ashikaverma13
Bunuel
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
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 approach5^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 handyNo 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.