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m01 #8

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m01 #8 [#permalink]

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New post 14 Dec 2008, 20:28
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Which expression has the greatest value?

(A) \(1^{999}\)
(B) \(2^{300}\)
(C) \(3^{200}\)
(D) \(4^{100}\)
(E) \(16^{50}\)

[Reveal] Spoiler: OA
C

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Re: m01 #8 [#permalink]

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New post 14 Dec 2008, 22:41
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A. 1^999 is what it is: 1.

B. 2^300

C. 3^200

D. 4^100: 4 can be written as 2^2. Therefore, the whole thing becomes (2^2)^100=2^200

E. 16^50: 16 can be written as 2^4. Accordingly, (2^4)^50=2^200.

It's either B or C. Logically guessing, if 2^3 is less than 3^2, then 2^300 must be less than 3^200. So I would go for C.

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Re: m01 #8 [#permalink]

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New post 17 Dec 2008, 20:54
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giantSwan wrote:
->How does this conversion work?

Which expression has the greatest value?

1^999
2^300
3^200
4^100
16^50



Yeah its C.

2^6 < 3^4
raise both side's power to 50
(2^6)^50 < (3^4)^50

2^300 < 3^200
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Re: m01 #8 [#permalink]

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New post 18 Dec 2008, 04:59
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I think it can be done in this way -

1^999 = 1
2^300 = (2^3)^100 = 8^100
3^200 = (3^2)^100 = 9^100
4^100 = 4^100
16^50 = (4^2)^50 = 4^100

by the rule a^x > b^x if a>b we can say C i.e. 3^200 is the greatest

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Re: m01 #8 [#permalink]

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New post 05 Feb 2009, 15:59
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C. I think you should take the nth root (say, ^100) of all numbers.
9>8>4=4>1

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Re: m01 #8 [#permalink]

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New post 12 Feb 2009, 11:25
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Simplify?

1^999=1
2^300/100=2^3=8
3^200/100=3^2=9
4^100/100=4^1=4
16^50/100=16^.5=Fraction 1/16

Is my logic correct here? C looks good

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Re: m01 #8 [#permalink]

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New post 15 Mar 2010, 05:22
giantSwan wrote:
Which expression has the greatest value?

(A) \(1^{999}\)
(B) \(2^{300}\)
(C) \(3^{200}\)
(D) \(4^{100}\)
(E) \(16^{50}\)

[Reveal] Spoiler: OA
C

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A is always 1
D & E can be reduced to 2^200
so either B or C.
B can be reduced to 2^300 = 4^150 = 8^100
C can be reduced to 3^300 = 9^150
hence C > B.
SO answer is C

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Re: m01 #8 [#permalink]

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New post 15 Mar 2010, 05:55
I think you meant \(3^{200}\) in C. Thus C is reduced to \(9^{100}\). The answer is still C, I just wanted to make things clearer.
bangalorian2000 wrote:

A is always 1
D & E can be reduced to 2^200
so either B or C.
B can be reduced to 2^300 = 4^150 = 8^100
C can be reduced to 3^300 = 9^150
hence C > B.
SO answer is C

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Re: m01 #8 [#permalink]

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New post 15 Mar 2010, 07:57
dzyubam wrote:
I think you meant \(3^{200}\) in C. Thus C is reduced to \(9^{100}\). The answer is still C, I just wanted to make things clearer.
bangalorian2000 wrote:

A is always 1
D & E can be reduced to 2^200
so either B or C.
B can be reduced to 2^300 = 4^150 = 8^100
C can be reduced to 3^300 = 9^150
hence C > B.
SO answer is C


Thanks for correcting the mistake.

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Re: m01 #8 [#permalink]

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New post 26 Mar 2012, 02:07
Its between B & C
just take random case
2^3 and 3^2
and you get answer........c it is

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Re: m01 #8 [#permalink]

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New post 26 Mar 2012, 06:55
just expand all the values to be a power of 100 and u will know that C = 9 to the power of 100 and thats the largest
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Re: m01 #8 [#permalink]

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New post 12 Mar 2013, 05:08
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A. 1.

B. 2^300 = 8^100

C. 3^200= 9^100

D. 4^100

E. 16^50 = 4^(2*50) = 4^100

Hence ans is C as 9^100 is the greatest

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Re: m01 #8 [#permalink]

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New post 22 Mar 2013, 06:06
C , with factorizing them all to 2s and 3s
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Re: m01 #8 [#permalink]

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New post 24 Mar 2013, 23:33
giantSwan wrote:
Which expression has the greatest value?

(A) \(1^{999}\)
(B) \(2^{300}\)
(C) \(3^{200}\)
(D) \(4^{100}\)
(E) \(16^{50}\)

[Reveal] Spoiler: OA
C

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C is correct. Here is my 2 cents.
1^999 = 1
2^300 = (2^3)^100 = 8^100
3^200 = (3^2)^100 = 9^100
4^100 = (2^2)^100 = 2^200 < 2^300
16^50 = (2^4)^50 = 2^200 < 2^300

So, 3^200 = 9^100 is greatest.
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Re: m01 #8 [#permalink]

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New post 18 Apr 2013, 01:40
Which expression has the greatest value?

(A) 1
(B) 8^100
(C) 9^100
(D) 2^200
(E) 2^200

Hence C

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Re: m01 #8   [#permalink] 18 Apr 2013, 01:40
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