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

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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...
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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
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By remainder theorem n=(p×q)+r, where n=number, p=divisor, q=quotient, r=remainder From given data x=28y+8, x=32y+12 y=(x-8)/28, y=(x-12)/32 (x-8)/28=(x-12)/32 8(x-8)=7(x-12) 8x-64=7x-84 x=-84+64 x=-20 The smallest number is -20
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the smallest number that will be devided by 28 and 32 is ,their LCM = 224 , but there remainder 8 when devide by 28 , so 28 - 8 = 20. remainder 12 when devide by 32 . so 32 - 12 = 20 . Now 224 - 20 = 204 answer.
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