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# M31-53

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Math Expert
Joined: 02 Sep 2009
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20 Jul 2015, 05:49
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Difficulty:

65% (hard)

Question Stats:

46% (01:41) correct 54% (01:18) wrong based on 54 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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20 Jul 2015, 05:49
1
2
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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17 Jul 2016, 03:20
I think this is a high-quality question and I agree with explanation. Great Trap on choice B! Excellent
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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
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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
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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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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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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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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.
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31 Oct 2018, 15:44
I think this the explanation isn't clear enough, please elaborate.
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31 Oct 2018, 22:42
bks170882 wrote:
I think this the explanation isn't clear enough, please elaborate.

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19 Nov 2018, 09:44
A quick and important tip for such questions!

ALWAYS REMEMBER IN DS questions statement 1 can never contradict statement 2!
So even in case u made mistake of taking a > b from statement 1, while doing statement 2 u realize$$a = b = 0$$.
You should immediately stop and check to see why statement 1 and 2 are contradicting.
Probably because u made some mistake.
This simple tip will help u avoid in a lot of DS questions.
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25 Jul 2019, 04:58
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

In order to understand St 2 better, you can consider the following approach

5^a = 35^b
=> 5^a = (7*5)^b = 7^b*5^b

Bring 5^b to LHS

=> 5^(a-b)=7^b

Now, the rule of cyclicity comes handy

No matter what the power of 7, it will never end in units digit being 5.
No matter what the power of 5, it will always end up in units digit being 5.

Hence, for LHS to be equal to RHS, the powers should be 0 so that both LHS = RHS = 1.

Now, a=b=0 therefore, we get a definite ans that a is not greater than b.
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Re: M31-53   [#permalink] 25 Jul 2019, 04:58
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# M31-53

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