The value of a as a whole is not necessarily relevant here. All we need to do is establish the units digit of a^36.

Whenever you see a units digit (UD) question involving exponents, the first thing to do is to establish the pattern of the UDs. For example, the pattern of UDs with powers of 3 is as follows:
3^1 = 3
3^2 = 9
3^3 = 7 (only worry about the UD)
3^4 = 1
3^5 = 3
3^6 = 9

You can see that the UD of powers of 3 repeat every 4 exponents. All UDs have similar patterns.

Statement (1) tells us that the UD of a^2 is 9. aman88 is right that we don't know what the UD of a is - it could be 3 or 7, both of which would have a UD of 9 when squared. But that doesn't mean that we don't know what the UD of a^36 is! For both 3 and 7, the UDs cycle every 4 exponents, so 3^4 has a UD of 1, and 7^4 has a UD of 1. So if the exponent is a multiple of 4, and we know that a is either 3 or 7, the UD has to be 1 in either case. The UD of a^36 = 1. SUFFICIENT

Statement (2) tells us that the UD of a^3 is 3. Unlike the first statement, this actually narrows a down to a single units digit: it has to be a UD of 7. If we know the UD of a, we can certainly find the UD of a^36. SUFFICIENT

varun289 wrote:What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 as the units digit


ans - D , while my answer is B , how ?

for this particular problem, there's also the following shorter approach:

* for statement 1, note that a^36 is (a^2)^18. therefore, according to the given information, a^36 is (something ending with 9)^18. whatever this is, it's going to be a unique value, so, sufficient.

* for statement 2, note that a^36 is (a^3)^12. therefore, according to the given information, a^36 is (something ending with 3)^12. whatever this is, it's going to be a unique value, so, sufficient.

so, done, (d).

--

of course, the approach given by ceilidh (above) is going to be more useful in the long run than this one, because it will apply to more cases.
still, more approaches = better.