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m and n are integers. [#permalink]
Expert's post 00:00

Question Stats: 76% (00:54) correct 23% (00:23) wrong based on 17 sessions
m and n are integers.

 Quantity A Quantity B $$(\sqrt{10^{2m}})(\sqrt{10^{2n}})$$ $$10^{mn}$$

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Kudos [?]: 16 , given: 20

Re: m and n are integers. [#permalink]
Can anyone explain this? thank you Intern Joined: 01 Sep 2017
Posts: 20
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Kudos [?]: 20  , given: 21

Re: m and n are integers. [#permalink]
1
KUDOS
Hello,

for this question it is best to try out different values.

For instance.

If m and n are both zero, both expressions will be equal to 1 (since anything exponent zero is one) => Answer C
If m and n are both 3, Qty A becomes 10^6 and Qty B becomes 10^9 => Answer B,

since the square root and exponent 2 cancel each other out, Qty A becomes 10^m * 10^n => 10^(m+n)= 10^(3+3)<10^(3*3)

Hence overall answer is D GRE Instructor Joined: 10 Apr 2015
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Re: m and n are integers. [#permalink]
2
KUDOS
Expert's post
Carcass wrote:
m and n are integers.

 Quantity A Quantity B $$(\sqrt{10^{2m}})(\sqrt{10^{2n}})$$ $$10^{mn}$$

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.

RULE: √x = x^(1/2)

So, √[10^(2m)] = [10^(2m)]^(1/2)
= 10^m [once we apply the Power of a Power rule]

Likewise, √[10^(2n)] = [10^(2n)]^(1/2)
= 10^n [once we apply the Power of a Power rule]

We get:

QUANTITY A: (10^m)(10^n)
QUANTITY B: 10^(mn)

Simplify to get:
QUANTITY A: 10^(m+n)
QUANTITY B: 10^(mn)

Let's test some values of m and n

m = 0 and n = 0
We get:
QUANTITY A: 10^(0+0) = 10^0 = 1
QUANTITY B: 10^[(0)(0)] = 10^0 = 1
In this case the two quantities are equal

m = 1 and n = 1
We get:
QUANTITY A: 10^(1+1) = 10^2 = 100
QUANTITY B: 10^[(1)(1)] = 10^1 = 10
In this case Quantity is greater

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