Tuesday, March 20, 2012

this week's MGRE Math Beast Challenge

From here:

Go to it! My attempt at an answer will eventually appear in the comments. (NB: The above image can be clicked to magnify.)

By the way, in case you're finding the mathematical expression difficult to read, I'm rewriting it here:

an = 2n - 1/(2n - 21)




Kevin Kim said...

I'm going to say that the answer is (A). This feels like a trick question whose trick is its simplicity. I may eat my words later on, but here's my reasoning:

It seems that, to compare sums, we can dispense with everything up to the 19th term, because Quantities A and B both begin at the same starting point, i.e., n-sub-1 (which I'll write as "n(1)" from now on; the comments don't allow subscript and superscript HTML).

If I'm right, then it's just a matter of comparing n(19) and n(20).

For n(19), then:

n(19) = 2^19 - (1/(2^(19 - 21))) =

2^19 - 1/(2^(-2)) =

2^19 - 4

Meanwhile, for n(20):

n(20) = 2^20 - (1/(2^(20 - 21))) =

2^20 - 1/(2^(-1)) =

2^20 - 2

Obviously, 2^20 is twice as large as 2^19. Adding or subtracting a 2 or a 4 makes almost no difference. That being the case, and given that the sum of n(1) to n(19) is the same for both Quantities A and B, I'd say it's safe to claim that Quantity A is larger-- significantly larger, in fact.

David Wester said...

However, much like the national debt, we have to take in consideration the accumulated "borrowing" of the first 18 instances of the equation. Does that quantity over ride the difference between line items 19 and 20?

mutzu501atgmail.com said...

sorry kevin, i screwed up on my last comment, just ignore it. i thought this was an accumulation of the series, but I see now that i read carefully its merely a question of the evaluation of n at 19 and n at 20. doh. so what they are really asking is if the right side of the equation drags down the value of n=20 enough to overcome 2 to the 20th power. What an insideous problem!

Kevin Kim said...


Right. I wondered about that, too, but as you implied in your second comment, there seems to be no getting around the fact that both quantities are equal up to n(19).

Here's a bit of trivia. There's a "zero point" that's neatly skipped in the series because, in order to get a zero result, n would have to be fractional (by which I mean, more precisely, not an integer). Before that zero point, all the terms of the series are negative; after that point, they're all positive. In an idle moment, I worked on this:

Set 2^n - (1/(2^(n - 21))) equal to zero and solve:

2^n - (1/(2^(n - 21))) = 0


2^n = (1/(2^(n - 21)))

2^n = 2^(21 - n) [via property of negative exponents]

Drop the bases (as if we were doing logarithms), and we see that

n = 21 - n

which becomes

2n = 21


n = 21/2, or 10.5

That's not a whole number, and there can be no "10.5th term" of a series, but it serves as a boundary: from n(1) to n(10), the results are always negative; beyond n(10), the results are always positive.

None of this is particularly relevant, but it is interesting in that it fleshes out the nature of the series we're dealing with. At the very least, it serves to confirm that, in our comparison of Quantities A and B, we can see that the sum of n(1) through n(19) is the same for both quantities.

Kevin Kim said...

An even quicker solution to this problem occurred to my slow brain:

We know that Quantities A and B have 19 out of 20 terms in common. We can therefore replace the sum of n(1) through n(19) with the value x, then compare quantities.

Quantity A: x + n(20)
Quantity B: x

So the only question is whether n(20) is positive. It's enough simply to calculate n(20), if that's even really necessary. If it's positive, then Quantity A is greater. If it's negative, then B is greater. If it's zero, then the answer is C, but as I mentioned in my previous comment, there's no zero term in this series because, for the expression to equal zero, n would have to equal 10.5, and there can be no "10.5th" term.

Wish I'd thought of all this earlier. 20/20 hindsight.