# Find all the pairs of odd integers 'A' and 'B' which satisfy the following question: A+128B = 3AB Pls Provide Working

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### 1 Answer

You need to isolate the term 128B to the left such that:

`128B = 3AB - A`

Factoring out A to the right yields:

`128B = A(3B - 1)`

Since the problem provides the information that A and B are odd, hence, 3B is odd, but `3B-1` is even.

Since `128B = A(3B - 1), ` then the even number `3B - 1` needs to be the divisor of 128 such that:

`D_128 = {+-1;+-2;+-4;+-8;+-16;+-32;+-64;+-128}`

You need to give values to `3B - 1` such that:

`3B-1 = 1 =>3B = 2 => B = 2/3 !in Z`

`3B-1= -1 =>3B =0 =>B = 0 in Z`

`3B-1=2 =>3B =3 =>B =1 in Z`

`3B-1= -2=>3B =-1 =>B = -1/3 !in Z`

`3B-1=4 => 3B =5 =>B = 5/3 !in Z`

`3B-1= -4 =>3B = -3 => B =-1 in Z`

`3B-1=8 => 3B =9 => B = 3 in Z`

`3B-1= -8 =>3B =-7 =>B = -7/3 !in Z`

`3B-1=16 =>3B = 17=>B = 17/3 !in Z`

`3B-1=-16 =>3B =-15 =>B =-5 in Z`

`3B-1=32 =>3B =33 =>B =11 in Z`

`3B-1=-32 =>3B = -31 =>B = -31/3 !in Z`

`3B-1 =64 =>3B = 65=>B = 65/3 !in Z`

`3B-1=-64 =>3B = -63 =>B = -21 in Z`

`3B-1=128 =>3B = 129 =>B =43 in Z`

`3B-1= -128 =>3B = -128/3 => B = -128/3 !in Z`

Notice that the following values of B satisfy the condition that B needs to be an odd integer.

`B in {-21,-5,-1,0,1,3,11,43}`

Since the relation `128B = A(3B - 1)` proves that `3B - 1` divides 128, hence, A needs to be one of divisors of B such that:

`B = -21 => A in {+-1 ; +-3 ; +-7}`

`B = -5 => A in {+-1 ; +-5} `

`B = 0 => Ain 0`

`B= 1 => Ain {+-1}`

`B = 3 => A in {+-1 ; +-3}`

`B = 11 => A in {+-1 ; +-11}`

`B = 43 => A in {+-1 ; +-43}`

**Hence, evaluating the odd integer values that satisfy the relation `128B = A(3B - 1)` yields `{-21,-5,-1,0,1,3,11,43}.` **