Exponents Lead to Cumbersome, Time-Consuming Calculations involving Large Numbers but it is Pattern-Driven

GMAT ExponentsThe GMAT's quantitative section is increasingly emphasizing problem solving skills over calculation abilities, and often does so in the form of "Number Properties" questions. The authors of the exam are also quite adept at recognizing "mathematical psychology", and creating questions that increase an examinee's anxiety by enough to make that process of problem solving a bit more difficult. One of the major themes that arises as a result is the use of exponents, which carry with them a number of properties extremely useful to the writers of the GMAT.


• Inspire fear (or at least apprehension) in test takers
• Lead to cumbersome, time-consuming calculations involving large numbers
• Are actually quite pattern-driven, and reward those who seek out those patterns rather than attempt to perform the extensive calculations

How can this help you on the exam?

If you embrace the pattern-driven quality of exponents, you can rest easy on exponent questions involving large numbers, knowing that you can test the pattern with small numbers, and simply extrapolate it to solve the overall question. Take, for example, a question that asks:

What is the sum of the digits of 10^25 - 37?

Listing out the numbers will be time consuming and contains the potential for error (counting to 25 when writing out the zeroes, then writing out that many digits in the difference, is a tedious process). But if you recognize that the first number will simply be "1" followed by 25 zeroes, and that the difference of the two will end in 63, preceded by a series of 9s, you can make quick work of this problem.

Try using a smaller exponent to see what the result will look like:

10^4 - 37:

- 37

If the exponent is 4, we end up with 4 digits in the answer: a 6, a 3, and the rest are 9s. Trying again with another exponent, we can see that the pattern holds for any exponent:

10^6 - 37:

- 37

Again, we have the same number of digits in the difference as the value of the exponent, and two of those are 6 and 3, with the others 9s.

So, we can conclude that 10^x - 37 will give us a solution in which we have (x-2) 9s, a 6, and a 3.

Because 6+3 is 9, the sum of the digits will be (x-2) * 9 + 9, or (x-1) * 9.

In the original example - 10^25 - 37, the exponent is 25, so we'll have (25-1) * 9, or 24*9, and the answer is 216.

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