Manhattan GMAT Challenge Problem of the Week – 14 Sept 2010

by Manhattan Prep, Sep 14, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.

As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

The sequence [pmath]S_n[/pmath] is such that, for every n where n > 1, [pmath]S_n=(S_{n-1}-1)^2[/pmath]. If [pmath]S_5[/pmath] = 100, what is [pmath]S_3[/pmath]?

(A) 10

(B) 11

(C) 11

(D) 11 + 1

(E) 101

Answer

This problem is much easier if we can decode the sequence [pmath]S_n=(S_{n-1}-1)^2[/pmath]

In other words, [pmath]S_n[/pmath] is just any term in the sequence (for instance, [pmath]S_3[/pmath] would be the 3rd term, [pmath]S_10[/pmath] would be the 10th term, etc.)

[pmath]S_{n-1}[/pmath] just means the term that comes right before [pmath]S_n[/pmath].

Therefore, we can rephrase [pmath]S_n=(S_{n-1}-1)^2[/pmath] as To get any term, take the term before it, subtract one, then square.

In this problem, however, weve been given [pmath]S_5[/pmath] and we want [pmath]S_3[/pmath]. That is, we must go backwards.

To go backwards in a sequence, do the opposite of each step, in the opposite order. That is, if, to go from [pmath]S_1[/pmath] to [pmath]S_2[/pmath] you would:

1) subtract 1

2) square

...then to go backwards, you would:

1) square root

2) add 1

So, if [pmath]S_5[/pmath] = 100, square root to get 10 and add 1 to get 11. Notice that we do not have to worry about the possibility of a negative square root, because every term is the square of some number, so no term can be negative.

If [pmath]S_4[/pmath] = 11, square root to get 11 and add 1 to get 11 + 1. The answer is D.

The problem could also be solved a bit more algebraically as follows:

[pmath]S_n=(S_{n-1}-1)^2[/pmath]

[pmath]S_5[/pmath] = 100

Therefore:

100 = [pmath](S_4-1)^2[/pmath]

10 = [pmath]S_4[/pmath] 1(Again, we drop the possibility of a negative root, because [pmath]S_4[/pmath] itself is a square.)

[pmath]S_4[/pmath] = 11

Now repeat the process:

11 = [pmath](S_3-1)^2[/pmath]

11 = [pmath]S_3[/pmath] 1

[pmath]S_3[/pmath] = 11 + 1

The correct answer is D.

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