MGMAT 700-800

X>0, x=int

x-1 a factor of 104?

Factors of 104 are 1, 2, 4, 8, 13, 26, 52, 104

1) X is divisible by 3

x=6; x is divisible by 3, but x-1=6-1=5 is not a factor of 104
x=9, x is divisble by 3 and x-1=9-1=8 is a factor of 104.

Therefore statement 1 is insufficient.

2) 27 is divisible by x, therefore x=1,3,9,27

x-1= 0,2,8,26. While 2,8,26 are factors of 104, 0 isn't. Therefore stmt 2 is insufficient.

Together X is divisible by 3 and 27 is divisible by x. Therefore x=3,9,27

x-1=2,8,26, all factors of 104.

Ans is C.

Zero is divisible by all numbers
Zero is neither negative nor positive

The question states that x is a positive number and therefore it is not zero


In Statement 2
Zero is not an option and all the other numbers suffice

IMO B

In statement 2, x=1, abut x-1 is 0. So I don't think you can count out zero in this case because x>0.

hrishikesh05 wrote:Zero is divisible by all numbers
Zero is neither negative nor positive

The question states that x is a positive number and therefore it is not zero


In Statement 2
Zero is not an option and all the other numbers suffice

IMO B

So Ian, what do you suggest.

STMT 2 is sufficient if we rule out x=1 i,e x-1 =0 to be a factor of 104?

Please let me know.

We cannot rule out x=1 because it satisfies all the conditions i.e

x>0, x=int and x is a divisor of 27. However, if x=1, then x-1=0 and 0 is not a factor of 104, so stmt 2 will never be sufficient by itself.

cubicle_bound_misfit wrote:So Ian, what do you suggest.

STMT 2 is sufficient if we rule out x=1 i,e x-1 =0 to be a factor of 104?

Please let me know.

i concur with iwanta700plus. the answer is C

I corrected it. Thanks for pointing it out and all your help on the forum.

Ian Stewart wrote:

cubicle_bound_misfit wrote:So Ian, what do you suggest.

STMT 2 is sufficient if we rule out x=1 i,e x-1 =0 to be a factor of 104?

Please let me know.

iwant700plus's solution above is perfect (except that '1' was left out of the list of factors of 104, though that doesn't affect the solution at all). I was only pointing out why you need to consider the possibility that x = 1 here, which is what makes Statement 2 insufficient. C is the answer.

I don't get it. Why is the answer C?

Let's look at the question again... 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.

For x - 1 to be a factor of 104, x -1 must be {1, 2, 4, 8, 13, 21, 52, 104}.
Therefore, x must be in {2, 3, 5, 9, 14, 22, 53, 105}.

(1) tells us that x is divisible by 3... so x could be 3, 9, 105... not sufficient.
(2) tells us that 27 is divisible by x... so x could be 1, 3, 9... not sufficient.
Together, (1) and (2) tell us that x could be either 3 or 9... not sufficient.

I think the answer should be E. If it isn't, can someone please tell me why?

Thanks.

If x is 3 or 9, then it would be sufficient because x-1 = 2, 8 which are factors of 104.

Anyway, in stmt 2, you left out 27, which is divible by 27. that also gives x-1=26 which is a factor of 104.

Remember to go back to the question, which is asking whether x-1 is a factor of 104.

avssrs wrote:I don't get it. Why is the answer C?

Let's look at the question again... 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.

For x - 1 to be a factor of 104, x -1 must be {1, 2, 4, 8, 13, 21, 52, 104}.
Therefore, x must be in {2, 3, 5, 9, 14, 22, 53, 105}.

(1) tells us that x is divisible by 3... so x could be 3, 9, 105... not sufficient.
(2) tells us that 27 is divisible by x... so x could be 1, 3, 9... not sufficient.
Together, (1) and (2) tell us that x could be either 3 or 9... not sufficient.

I think the answer should be E. If it isn't, can someone please tell me why?

Thanks.

Ian Stewart wrote:

hrishikesh05 wrote:Zero is divisible by all numbers

Let's look at zero in more detail:

- Zero is divisible by all integers except zero.
- Zero is *not* a divisor of any integer.

In other words, you can divide zero by almost anything, but you can't divide anything by zero. In this question, we aren't dividing zero by anything, so the first fact above is not relevant to this question. It's the second fact that's important: we want to know whether 0 is a divisor of 104, and the answer is 'no'.

hrishikesh05 wrote: The question states that x is a positive number and therefore it is not zero

Yes, x cannot be zero, but we want to know whether x-1 is a divisor of 104, and not whether x is a divisor of 104. If x = 1, then x is positive, and x - 1 = 0.

Is zero the multiple of anything? (besides itself) My understanding of a multiple is that y is a multiple of x iff x times another integer equals y.

0 is an integer, and x times 0 equals 0. Does this make 0 the multiple of x??

Good point. I failed to see that this question's answer is a yes/no as opposed to a single value. Thanks for the clarification!