TomV's answer is excellent in its simplicity. Just note a and b are integers between 1 and 9, inclusive.
Let's try a somewhat different approach though. Let us prove that no solution exists. We will need to prove two statements:
(a) For any nonnegative integer z, 9 divides 10^z - 1. (b) Suppose that k is an integer. If m is an integer whose digits are determined by transposing two of those from k, then k - m is divisible by 9.
Proof of (a). Clearly, for z = 0, the statement is true as 10^0 - 1 = 0 = 9(0). Suppose, then, for integers 1, 2, 3, ..., k, that 9 divides 10^k - 1. Then there is an integer s such that 9s = 10^k - 1 --> 10^k = 9s + 1. So,
and 9 divides 10^(k + 1) - 1. So, by induction, 9 divides 10^z - 1 for every nonnegative integer z.
Proof of (b): Let k = a_n 10^n + a_(n-1)10^(n-1) + ... +a_s 10^s + ... + a_t 10^t + ... + a_1 (10) + a_0 where the a_i's are integers between 0 and 9. Then, the integer m is of the form m = a_n 10^n + a_(n-1)10^(n-1) + ... +a_t 10^s + ... + a_s 10^t + ... + a_1 (10) + a_0. So,
Clearly, if s = t, then no transposition occurs and k - m = 0, which is clearly divisible by 9. So, suppose that s > t (we will omit the last case of s < t for it is very similar to this one). Then,
But, s - t > 0 --> 9 divides 10^(s - t) - 1. So, 9 must divide the difference k - m. In the case s < t, we get the same result. Therefore 9 divides any such difference k - m (a difference between numbers whose digits are transposed). _____________________________
To finish this problem, note that 9 does not divide 30. But, you claim that the difference between a number and one of its transposes is 30. So, 9 divides 30. This is obviously false (Since 30 = 9(3) + 3)! As a result, this problem is unsolvable.