x≡12mod32. There is a formula when the moduli are coprime. We'll reduce the problem to this case. Any solution has to be divisible by 44, so we'll set x=4yx=4y. The congruences can be written as {4y≡8mod284y≡12mod32⟺{y≡2mod7y≡3mod8 {4y≡8mod284y≡12mod32⟺{y≡2mod7y≡3mod8 Now a Bézout's relation between 77 and 88 is 8−7=18−7=1, hence the solutions for yy are y≡2⋅8−3⋅7=−5mod56, y≡2⋅8−3⋅7=−5mod56, whence x=4y≡−20mod224x=4y≡−20mod224. So the smallest positive value is x=204x=204. Added: More generally, one shows a system of linear congruences x≡aimodmi(i=1,…,r) x≡aimodmi(i=1,…,r) where the mimi are not necessarily mutually coprime, has a solution if and only if ∀i∀j,ai≡ajmodgcd(mi,mj) ∀i∀j,ai≡ajmodgcd(mi,mj) and in this case, the solution is unique modulo lcm(m1,…,mr)lcm⁡(m1,…,mr). read less