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10 May 2022, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:22) correct
68%
(01:48)
wrong
based on 68
sessions
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If \(a^x = b^y\) where a, b are positive integers and x and y are non-negative integers, is x divisible by y?
(1) a < b
(2) y is an odd number
(1) a < b
(2) y is an odd number
Archit3110
01 Jun 2022, 22:31
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given a,b are +ve integers and x,y are non -negative integers i.e. they can be '0' as well
target is x divisible by y
#1
a<b
case 1
2<4
2^2=4^1
yes
case 2
1^0=3^0
not defined
insufficient
#2
y is odd number ;
case 1
a=4 ; x=1 and b=2 y= 2
no
case 2
a=2 ; x=2 and b = 4 y=1
yes
insufficient
from 1 &2
a<b and y is odd
so given condition \(a^x = b^y\) would be possible for even power of prime numbers such that x would always be even and b would be square of prime number
sufficient
option C
target is x divisible by y
#1
a<b
case 1
2<4
2^2=4^1
yes
case 2
1^0=3^0
not defined
insufficient
#2
y is odd number ;
case 1
a=4 ; x=1 and b=2 y= 2
no
case 2
a=2 ; x=2 and b = 4 y=1
yes
insufficient
from 1 &2
a<b and y is odd
so given condition \(a^x = b^y\) would be possible for even power of prime numbers such that x would always be even and b would be square of prime number
sufficient
option C
Bunuel
02 Jun 2022, 01:40
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If a^x = b^y, (a,b) > 0, (x,y)>= 0, and (a,b,x,y) are integers,
Then for every value except when x = y = 0(since 0/0 is not defined), either y will be divisible to x (if a>b) or x will be divisible to y (if a<b).
From statement 1, since a<b, we will know that x will be divisible by y everywhere except at x = y = 0.
So, this statement is not sufficient.
From statement 2, since y is an odd number(or y>0), we know that either y will be divisible to x (if a>b) or x will be divisible to y (if a<b). But without the relationship of a and b, this is also not sufficient.
If both the statements are used together, then x will be divisible by y.
Then for every value except when x = y = 0(since 0/0 is not defined), either y will be divisible to x (if a>b) or x will be divisible to y (if a<b).
From statement 1, since a<b, we will know that x will be divisible by y everywhere except at x = y = 0.
So, this statement is not sufficient.
From statement 2, since y is an odd number(or y>0), we know that either y will be divisible to x (if a>b) or x will be divisible to y (if a<b). But without the relationship of a and b, this is also not sufficient.
If both the statements are used together, then x will be divisible by y.
