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

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If a and b are non-negative integers, is a > b? [#permalink]

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New post 24 Jun 2015, 07:16
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45% (01:00) correct 55% (00:56) wrong based on 139 sessions

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Re: If a and b are non-negative integers, is a > b? [#permalink]

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New post 24 Jun 2015, 07:39
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Using statement 1 alone :
6^a = 36^b
-> 6^a = 6^2b
-> a = 2b
-> a>b ------------------------ Case 1.
But, at the same time it is possible that:
a = b = 0 --------------------- Case 2.
Also, 6^-2 = 36^-1
a<b ---------------------------- Case 3.
Hence Insufficient.

Using statement 2 alone :
5^a = 7^b * 5^b
5^(a-b) = 7^b
Now this is possible only if a = b = 0.
Hence Sufficient !!
Answer is B !!

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Re: If a and b are non-negative integers, is a > b? [#permalink]

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New post 24 Jun 2015, 08:21
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B

From I -> a,b = 0,0 /2, 1/4,2/ ...
Hence Not Sufficient

From II -> a,b = 0,0
Hence Sufficient

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Re: If a and b are non-negative integers, is a > b? [#permalink]

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New post 24 Jun 2015, 10:31
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(1) 6^a = 36^b
6^a = 6^2b
Or, a=2b

Taking b=1, then a=2.... In this case a> b
Taking b=-1, then a=-2 ..In this case a<b

Hence, insufficient

(2) 5^a = 35^b
5^a = (5*7)^b
5^a = 5^b * 7^b
5^(a-b) = 7^b
Now, 5 raise to power something (other than zero) can never be equal to 7 raise to power something because 5 raise to something always ends with unit digit as 5 whereas 7 raise to power something will end with unit as {7,9,3,1,7.....}

So, only one value can satisfy the above equation 5^0 = 7^0.


That means b=0 and a-b=0, a=0....a is not greater than b

Sufficient
Ans B

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Re: If a and b are non-negative integers, is a > b? [#permalink]

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New post 25 Jun 2015, 10:05
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A ad b are integers, is a>b?
1) 6^a = 36^b =6^2b therefore a=2b; when b=-1,a=-2 then a<b but when b=1, a=2 then a>b; hence insufficient
2) 5^a = 35^b=5^b * 7^b therefore 5^( a-b) = 7^b *5^0 therefore a-b=0 and a=b; hence ais not greater than b; hence Sufficient.

Hence answer is B
Thanks,

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Re: If a and b are non-negative integers, is a > b? [#permalink]

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New post 20 Jul 2015, 04:51
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Bunuel wrote:
If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

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

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

Kudos for a correct solution.

M31-53


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,


Answer: B.
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Re: If a and b are non-negative integers, is a > b? [#permalink]

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Re: If a and b are non-negative integers, is a > b?   [#permalink] 27 Nov 2017, 09:29
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