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If n is greater than 20, what is the number closest to n^100 - n^90?

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haardiksharma
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If n is greater than 20, what is the number closest to n^100 - n^90? [#permalink] New post   Updated on: 24 Jun 2017, 06:37
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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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B

Originally posted by haardiksharma on 24 Jun 2017, 05:50.
Last edited by haardiksharma on 24 Jun 2017, 06:37, 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? [#permalink] New post   24 Jun 2017, 06:42
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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


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Re: If n is greater than 20, what is the number closest to n^100 - n^90? [#permalink] New post   24 Jun 2017, 05:58
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It can be deduced to
n^90(n^10-1)
Since n is greater than 20 so n^10-1~ n^10.
So it will be closest to n^100.
B

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Re: If n is greater than 20, what is the number closest to n^100 - n^90? [#permalink] New post   24 Jun 2017, 06:58
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Solution:

n>20.
n^100-n^90 .Take n to be 100 for example. 100^100-100^90.
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? [#permalink] New post   24 Jun 2017, 07:05
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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 [#permalink] New post   08 Jul 2018, 12:05
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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? [#permalink] New post   27 Jul 2018, 22:03
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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 \(Y^x-z^q\) :-) for example \(6^{18}-5^8\) how should I solve this :?


Hey dave13

Unfortunately, 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? [#permalink] New post   28 Jul 2018, 04:27
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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 \(Y^x-z^q\) :-) for example \(6^{18}-5^8\) how should I solve this :?


Hey dave13

Unfortunately, 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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Re: If n is greater than 20, what is the number closest to n^100 - n^90? [#permalink] New post   24 Jun 2017, 06:43
Bunuel
Thank you so much
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If n is greater than 20, what number is closest to n^100 - n^90 [#permalink] New post   07 Jul 2018, 20:37
\(n^{100}-n^{90}\) 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)

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Re: If n is greater than 20, what is the number closest to n^100 - n^90? [#permalink] New post   11 Jul 2018, 22:43
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? [#permalink] New post   16 Jul 2018, 09:22
\(n^{90}\) * \((n^{10}-1)\) 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? [#permalink] New post   19 Jul 2018, 12:36
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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? [#permalink] New post   27 Jul 2018, 09:42
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 \(Y^x-z^q\) :-) for example \(6^{18}-5^8\) how should I solve this :?
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