## question from mba.com CAT

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### question from mba.com CAT

by saritalr » Fri Jul 30, 2010 11:46 am
Hi, this is my first question post. The question comes from the first CAT from the mba.com site. I don't believe there are explanations out there for these problems, but please correct me if I'm wrong. Similarly, I had trouble searching the forum to see if my question (or a similar one) had already been posted. Is there a recommended way to search? I'd hate post a bunch of repeats.

Which of the following is equal to the value of 2^5 + 2^5 + 3^5 + 3^5 + 3^5.

1. 5^6
2. 13^5
3. 2^6 + 3^6
4. 2^7 + 3^8
5. 4^5 + 9^5

The official answer is option 3. I recognize now that this is the case, but I haven't identified a non-calculation intensive method to solve the problem. What am I missing?

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by RumpelThickSkin » Fri Jul 30, 2010 12:06 pm
saritalr wrote:
Which of the following is equal to the value of 2^5 + 2^5 + 3^5 + 3^5 + 3^5.

1. 5^6
2. 13^5
3. 2^6 + 3^6
4. 2^7 + 3^8
5. 4^5 + 9^5

The official answer is option 3. I recognize now that this is the case, but I haven't identified a non-calculation intensive method to solve the problem. What am I missing?

2^5 + 2^5 + 3^5 + 3^5 + 3^5

take 2^5 and 3^5 common you will get

2^5 (1+1) + 3^5 (1+1+1)
= 2^5 (2) + 3^5 (3)
= 2^6 + 3^6

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by saritalr » Fri Jul 30, 2010 12:13 pm
RumpelThickSkin wrote:
saritalr wrote:
Which of the following is equal to the value of 2^5 + 2^5 + 3^5 + 3^5 + 3^5.

1. 5^6
2. 13^5
3. 2^6 + 3^6
4. 2^7 + 3^8
5. 4^5 + 9^5

The official answer is option 3. I recognize now that this is the case, but I haven't identified a non-calculation intensive method to solve the problem. What am I missing?

2^5 + 2^5 + 3^5 + 3^5 + 3^5

take 2^5 and 3^5 common you will get

2^5 (1+1) + 3^5 (1+1+1)
= 2^5 (2) + 3^5 (3)
= 2^6 + 3^6
Thanks for your response. Would you mind elaborating a little more? How did you determine that both powers would be 6?

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by RumpelThickSkin » Fri Jul 30, 2010 11:52 pm
= 2^5 (2) + 3^5 (3)

multiply 2^5(2) = 2^6 + multiply 3^5 (3) = 3^6

answer 2^6 + 3^6 .. that's all there is to it!

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by saritalr » Sat Jul 31, 2010 5:45 am
RumpelThickSkin wrote:= 2^5 (2) + 3^5 (3)

multiply 2^5(2) = 2^6 + multiply 3^5 (3) = 3^6

answer 2^6 + 3^6 .. that's all there is to it!
Okay, thanks for clarifying. I'd originally considered that method to be more calculation intensive, but I suppose if I memorized 2^2 through 2^10 I'd have a clearer idea of the right answer straight off the bat.

Mostly I was concerned that there was a simpler method that I was missing - but I guess that isn't the case. Thanks again!

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by deeyah » Mon Aug 02, 2010 5:52 pm
Or if you take 2^5 as "a" and 3^5 as "b"

you get a+a+b+b+b
2a+3b

Subsitute the value of a and b you get
2x2^5+3x3^5

2^(1+5)+3^(1+5)

=2^6+3^6

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by Abhishek009 » Wed Aug 04, 2010 9:19 am
saritalr wrote:Hi, this is my first question post. The question comes from the first CAT from the mba.com site. I don't believe there are explanations out there for these problems, but please correct me if I'm wrong. Similarly, I had trouble searching the forum to see if my question (or a similar one) had already been posted. Is there a recommended way to search? I'd hate post a bunch of repeats.

Which of the following is equal to the value of 2^5 + 2^5 + 3^5 + 3^5 + 3^5.

1. 5^6
2. 13^5
3. 2^6 + 3^6
4. 2^7 + 3^8
5. 4^5 + 9^5

The official answer is option 3. I recognize now that this is the case, but I haven't identified a non-calculation intensive method to solve the problem. What am I missing?
Normally CAT test takers follow an approach of Cyclicity of numbers.

