ri2007 wrote:If x is a positive integer, is x – 1 a factor of 104?

(1) x is divisible by 3.

(2) 27 is divisible by x.

Is it B?

1 - insufficient.
Take x = 3; 2 is a factor of 104.
Take x = 6; 5 is not a factor of 104.

2 - sufficient. Only values of x are 3 and 9.
2 and 8 are factors of 104

Ans given below - it is a copy of a reply from Vbhup2 from another forum

[spoiler]1) INSUFFICIENT.

x is divisible by 3 means x = {3,6,9,15,...}
x-1 = {2,5,8,14,...} etc.

2 is a factor of 104 but not 5. Hence, INSUFFICIENT.

(2) INSUFFICIENT.
27 is divisible by x.
Therefore x = {1,3,9,27}
x-1 = {0,2,8,26}

2,8, and 26 are factors. But, 0 is not a factor. Hence, INSUFFICIENT.

Combining (1) and (2), SUFFICIENT.

The intersection of the solution sets of x gives:

x= {3,9,27}
x-1 = {2,8,26}

All are factors of 104.

Hence, C.[/spoiler]

1.x=3*k where k is an integer. if x=3 then x-1=2 and x-1 is a factor of 104. if x=6 then x-1=5 and x-1 not a factor of 104. thus insufficient.
2. 27=x*m where m an integer. so x=27/m and since x is an integer positive, then x=1 or x=3 or x=9 or x=27. if x=1 then x-1 is not a factor of 104 but for the other cases x-1 is a factor of 104. insufficient.

1+2 together: sufficient since x cannot be 1 because x=3k from the first one (k=integer) so for all the other cases x-1 is a factor of 104. ANSWER C