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If n is greater than 20, what is the number closest to n^100  n^90?
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Updated on: 24 Jun 2017, 05:37
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If n is greater than 20, what is the number closest to n^100  n^90? A) n^90 B) n^100 C) n^99 D) n^190 E) n^10
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Originally posted by haardiksharma on 24 Jun 2017, 04:50.
Last edited by haardiksharma on 24 Jun 2017, 05:37, edited 3 times in total.




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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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24 Jun 2017, 05:42




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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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24 Jun 2017, 04:58
It can be deduced to n^90(n^101) Since n is greater than 20 so n^101~ n^10. So it will be closest to n^100. B Sent from my Moto G (5) Plus using GMAT Club Forum mobile app



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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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24 Jun 2017, 05:43



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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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24 Jun 2017, 05:58
Solution:
n>20. n^100n^90 .Take n to be 100 for example. 100^100100^90. Simplifying, 100^90{(10^10) 1} Comparing to the value of 10^10, 1 is negligible.
Therefore, the answer is Option B.



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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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24 Jun 2017, 06:05
haardiksharma wrote: If n is greater than 20, what is the number closest to n^100  n^90?
A) n^90 B) n^100 C) n^99 D) n^190 E) n^10 n^100  n^90 n^90 (n^10  1) Since n is greater than 20, n^20 would be a very large number. Thus we may we write n^90 (n^10  1) = n^90 (n^10) = n^100 Answer B
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If n is greater than 20, what number is closest to n^100  n^90
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07 Jul 2018, 19:37
\(n^{100}n^{90}\) can be written as \(n^{90}(n^{10}1)\) \((n^{10}1)\) can be approximated to \(n^{10}\) since 1 is negligible in comparison to \(n^{10}\) when n>20. So our expression becomes \(n^{90}*n^{10}=n^{100}\). Ans. (B) Posted from my mobile device
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If n is greater than 20, what number is closest to n^100  n^90
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08 Jul 2018, 11:05
Karthik200 wrote: If n is greater than 20, what number is closest to \(n^{100}  n^{90}\)
A. \(n^{90}\) B. \(n^{100}\) C. \(n^{99}\) D. \(n^{190}\) E. \(n^{10}\) Since n is a number greater than 20, for simplicity sake, let's assume this number to be \(25\) or \(5^2\) \((5^2)^{100}  (5^2)^{90} = 5^{200}  5^{180} = 5^{180}(5^{20}  1) = 5^{180}(5^{20}) = 5^{180+20} = 5^{200} = (5^2)^{100}\) This is because \(3125*3125*3125*3125  1\) is almost the same as \(3125*3125*3125*3125\). So, we can extrapolate this result for n = 25 to show that \(n^{100}  n^{90} = n^{100}\) (Option B)
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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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11 Jul 2018, 21:43
Karthik200 wrote: If n is greater than 20, what number is closest to \(n^{100}  n^{90}\)
A. \(n^{90}\) B. \(n^{100}\) C. \(n^{99}\) D. \(n^{190}\) E. \(n^{10}\) \(n^{90} (n^{10}1)\) \(n^{90+10}\) (1 can be neglected) B
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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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16 Jul 2018, 08:22
\(n^{90}\) * \((n^{10}1)\) is as good as \(n^{90}*n^{10} = n^{100}\) Hence option B is the correct answer.
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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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19 Jul 2018, 11:36
haardiksharma wrote: If n is greater than 20, what is the number closest to n^100  n^90?
A) n^90 B) n^100 C) n^99 D) n^190 E) n^10 Factoring n^90 from both terms, we have: n^90(n^10  1). Now, since n is relatively large, we see that (n^10  1) is essentially equal to n^10. We can thus simplify the expression n^90(n^10  1) to n^90(n^10) = n^100. Answer: B
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If n is greater than 20, what is the number closest to n^100  n^90?
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27 Jul 2018, 08:42
pushpitkc wrote: Karthik200 wrote: If n is greater than 20, what number is closest to \(n^{100}  n^{90}\)
A. \(n^{90}\) B. \(n^{100}\) C. \(n^{99}\) D. \(n^{190}\) E. \(n^{10}\) Since n is a number greater than 20, for simplicity sake, let's assume this number to be \(25\) or \(5^2\) \((5^2)^{100}  (5^2)^{90} = 5^{200}  5^{180} = 5^{180}(5^{20}  1) = 5^{180}(5^{20}) = 5^{180+20} = 5^{200} = (5^2)^{100}\) This is because \(3125*3125*3125*3125  1\) is almost the same as \(3125*3125*3125*3125\). So, we can extrapolate this result for n = 25 to show that \(n^{100}  n^{90} = n^{100}\) (Option B)
pushpitkc what is formula of exponents when i need to perform subtraction \(Y^xz^q\) for example \(6^{18}5^8\) how should I solve this



