Typo Error

Obviously neither $a$ nor $b$ can be large because the right-hand side is bounded above by $3000$ and the left-hand side grows exponentially. Observe that the only powers of $9$ to consider are $9^{0} = 1, 9^{1} = 9, 9^{2} = 81$ , and $9^{3} = 729$ . Since $9^{4} = 6561$ , the number $(2^{a})(9^{b})$ = $\overline{2a9b}$ cannot be a power of $9$ , so $a$ cannot be $0$ and $b$ must be even.

Now consider cases:

• $b = 0$ gives a contradiction since $2^{a} = \overline{2a90}$ is impossible because no power of $2$ ends in $0$ .

• $b = 2$ gives $(81) (2^{a}) = \overline{2a92}$ , so that $\overline{2a92}$ must be divisible by $9$ . This requires that $2 + a + 9 + 2 = a + 13$ be a multiple of $9$ . The only value of $a$ in the required interval $[0, 1, 2, 3...9]$ is $a = 5$ . In this case, $2^{5} \cdot 9^{2} = 2592$ . Thus $a = 5, b = 2$ and $\boxed{a + b = 7}$ .

We note that the number must be a multiple of 9. So a+b can only be 7 or 16 so that the sum of digits of the number is a multiple of 9. a+b can't be 16 as the minimum value of the number in that case is 65536 where a=16, b=0 which is a 5 digit number and other values of a and b for which a+b is 16 will give even bigger values. So a+b is obviously 7.

Since $9^4 > 3000$ , $b < 4$ . Moreover, $b$ must be even because there is no power of 9 of the given form. Also, $b$ can't be zero because the number does not contain factors 5. This leaves $b = 2, 9^b = 81$ .

Now we know that the calculation is $2^a \cdot ...1 = ...2$ , which shows that $2^a$ ends in the digit 2. Candidates are $2^1 = 2$ (too small), $2^5 = 32$ (just right), or $2^9 = 512$ (too large).

Thus $a + b = 5 + 2 = 7$ .