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mba.com cat question: 2+2+2^2....

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by saritalr » Mon Nov 01, 2010 3:43 pm
Hi everyone. I'm still running into trouble with exponent problems. Here is an exponent problem from the online CAT at MBA.com.


2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

a)2^9
b)2^10
c)2^16
d)2^35
e)2^37

It's sort of hard to look at when the exponents don't display correctly. I suspected that the solution might involve factoring - but alas, that was about as far as I got. The correct answer is A (I guessed D). Any thoughts on a good approach for the problem?

Thank you in advance!
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by Brian@VeritasPrep » Mon Nov 01, 2010 4:38 pm
Hey saritalr:

Great question...I love this one! I actually wrote about this a while back on our blog:

https://www.veritasprep.com/blog/2010/09 ... many-twos/

Here's the text - and you were exactly right on about factoring! It just takes a few steps:
Veritas Prep Blog - GMAT Tip of the Week: Too Many Twos

This question brings up an important point about exponents - we only have a few "core competencies" when it comes to performing with algebra, and those are:

-Multiplying/dividing exponents with common bases
-Finding patterns (units digits, relationships between adding/subtracting common terms, etc.)
-Setting common bases equal to equate exponents

Outside of that, there's very little that we can do without the use of a calculator. So, in order to take advantage of what we do well, we should find ways when we see exponents to:

-Find common bases
-Multiply (using factorization to turn addition/subtraction into multiplication)

Here, we're asked to add several terms together...that's not something that we do well with exponents. However, by blending our abilities to factor terms (to get to multiplication) and to see patterns, we can attack this question relatively efficiently:

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Combine the 2s to be 4, or 2^2, and you have:

2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Now we can add them together, and we have 2(2^2), or 2^3, simplifying the entire statement to:

2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Notice that we'll be able to combine two more terms, the two 2^3 terms, to be 2(2^3) or 2^4, leaving:

2^4 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

By now hopefully you've seen a pattern (patterns come up frequently in exponent questions) - the first two terms will add to the third, and then adding those will add to the fourth:

2^4 + 2^4 (the first two) = 2^5 (so now we have two of the third term):

2^5 + 2^5 + 2^6 + 2^7 + 2^8

Do that again and we'll have:

2^6 + 2^6 + 2^7 + 2^8

If we repeat the pattern, we'll end up with:

2^8 + 2^8 = 2(2^8) = 2^9. Therefore, the correct answer is A.

When approaching exponent problems, keep your core competencies in mind: factor, multiply, find common bases, and look for patterns. These strategies will help you turn complicated problems into efficient solutions.
Brian Galvin
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Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
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by Rahul@gurome » Mon Nov 01, 2010 10:52 pm
One more method:

Let S = 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8.

So S - 2 = 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

Now 2*(S-2) = 2^2+2^3+2^4+2^5+2^6+2^7+2^8 + 2^9.

So 2*(S-2) - (S-2) = (2^2+2^3+2^4+2^5+2^6+2^7+2^8 + 2^9) - (2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)

= 2^9 - 2.

Or S-2 = 2^9 - 2
Or S = 2^9.

The correct answer is a).
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by saritalr » Tue Nov 02, 2010 9:56 am
Thanks to both of you. Brian - I believe I understand your explanation, combining the like terms one at a time.

Rahul - I'm not as clear on the explanation you posted. I see that in the firs step you subtracted 2 from both sides. But you lost me on the 2nd and 3rd step. It looks like you multiplied by two? then?

Thanks,
Sarah
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by Rahul@gurome » Tue Nov 02, 2010 7:02 pm
Rahul - I'm not as clear on the explanation you posted. I see that in the firs step you subtracted 2 from both sides. But you lost me on the 2nd and 3rd step. It looks like you multiplied by two? then?


Yes, I have multiplied by 2.
Now 2*(S-2) = 2^2+2^3+2^4+2^5+2^6+2^7+2^8 + 2^9. .................(1)
(S-2) = 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8.............(2).
If you subtract (2) from (1) , all terms will cancel except for 2^9 - 2.

So Left hand side is 2*(S-2) - (S-2) = (S-2) and right hand side is 2^9 - 2.
Or S - 2 = 2^9 - 2.
Or S = 2^9.

Hope its clear.
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