Data Sufficiency

When dealing with UNITS DIGITS and exponents, we must look for a pattern.

If \(a\) is an integer, what is the units digit of \(a^{18}?\)

(1) \(a^2\) has a units digit of 9.

We can express \(a^{18}\) as \((a^2)^9\) . If \(a^2\) has a units digit of \(9\), then \(a^{18}\) will have the same units digit as \(9^9\).

We don't even have to calculate - we know that \(9^9\) will have a knowable units digit. Just as a reminder, though, units digits will always follow a pattern. Here is the pattern for the units digit of 9:
\(9^1\) = 9
\(9^2\) = ...1
\(9^3\) = ...9
\(9^4\) = ...1

9 raised to any odd exponent will end in 9, and to any even exponent will end in 1. Thus, \(9^9\) will end in 9.

(2) \(a^7\) has a units digit of 3.

We cannot easily relate \(a^7\) to \(a^{18}\). So, we must look for numbers that have a units digit of 3 when raised to the 7th power. You might already know that 3 and 7 are the only units digits that can yield a 3. (No even digits when raised to a power will ever end in an odd digit, and \(1^x\) and \(5^x\) will always end in 1 and 5, respectively. As shown above, \(9^x\) can only end in 9 or 1).

Test the powers of 3:
\(3^1\) = 3
\(3^2\) = 9
\(3^3\) = ...7
\(3^4\) = ...1
\(3^5\) = ...3
\(3^6\) = ...9
\(3^7\) = ...7
This doesn't fit.

Try 7:
\(7^1\) = 7
\(7^2\) = ...9
\(7^3\) = ...3
\(7^4\) = ...1
\(7^5\) = ...7
\(7^6\) = ...9
\(7^7\) = ...3

This fits! We know that the units digit of \(a\) must be 7. This is sufficient to know what the units digit of \(a^{18}\) would be.

Again, don't waste your time on test day figuring out what the units digit actually is. But for reference, if \(a^{18}\) is \(...7^{18}\), we can use the fact 7's repeat every 4 powers (see above: \(7^1\) has the same units digit as \(7^5\).) Every \(7^{4x}\) will end in a 1. Thus,
\(7^{16}\) = ...1
\(7^{17}\) = ...7
\(7^{18}\) = ...9

The answer is D.