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

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

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Updated on: 24 Jun 2017, 06:37
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Question Stats:

57% (01:17) correct 43% (01:41) wrong based on 283 sessions

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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

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]

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24 Jun 2017, 06:42
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Re: If n is greater than 20, what is the number closest to n^100 - n^90?  [#permalink]

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24 Jun 2017, 05:58
1
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]

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24 Jun 2017, 06:43
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Thank you so much
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Re: If n is greater than 20, what is the number closest to n^100 - n^90?  [#permalink]

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24 Jun 2017, 06:58
1
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]

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24 Jun 2017, 07:05
1
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

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If n is greater than 20, what number is closest to n^100 - n^90  [#permalink]

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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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If n is greater than 20, what number is closest to n^100 - n^90  [#permalink]

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08 Jul 2018, 12:05
1
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]

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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]

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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]

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19 Jul 2018, 12:36
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.

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

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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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Re: If n is greater than 20, what is the number closest to n^100 - n^90?  [#permalink]

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27 Jul 2018, 22:03
1
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]

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28 Jul 2018, 04:27
1
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] 28 Jul 2018, 04:27
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