If X and Y are positive integers, is ( 10^Y -X )/3 an integer? (1)( X-1)/3 is an integer (2) (Y-1)/3 is an integer
If the remainder is 2 when x is divided by 5, then is x divisible by 7? (1)x is a prime number (2)x+3 is a multiple of 10
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neelgandham
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If X and Y are positive integers, is (10^Y -X )/3 an integer?
Assuming that (10^Y -X )/3 is same as ((10^Y)-(X))/3
(1) (X-1)/3 is an integer
(X-1)/3 = K => X = 3K+1
Then ((10^Y)-(X))/3 = ((10^Y)-(3K+1))/3 = (((10^Y)-1)+(3K))/3
((10^Y)-1) can be 9,99,999,9999,99999,999999 ... and is always divisible by 3
(3K) is divisible by 3 as K is an integer.
So Sufficient
(2) (Y-1)/3 is an integer
((10^Y)-(X))/3 = ((10^(3K+1))-(X))/3
If X = 1 then ((10^Y)-(X))/3 is an integer
if X = 2 then ((10^Y)-(X))/3 may be an integer or may be not (Y = 1, x = 2 leaves a remainder 2;Y = 1, x = 1 leaves a remainder 0 )
Insufficient
Answer A
If the remainder is 2 when x is divided by 5, then is x divisible by 7?
(1)x is a prime number
X = 7 then X is divisible by 7
X = 17, then X is not divisible by 7
Insufficient !
(2)x+3 is a multiple of 10
X + 3 = 10 then X = 7 and is divisible by 7
X + 3 = 20 then X = 17 and is not divisible by 7
Insufficient !
IMO Answer E
Assuming that (10^Y -X )/3 is same as ((10^Y)-(X))/3
(1) (X-1)/3 is an integer
(X-1)/3 = K => X = 3K+1
Then ((10^Y)-(X))/3 = ((10^Y)-(3K+1))/3 = (((10^Y)-1)+(3K))/3
((10^Y)-1) can be 9,99,999,9999,99999,999999 ... and is always divisible by 3
(3K) is divisible by 3 as K is an integer.
So Sufficient
(2) (Y-1)/3 is an integer
((10^Y)-(X))/3 = ((10^(3K+1))-(X))/3
If X = 1 then ((10^Y)-(X))/3 is an integer
if X = 2 then ((10^Y)-(X))/3 may be an integer or may be not (Y = 1, x = 2 leaves a remainder 2;Y = 1, x = 1 leaves a remainder 0 )
Insufficient
Answer A
If the remainder is 2 when x is divided by 5, then is x divisible by 7?
(1)x is a prime number
X = 7 then X is divisible by 7
X = 17, then X is not divisible by 7
Insufficient !
(2)x+3 is a multiple of 10
X + 3 = 10 then X = 7 and is divisible by 7
X + 3 = 20 then X = 17 and is not divisible by 7
Insufficient !
IMO Answer E
Anil Gandham
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