Bunuel wrote:

What is the 101st digit after the decimal point in the decimal representation of \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}\)?

A. \(0\)

B. \(1\)

C. \(5\)

D. \(7\)

E. \(8\)

An alternative approach: This one is for people (like me) who have some number sense but are not as studied or quick on the draw as a math expert. The first three fractions would be easy enough to combine, but that forth denominator throws things off. What about the individual decimals instead?

\(\frac{1}{3}\) = 0.333333...

\(\frac{1}{9}\) = 0.111111... (this is one of the more well-known properties of dividing by 9)

\(\frac{1}{27}\) = 0.037037... (via long division and pattern recognition: once you hit the second 3, you can see that the next digit would be 7)

\(\frac{1}{37}\) = 0.027027... (again, long division and pattern recognition)

At this point, you can chunk or cluster some of the repeating decimals by finding their sum. I split the four decimals into the first two and second two:

0.333333 + 0.111111 = 0.444444 and

0.037037 + 0.027027 = 0.064064

Adding these chunks to each other yields 0.508508...

Now, we are getting really close. That is,

every third digit will be an 8, so we just need to figure out on which digit the 101st spot will land. Since 33 repetitions of 3 gets us to 99, we can pick up the pattern there:

99th digit: 8

100th digit: 5

101st digit: 0

Of course, the 102nd digit would loop back to an 8, the 103rd to a 5, and so on, but the underlying principle is that whether we are looking for the 101st or the 1001st digit, we can use pattern recognition to arrive the correct response fairly efficiently and crack a tough question.

Note: I am not saying that this method is better than the official solution proposed above, just that it can help some test-takers mitigate a panic response or a compulsion to guess at random.

Good luck with your studies.

- Andrew