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Re M3153
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20 Jul 2015, 04:49
Official Solution:If \(a\) and \(b\) are nonnegative 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, Answer: B
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Re M3153
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17 Jul 2016, 02:20
I think this is a highquality question and I agree with explanation. Great Trap on choice B! Excellent



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04 Jun 2017, 07:31
Very good question. I made a mistake assuming that a = 2b => a> b without considering b =0



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Re: M3153
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09 Jul 2017, 02:23
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)
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09 Jul 2017, 02:29



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Re: M3153
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26 Aug 2017, 03:15
Bunuel wrote: Official Solution:
If \(a\) and \(b\) are nonnegative 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,
Answer: B 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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Re: M3153
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26 Aug 2017, 04:06
ashikaverma13 wrote: Bunuel wrote: Official Solution:
If \(a\) and \(b\) are nonnegative 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,
Answer: B 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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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, 14:44
I think this the explanation isn't clear enough, please elaborate.



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31 Oct 2018, 21:42



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Re: M3153
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19 Nov 2018, 08:44
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.










