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Let's simplify each of the options if we can: A. \(999^{12} \approx 1000^{12} = 10^{36}\) B. \(10^{30} = 1000^{10}\) C. \(777^{10}\) - it's obviously smaller than B as the bases differ but the power is the same D. \((-20)^{24} = 20^{24} = 2^{24}*10^{24} \approx 2^4*10^6*10^{24} = 16*10^{30}\) (we used this approximation: \(2^{10} \approx 10^3\)) E. \((\sqrt{15})^{40} = 15^{20}\) - this one is for sure smaller than D

We are left with A, B and D. B is approximately 16 times smaller than D. Although we used approximation for A, it should still be greater than D. A is the greatest. _________________

Let's simplify each of the options if we can: A. \(999^{12} \approx 1000^{12} = 10^{36}\) B. \(10^{30} = 1000^{10}\) C. \(777^{10}\) - it's obviously smaller than B as the bases differ but the power is the same D. \((-20)^{24} = 20^{24} = 2^{24}*10^{24} \approx 2^4*10^6*10^{24} = 16*10^{30}\) (we used this approximation: \(2^{10} \approx 10^3\)) E. \((\sqrt{15})^{40} = 15^{20}\) - this one is for sure smaller than D

We are left with A, B and D. B is approximately 16 times smaller than D. Although we used approximation for A, it should still be greater than D. A is the greatest.

It is indeed a tricky question. Can you please advise how it is possible to reach approximation (highlighted in red) without a calculator? Thank you!

This question is indeed tricky. I used this approximation because I knew that \(2^{10}=1024\). If you encounter a question similar to this one you might want to search for another approximation you can use in that particular situation. However, some numbers people just choose to remember (like squares of numbers up to 20) as it might turn out very handy during GMAT.

2 taken to powers up to 10 might be useful too, but you need to decide if you're that kind of a person.

Let's simplify each of the options if we can: A. \(999^{12} \approx 1000^{12} = 10^{36}\) D. \((-20)^{24} = 20^{24} = 2^{24}*10^{24} \approx 2^4*10^6*10^{24} = 16*10^{30}\) (we used this approximation: \(2^{10} \approx 10^3\))

I would solve it differently - don't think we need to know that \(2^{10} = 1024\)

We can easily compare A to D by transforming D from \((-20)^{24}\) to \(((-20)^2)^{12}\)

\(999^{12} > 400^{12}\) A is greater and D is smaller.

You don't have to get EVERY answer choice to the same base or same power, as long as you can get two to the same base/power - you can eliminate one. _________________

If you had to compare between 10^36 and 15^20 how would you compare and find which one is greater?This is where I had goten stuck...Please explain! _________________

Let's simplify each of the options if we can: A. \(999^{12} \approx 1000^{12} = 10^{36}\) B. \(10^{30} = 1000^{10}\) C. \(777^{10}\) - it's obviously smaller than B as the bases differ but the power is the same D. \((-20)^{24} = 20^{24} = 2^{24}*10^{24} \approx 2^4*10^6*10^{24} = 16*10^{30}\) (we used this approximation: \(2^{10} \approx 10^3\)) E. \((\sqrt{15})^{40} = 15^{20}\) - this one is for sure smaller than D

We are left with A, B and D. B is approximately 16 times smaller than D. Although we used approximation for A, it should still be greater than D. A is the greatest.

I will approach this question slightly different - I will divide all powers by a common integer, hence to make the calculations rather easier. I chose always 10 when powers are two digits. Hence this gives, 1.2, 3, 1, 2.4 & 2. Therefore: Answer A is the highest - no?

I looked at this question like, "What is this and where do I start?" I did figure that the bases needed to be simplified. Lesson learned: estimate!!! _________________

Let's simplify each of the options if we can: A. \(999^{12} \approx 1000^{12} = 10^{36}\) B. \(10^{30} = 1000^{10}\) C. \(777^{10}\) - it's obviously smaller than B as the bases differ but the power is the same D. \((-20)^{24} = 20^{24} = 2^{24}*10^{24} \approx 2^4*10^6*10^{24} = 16*10^{30}\) (we used this approximation: \(2^{10} \approx 10^3\)) E. \((\sqrt{15})^{40} = 15^{20}\) - this one is for sure smaller than D

We are left with A, B and D. B is approximately 16 times smaller than D. Although we used approximation for A, it should still be greater than D. A is the greatest.

I will approach this question slightly different - I will divide all powers by a common integer, hence to make the calculations rather easier. I chose always 10 when powers are two digits. Hence this gives, 1.2, 3, 1, 2.4 & 2. Therefore: Answer A is the highest - no?

this qxn goes to lugo, if by using your method we arrive at 1.2 as answer for A, how then do you end up by saying that makes it the correct answer ( meaning it is the greatest number ryt?) but then how about the value 3? shouldn't that be the greatest number? well, am confused _________________

if someone was able to do it, this proves it can be done, and so is within reach = I CAN TOO

I think you already refreshed your knowledge of basic exponents rules. If you answer the same question incorrectly twice, you probably need to analyze your mistakes closer. Make sure you use an Error log to track down your weak areas. When you know your weaknesses, you have to go back to basics and study the concepts behind the problems you got wrong. After you refresh the concepts, you have to practice and practice and practice some more until you feel confident about your once weak areas.

I think you already refreshed your knowledge of basic exponents rules. If you answer the same question incorrectly twice, you probably need to analyze your mistakes closer. Make sure you use an Error log to track down your weak areas. When you know your weaknesses, you have to go back to basics and study the concepts behind the problems you got wrong. After you refresh the concepts, you have to practice and practice and practice some more until you feel confident about your once weak areas.

You'll find a bunch of links to theoretic material there.

I hope it helps in some way .

gottabwise wrote:

Answered incorrectly twice. Any recommendations for reviewing and practicing how to manipulate bases and powers?

dzyubam: Thanks for the links; will check them out. On the error log, I am using one; just haven't addressed it yet. In fact, I'm still reviewing concepts; been away from math for a while. In about a week or so, I plan to take another practice test and start addressing the error log. I'm going through club problems in the interim to keep my mind abreast of potential question types before digging into the OG. _________________

If you had to compare between 10^36 and 15^20 how would you compare and find which one is greater?This is where I had goten stuck...Please explain!

you need compare apple with apple, so you try replace 15 with 10^n. Because it it easy to compare 10^36 with 15^20=(10^n)^20=10^(n*20). So we go like this 15=10^n, what is n ? definitely n<2 because 10^2 is 100, definitely n>1 because 10^1 is only 10. Let's check the result for 10^1.5 approximately is 10^1*10^0.5=10*sqrt(10)= near 33 (root 10 is around 3.3) So 10^1.5 is equal near 33 which is more than 15, so for 15=10^n, n will be even less than 1.5. Let's put 1.5 to our original formula 10^(1.5*20)=10^30 which is already less than 10^36, and we remember that n will be even less than 1.5. the answer is 10^36 > 15^20