I am providing a link for the same : https://takshzilabeta.com/index.php?opti ... -&Itemid=5

This is just for reference but I personally feel that the solution provided by deeyah is the best and quickest method for solving this problem.
Abhishek

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by saritalr » Thu Aug 05, 2010 10:21 am
saritalr wrote:Which of the following is equal to the value of 2^5 + 2^5 + 3^5 + 3^5 + 3^5.

1. 5^6
2. 13^5
3. 2^6 + 3^6
4. 2^7 + 3^8
5. 4^5 + 9^5

The official answer is option 3. I recognize now that this is the case, but I haven't identified a non-calculation intensive method to solve the problem. What am I missing?
I just read something in my number properties review book that made the problem above snap into place. RumpleThickSkin - it may have been what you were trying to explain but I just didn't understand because I didn't know the exponent rule.

Exponent rule: a^x + a^x + a^x = 3a^x

Example: 3^4 + 3^4 + 3^4 = 3 "¢ 3^4 = 3^5

Example 2: 3^x + 3^x + 3^x = *3 "¢ 3^x = 3^1 "¢ 3^x* = 3^(x +1)

Example 3: 2^3 + 2^3 = 2(2^3) = 2^4

*note that any number that does not have an exponent implicitly has an exponent of 1.
3 "¢ 3^x = 3^1"¢ 3^x = 3^(x + 1)*

Thanks everyone!

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by [email protected] » Thu Aug 05, 2010 12:09 pm
saritalr wrote:
RumpelThickSkin wrote:= 2^5 (2) + 3^5 (3)

multiply 2^5(2) = 2^6 + multiply 3^5 (3) = 3^6

answer 2^6 + 3^6 .. that's all there is to it!
Okay, thanks for clarifying. I'd originally considered that method to be more calculation intensive, but I suppose if I memorized 2^2 through 2^10 I'd have a clearer idea of the right answer straight off the bat.

Mostly I was concerned that there was a simpler method that I was missing - but I guess that isn't the case. Thanks again!
Hi,

you don't need to do any calculations with the method provided - you just need to understand how exponents work.

An exponent signifies how many times to multiply a number by itself.

For example, 2^5 = 2*2*2*2*2

Accordingly, when we multiply two terms with the same base, we add the exponents:

2^5 * 2^6 = (2*2*2*2*2)*(2*2*2*2*2*2) = 2^(5+6) = 2^11

So, back to the question you posted:

2^5 + 2^5 = 2 * 2^5 = 2*(2*2*2*2*2) = 2^6

(As another poster noted, a non-visible exponent is really an exponent of 1, so 2^1 * 2^5 = 2^(1+5) = 2^6.)

Similarly:

3^5 + 3^5 + 3^5 = 3 * 3^5 = 3*(3*3*3*3*3) = 3^6

(Or, again, 3^1 * 3^5 = 3^(1+5) = 3^6.)

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by [email protected] » Thu Aug 05, 2010 12:15 pm
Great thread, everyone! You've all worked the problem out nicely, but it also brings up a pretty key takeaway that's worth mentioning.

I talk to my students about "finding opportunities to do what you do well", kind of like the business concept of "core competencies" (if you're, like Amazon.com, great at distribution of media, then find opportunities like the Kindle to continue to do it even better; Amazon, however, isn't great at brick-and-mortar setups, so you won't see them partnering with Starbucks to create in-person stores with cafes, etc.).

On the GMAT, there are things that you should be good at, and some things that you can probably recognize right away that you're not good at (at least not in ~2 minutes per question without a calculator).

This is a great example of that - we're awful at adding and subtracting exponents, but we're pretty good at multiplying them. When you see a series of exponents added together, you probably need to find a way to use your "core competency" of multiplication. The easiest way to do that is typically to factor, which allows you to turn addition into multiplication.

When you see complex problems on the GMAT, ask yourself how to turn what you see into something that you do well. With core competencies like prime factorization, factoring, using number properties, etc., you should be able to solve any problem as long as you can get that problem on your terms.
Brian Galvin
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