In the last math post, we dealt with exponents and really big numbers in quantitative comparison. Let’s now try the same but in the problem solving context.

The general approach is not too different—you want to look for a pattern. However this time you will have to be exact, instead of simply saying which side is bigger. One tactic the GRE employs is to make a problem look impossible to solve in 2 minutes. But don’t despair. All problems are solvable once you unlock them. So if your approach is taking too much time, step back and look at the problem from a new angle.

Try the following and see if you can solve it in fewer than 2 minutes.

What is the sum of the digits of integer x, where x = 4^10 x 5^13?

(A) 13

(B) 11

(C) 10

(D) 8

(E) 5

If you spotted the pattern early on, you may have only required a minute. But if you couldn’t unlock the problem, here is how you do so. Notice that 4^10 can be rewritten as 2^20. We can now express x as 2^20 x 5^13. The logic here is that 2 x 5 = 10. That is, 10 to any integer power greater than 1,will be a 1 followed by zeroes

So now let’s rewrite the problem again so we get 2^7 x 2^13 x 5^13. Combine 2^13 x 5^13 and we get 10^13. That is, we get 1 followed by 13 zeroes. If you are taking the sum, it’s straightforward: 1 13 zeroes is 1. We are not done yet as we have the 2^7. When you multiply this out, you get 128. 128 x 1,000 thirteen zeroes is equal to 128 followed by the thirteen zeroes. Ignore the zeroes and we get 1 + 2 + 8, which equals 11. Answer B