# Thread: Mgmat Ds

1. Good post? |

## Mgmat Ds

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.

2. Good post? |
Let's start by writing the prime factors of 104. 13 * 2 * 2 * 2. Therefore, x-1 must fall in the combination of these factors.

(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.

3. Good post? |
A side NOTE : 0 can be counted as a multiple of a number but not as a factor.

4. Good post? |
stmt. 1 x divisible by 3...
say x = 3 => 3-1 = 2 factor of 104
say x=108=>108-1=107 /..not a factor of 104 ..hence insuff.
stmt 2.
27 is divisible by x...so x = 1 or 3 or 9 or 27
x-1 = 0 or 2 or 8 or 26
everything but 0 can be said as factor of 104
hence insuff.
combining 2 stmt. ...option of x=1 is eliminated...hence x = 3 or 9 or 27
x-1 will be a factor of 104...Hence C/
wats the Official Answer?

5. Good post? |
My vote for C

6. Good post? |
Official Answer is C

7. Good post? |
c is the answer

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