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here it goes--In the infinite sequence S, where S1 = 25, S2 = 125, S3 = 225, ..., Sk = 100*(k-1) + 25, is odd integer x a divisor of every member of S?
(1) For all values of k ≥ 3, x is a divisor of Sk.
(2) x is a divisor of 500.
force5 wrote:here it goes--In the infinite sequence S, where S1 = 25, S2 = 125, S3 = 225, ..., Sk = 100*(k-1) + 25, is odd integer x a divisor of every member of S?
(1) For all values of k ≥ 3, x is a divisor of Sk.
(2) x is a divisor of 500.
S1= 25 then its divisors can be - 1,5,25
s2= 125 then its divisors can be - 1,5,25,125
s3 = 225 then its divisors can be - 1,3,5,9,25
s4= 325 then its divisors can be - 1,5,25,13
s5= 425 then its divisors can be - 1,5,25,17
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statement 1 says x can be 1,5,25 ( since its common in all Sk) since these( 1,5,25) are divisors of every member of S this answers our question.
Statement 2 - x can be 1,2,4,5,25,125 hence not sufficient.
IMO- A
Be careful here...srcc25anu wrote:@ force5: just one small question:
from stat 1 we know x can be 1,5,25 (odd and factors of S) but we cannot identify one odd integer (as the Q asks)
also from stat 1 we know 13,17 (odd and NOT divisors of S)
hence should we not mark it as INSUFFICIENT since we have multiple values of X that satisfy the solution in first case?
is odd integer x a divisor of every member of S?
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