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# M31-53

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Math Expert
Joined: 02 Sep 2009
Posts: 46129

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20 Jul 2015, 05:49
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Difficulty:

65% (hard)

Question Stats:

46% (01:03) correct 54% (00:49) wrong based on 63 sessions

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If $$a$$ and $$b$$ are non-negative integers, is $$a > b$$?

(1) $$6^a = 36^b$$

(2) $$5^a = 35^b$$

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Math Expert
Joined: 02 Sep 2009
Posts: 46129

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20 Jul 2015, 05:49
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Official Solution:

If $$a$$ and $$b$$ are non-negative integers, is $$a > b$$?

(1) $$6^a = 36^b$$. Simplify: $$6^a = 6^{2b}$$. Bases are equal, hence we can equate the powers: $$a=2b$$. If $$a=b=0$$, then $$a$$ is NOT greater than $$b$$ but if $$a=2$$ and $$b=1$$, then $$a$$ IS greater than $$b$$. Not sufficient.

(2) $$5^a = 35^b$$. If both $$a$$ and $$b$$ are positive integers, then we'd have that $$5^a$$ is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both $$a$$ and $$b$$ must be 0, giving a NO answer to the question whether $$a$$ is greater than $$b$$. Sufficient,

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Joined: 31 Mar 2016
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Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
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17 Jul 2016, 03:20
I think this is a high-quality question and I agree with explanation. Great Trap on choice B! Excellent
Intern
Joined: 24 Aug 2013
Posts: 11
GMAT 1: 710 Q50 V36

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04 Jun 2017, 08:31
Very good question. I made a mistake assuming that a = 2b => a> b without considering b =0
Intern
Joined: 12 Mar 2016
Posts: 5

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09 Jul 2017, 03:23
Hi Bunuel,

Since it is given a & b are positive, how can we assume a=0 from statement 1? (0 is neither -ve nor +ve)

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 46129

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09 Jul 2017, 03:29
petal123 wrote:
Hi Bunuel,

Since it is given a & b are positive, how can we assume a=0 from statement 1? (0 is neither -ve nor +ve)

Thanks

We are not given that a and b are positive, we are given that a and b are non-negative. Non-negative, means 0 or positive (so those which are not negative).
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Joined: 19 Aug 2016
Posts: 153
Location: India
GMAT 1: 640 Q47 V31
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26 Aug 2017, 04:15
Bunuel wrote:
Official Solution:

If $$a$$ and $$b$$ are non-negative integers, is $$a > b$$?

(1) $$6^a = 36^b$$. Simplify: $$6^a = 6^{2b}$$. Bases are equal, hence we can equate the powers: $$a=2b$$. If $$a=b=0$$, then $$a$$ is NOT greater than $$b$$ but if $$a=2$$ and $$b=1$$, then $$a$$ IS greater than $$b$$. Not sufficient.

(2) $$5^a = 35^b$$. If both $$a$$ and $$b$$ are positive integers, then we'd have that $$5^a$$ is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both $$a$$ and $$b$$ must be 0, giving a NO answer to the question whether $$a$$ is greater than $$b$$. Sufficient,

Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks
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Math Expert
Joined: 02 Sep 2009
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26 Aug 2017, 05:06
1
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:

If $$a$$ and $$b$$ are non-negative integers, is $$a > b$$?

(1) $$6^a = 36^b$$. Simplify: $$6^a = 6^{2b}$$. Bases are equal, hence we can equate the powers: $$a=2b$$. If $$a=b=0$$, then $$a$$ is NOT greater than $$b$$ but if $$a=2$$ and $$b=1$$, then $$a$$ IS greater than $$b$$. Not sufficient.

(2) $$5^a = 35^b$$. If both $$a$$ and $$b$$ are positive integers, then we'd have that $$5^a$$ is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both $$a$$ and $$b$$ must be 0, giving a NO answer to the question whether $$a$$ is greater than $$b$$. Sufficient,

Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks

If a and b are not both 0, then we'd have that {not a multiple of 7} = {a multiple of 7}, which cannot be true, thus both $$a$$ and $$b$$ must be 0.
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Joined: 13 Oct 2017
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26 Feb 2018, 05:00
Hi Bunuel,

I worked the second statement out in the following way:

5^a=5^b*7^b

Since exponents with the same bases can be equated...I then got a=b (by equating the exponents of the two 5s)
If a=b, a cannot be greater than b...hence sufficient.

I was still uncomfortable with my working out because I wasn't sure whether ignoring the 7^b was correct...was my approach correct?

Thanks.
M31-53   [#permalink] 26 Feb 2018, 05:00
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# M31-53

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