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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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27 Jul 2018, 21:03
dave13 wrote: pushpitkc wrote: Karthik200 wrote: If n is greater than 20, what number is closest to \(n^{100}  n^{90}\)
A. \(n^{90}\) B. \(n^{100}\) C. \(n^{99}\) D. \(n^{190}\) E. \(n^{10}\) Since n is a number greater than 20, for simplicity sake, let's assume this number to be \(25\) or \(5^2\) \((5^2)^{100}  (5^2)^{90} = 5^{200}  5^{180} = 5^{180}(5^{20}  1) = 5^{180}(5^{20}) = 5^{180+20} = 5^{200} = (5^2)^{100}\) This is because \(3125*3125*3125*3125  1\) is almost the same as \(3125*3125*3125*3125\). So, we can extrapolate this result for n = 25 to show that \(n^{100}  n^{90} = n^{100}\) (Option B)
pushpitkc what is formula of exponents when i need to perform subtraction \(Y^xz^q\) for example \(6^{18}5^8\) how should I solve this Hey dave13Unfortunately, there is no formula to solve \(6^{18}  5^8\) as the bases have no common factor However, if the expression we were asked to find the value for was \(6^{18}  3^8\), we could simplify it as follows  \(3*2^{18}  3^8\) = \(3^8(3^{10}*2^{18}  1)\) P.S there must be a common factor in Y and z(which I have highlighted in your post) for us to make this simplification. Hope this clears your confusion!
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Re: If n is greater than 20, what is the number closest to n^100  n^90?
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28 Jul 2018, 03:27
pushpitkc wrote: dave13 wrote: pushpitkc wrote: If n is greater than 20, what number is closest to \(n^{100}  n^{90}\)
A. \(n^{90}\) B. \(n^{100}\) C. \(n^{99}\) D. \(n^{190}\) E. \(n^{10}\)
Since n is a number greater than 20, for simplicity sake, let's assume this number to be \(25\) or \(5^2\)
\((5^2)^{100}  (5^2)^{90} = 5^{200}  5^{180} = 5^{180}(5^{20}  1) = 5^{180}(5^{20}) = 5^{180+20} = 5^{200} = (5^2)^{100}\)
This is because \(3125*3125*3125*3125  1\) is almost the same as \(3125*3125*3125*3125\).
So, we can extrapolate this result for n = 25 to show that \(n^{100}  n^{90} = n^{100}\)(Option B)
pushpitkc what is formula of exponents when i need to perform subtraction \(Y^xz^q\) for example \(6^{18}5^8\) how should I solve this Hey dave13Unfortunately, there is no formula to solve \(6^{18}  5^8\) as the bases have no common factor However, if the expression we were asked to find the value for was \(6^{18}  3^8\), we could simplify it as follows  \(3*2^{18}  3^8\) = \(3^8(3^{10}*2^{18}  1)\) P.S there must be a common factor in Y and z(which I have highlighted in your post) for us to make this simplification. Hope this clears your confusion! You can factor \(6^{18}  5^8\) by applying a^2  b^2 = (a  b)(a + b): \(6^{18}  5^8=(6^{9})^2  (5^4)^2=(6^9  5^4)(6^9 + 5^4)\). But the easiest approach to solve this problem is given HERE or in any of the links given in this post.
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