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

Question

If  a  and  b  are non-negative integers, is  a > b ?
 
(1)  6^a = 36^b 
 
(2)  5^a = 35^b

M31-53

NickHalden · Answer

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 !!

mohamedfaras · Answer

B

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

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

karant · Answer

(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

lipsi18 · Answer

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,

Bunuel · Answer

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.

blitzkriegxX · Answer

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.

Neha2050 · Answer

Taking b=-1, then a=-2 ..In this case a<b
why this when its already given that a and b are non negative integers

ManjariMishra · Answer

Here why are we considering negative numbers for a and b when it already mentioned non negative numbers.?

Basshead · Answer

If  a  and  b  are non-negative integers, is  a > b ?
 
(1)  6^a = 36^b 
 
 6^a = 6^2b

a = 2b

However, we don't know if either is 0. Insufficient.

(2)  5^a = 35^b 
 5^a = 5^b7^b 
 5^{a-b} = 7^b 
Only possible if  a = b = 0

Sufficient.

Answer is B.

bumpbot · Answer

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

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