Find all pairs of odd integers a and b which satisfy the following equation: a + 128b = 3ab

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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For starters, try distributing the following:




Now, if a and b satisfy the equation, then `-3ab+a+128b=0`

So we have:

`(-3a+128)(3b-1) = 3(0)-128=-128`

So, the numbers `-3a+128` and `3b-1` multiply to `-128`

Now, we are told that a is odd, so -3a is odd, and -3a+128 is odd.

But -128 has the divisors 1,2,4,8,16,32,64,128 (and their negatives)

All of these numbers are even, except 1.

So we must have that `-3a+128 = +- 1`

So, either:

`-3a+128=-1` AND `3b-1=128`


`-3a+128=1` AND `3b-1=-128`

Solving the first pair of equations, we get `a=43` and `b=43`

Solving the second pair of equations, we get `a=42.33` and `b=-42.33`

But we are told that a and b are integers, so we must only have the solution

`a=43` and `b=43`

PS: How to derive `(-3a+128)(3b-1)`

We start with `a+128b-3ab=0`

We want to "factor" this somehow. Set up:


We want to multiply b by 128, so we have:


We want to multiply ab by -3, so we have:


We want to multipy a by 1, so we have:


We figure out what to make the last constant by multiplying everything out:


So we have:


Now, if `a+128b-3ab=0`, then:


In other words,


Multiply both sides by 3 to simplify the expression. You can choose to multiply either the (-3a+128) or the (b-1/3) by 3, but if you choose the second, you can get rid of all the fractions, and you get:


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