Let AB and BA be two-digit numbers, such as 23 and 32. If AB – BA=45, find the sum of all possible values of AB.

We can write:

BA = B*10 + A

AB = A*10 + B

where A and B are integers.

And we know that:

AB - BA = 45

A*10 + B - B*10 - A = 45

A*9 - B*9 = 45

(A - B)*9 = 45

(A - B) = 45/9 = 5

Then we must have that A = B + 5.

then the possible values of AB are:

if B = 0, A = 0 + 5 = 5

AB = 50

if B = 1, A = 5  + 1 =6

AB = 61

if B = 2, A = 5 + 2 = 7

AB = 72

if B = 3, A = 3 + 5  = 8

AB = 83

if B = 4, A = 4 + 5 = 9

AB = 94

And we do not have more possible numbers of two digits with these conditions:

The summ of all of them will be:

50 + 61 + 72 + 84 + 94 = 360.

This is the case where AB is a positive number. (i suppose that this is what you wanted)

But let's be more complete, just because we can:

If AB can also be negative, then the possible values of AB negatives will be equal to -BA for each case of the ones shown above:

AB = -05 (this is not a two digit number, so we can discard this)

AB = -16 AB - BA = -16 - (-61) = 45

AB = -27 AB - BA = -27 - (-72) = 45

AB = -38 AB - BA = -38 - (-83) = 45

AB = -49 AB - BA = -49 - (-94) = 45

The sum if we also include these will be:

360 - 16 - 27 - 38 - 49 = 230

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