Exact Powers

[haoyang_qu]

The following was a Q from my CAT exam. The red is my question within the explanation. Thank you.

Q:
Is x > 10^10 ?

(1) x > 2^34

(2) x = 2^35


Explanation given:
1) SUFFICIENT: Statement(1) tells us that x > 2^34, so we want to prove that 2^34 > 1010. We'll prove this by manipulating the expression 2^34.

2^34 = (2^4)(2^30)
2^34 = 16(2^10)^3

Now 2^10 = 1024, and 1024 is greater than 10^3. Therefore:

2^34 > 16(10^3)^3
Where did the 16(10^3)^3 come from? Are we suppose to see if 2^34 is > than 10^10?
2^4 = 16.
10^10 = 2^10 x 5^10.
16(10^3)^3 = 2^4(10^9)
2^34 > 16(10^9)
2^34 > 1.6(10^10).

Since 2^34 > 1.6(10^10) and 1.6(10^10) > 10^10, then 2^34 > 10^10.

(2) SUFFICIENT: Statement (2) tells us that that x = 2^35, so we need to determine if 2^35 > 10^10. Statement (1) showed that 2^34 > 10^10, therefore 2^35 > 10^10.

The correct answer is D.

[JonathanSchneider]

Note that 2^10 = 1,024. Note also that 10^3 = 1,000. Of course, 1,024 > 1,000, so we know that 2^10 > 10^3. This is a handy method for comparing the exponential expressions in the given problem, as we wish to compare some number of 2's to some number of 10's. Because we can show that 2^34 = (1024)(1024)(1024)(16), we can show that it is greater than 10^10.

[haoyang_qu]

Thank you. Your last sentence sums it up perfectly.

[esledge]

On a more general note, it's a good idea to have certain "magic numbers" up your sleeve on the GMAT. You should review the first few powers of the first 10 integers, at least through the 3-digit numbers. For example:

2: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, etc.
3: 3, 9, 27, 81, 243, 729, etc.
4: 4, 16, 64, 256, 1024, etc.

1024's a neat one, as you have seen, because it is just above 1000. It's also handy because it's a power of both 2 and 4.

It's not necessary to memorize these, exactly, but at least be able to recognize them as special. When I see 243 on the test, I may not recall exactly which power of 3 it is, but I know there is something I can do with it, giving me a place to start.

[manochsa]

I found the official answer to this question somewhat difficult to follow. Could someone please tell me if I would be correct in analyzing Statement 1 as follows:

Is 2^34 > 10^10?
Is (2^10 * 2^10 * 2^10 * 2^4) > (10^3 * 10^3 * 10^3 * 10)?

Since each 2^10 (1024) factor is greater than each 10^3 (1000) factor, we can simplify:
Is 16>10. Yes. Statement 1 is sufficient.

Thanks!

[Ben Ku]

I found the official answer to this question somewhat difficult to follow. Could someone please tell me if I would be correct in analyzing Statement 1 as follows:

Is 2^34 > 10^10?
Is (2^10 * 2^10 * 2^10 * 2^4) > (10^3 * 10^3 * 10^3 * 10)?

Since each 2^10 (1024) factor is greater than each 10^3 (1000) factor, we can simplify:
Is 16>10. Yes. Statement 1 is sufficient.

Thanks!

Yes, your approach is good.

[griffin.811]

I follow the logic on this perfectly fine, and understand the explanation.

My question, however, is how do we know where to start? I get that 2^34=16(2^10)^3, but it also equals a bunch of other combinations; why use this one?

Same for the 10^10 experssion; what should we key in on to recognize that we need to express this as (10^3)(10^3)(10^3)(10), instead of some other combination?

Thanks

[jlucero]

I follow the logic on this perfectly fine, and understand the explanation.

My question, however, is how do we know where to start? I get that 2^34=16(2^10)^3, but it also equals a bunch of other combinations; why use this one?

Same for the 10^10 experssion; what should we key in on to recognize that we need to express this as (10^3)(10^3)(10^3)(10), instead of some other combination?

Thanks

That's a good question and one that has no perfect explanation. Solving GMAT problems isn't always a science, so having mental flexibility is a key component to achieving a great score. There's a lot of ways to get an answer to this problem (I would cross out the 2s and then compare the similar values of 5^3 = 125 and 2^7 = 128). But more importantly, to me, this question tests a student's ability to compare things that aren't perfectly alike. While most exponent questions base themselves on canceling things out and finding things that are equal, this would puzzle the student who tries to do the same thing. But the student who can recognize that this method won't work here will be able to try something else out without getting discouraged. To me, the student who can write out the first few multiples of 2, 5, or 10 and find a similar enough comparison is going to be able to do this question, while the student who gets stuck and feels like he/she is missing something is going to take too long on this problem.

Moral of the story: find methods, get great at those methods, but be ready to abandon those same methods and try something else when the methods don't work.

[griffin.811]

Thanks for the quick response!

[tim]

:)

[RonPurewal]

I follow the logic on this perfectly fine, and understand the explanation.

My question, however, is how do we know where to start? I get that 2^34=16(2^10)^3, but it also equals a bunch of other combinations; why use this one?

You don't need to recognize this. Take a look at the second post on the thread (from Jonathan Schneider), who basically just multiplied together a bunch of 2's until he was like, "Oh hey, that's a little over a thousand". At which point he had a ready point of comparison with powers of 10.

Whenever there are "clever" approaches to problems, there will ALWAYS be less clever, more "grind"-y approaches, too. I don't think I've ever seen an official problem that REQUIRES a huge helping of "cleverness".

[griffin.811]

Thanks Ron.

I stuck with this problem and finally found an approach that works for me, and I think would transfer well to similar problems. Figured I'd post here in case it helps others.

in statement 1 we need to compare 2^34 to 10^10. Let's square both sides to 2^17 and 10^5. 2^17=2^10 * 2^7. This becomes 1024*128. 128 is ~100, so call it 102,400. We know 10^5=100,000, so since 2^17 is greater than 10^5, 2^34 is greater than 10^10 therefore, since x>2^34, x>10^10.

Statement 2 is self explanatory, if 2^34 is greater than 10^10, then so is 2^35.

Both sufficient alone.

Best

[tim]

Simplification and estimation definitely help on this one!

[RonPurewal]

2^17=2^10 * 2^7. This becomes 1024*128. 128 is ~100, so call it 102,400.

The way you did this is ... interesting.

Basically, you could have chosen to estimate either of these two quantities.

I.e., you have two options:

1/
You can round 128 down to 100 (that's a reduction of more than 20% from the original value), and leave 1024 alone.

2/
You can round 1024 down to 1000 (that's a reduction of only about 2 percent), and leave 128 alone.

You clearly understand what you're doing, and, moreover, the reasoning is identical between the two options. So, it's, er, interesting that you'd choose option #1, considering that its relative error is about 10 times as big.

[RonPurewal]

Further notes on the above:

• In this problem, it doesn't matter which option you choose, because the point is that the product is greater than 100,000—a conclusion that can be reached by either pathway. But clearly you can see how a tenfold difference in accuracy could matter!

• The direction of rounding matters, too.
In this problem, both options involve rounding down, so you're indifferent between them. But let's say the product was, I don't know, 92 x 1111.
– If you made this into 92 x 1000, which was still bigger than some other value, then your original 92 x 1111 is also bigger than that value.
– However, if you made this into 92 x 1000, which was then smaller than some other value, that result would be inconclusive as far as a comparison between 92 x 1111 and that value.
– In both cases, the reverse would be true for rounding up to 100 x 1111.