Is the integer x divisible by 6?

[arielle.bertman]

Is the integer x divisible by 6?

(1) x + 3 is divisible by 3
(2) x + 3 is an odd number

I got this question right by creating a table with various numbers and testing values (x = 3, 6, 9, 12...), although this process was a bit time consuming and I was less than confident with my answer. I know these are common questions on the GMAT - was wondering if there is a faster way or more methodical way to solve using theory. If not, what are your suggestions for the best way to select values to test each condition?

Thanks so much for your help!

Arielle

[venkat_yj]

Try this:

for x to be divisible by 6, it has to be divisible by factors of 6-> 3 and 2.

From (1), we can conclude that x is divisible by 3 (if x+3 is divisible by 3, it means x+3 is a multiple of 3 or x is a multiple of 3)

From (2), we can conclude that x is divisible by 2 (if x+3 is an odd number, x+2 has to be an even number, meaning x itself is even divisible by 2)

We need both conditions to conclusively say x is divisible by both 3 and 2. So answer should be C (now I hope thats the case :-)

[Ben Ku]

Thanks, venkat_yj, for providing a clear and straightforward answer. I agree with the answer (C).

[cwhardie]

Couldn't x also be zero?

0 +3 is divisible by 3 and
0 + 3 is an odd number

zero fit both categories but is not divisible by 6 so I thought the answer to this problem was E (stmts together were not sufficient)

Thanks in advance,
Carl

[abhyuday.c]

I agree with cwhardie. What if x = 0?

Shoudn't the answer be E?

[debmalya.dutta]

The answer should be E . 0 is an integer and hence it fits.

[sanyalpritish]

0 is divi by 6 ...

0 is multiple of all no.
0 is not a factor of any no.

[cwhardie]

Sanyalpritish,

I'm sorry but I don't understand the answer you posted. Could you explain it a little more?

Thanks

[sanyalpritish]

Well I meant that

0 is a factor of all Numbers (1,2,3.....)
[editor: this should say "multiple", not "factor".
a factor is the same thing as a divisor, so this statement, as written, contradicts the following one.]

0 is not a divisor of any number (1/0,2/0....)

[RonPurewal]

Couldn't x also be zero?

0 +3 is divisible by 3 and
0 + 3 is an odd number

zero fit both categories but is not divisible by 6 so I thought the answer to this problem was E (stmts together were not sufficient)

Thanks in advance,
Carl

actually, zero is divisible by 6. in fact,
* zero is divisible by any positive integer
* zero is a multiple of any positive integer

the above are facts -- which, given this problem, it's apparently important that you know -- but this really doesn't seem to be the sort of thing that the gmat would test. (on EVERY other official question that i have seen, the test has absolutely gone out of its way to exclude "tricky" cases such as this one.)

to all posters:
has anybody seen this actual problem on the software? are we sure that the problem statement doesn't say "POSITIVE integer", not just "integer"?

[askhoman4]

to all posters:
has anybody seen this actual problem on the software? are we sure that the problem statement doesn't say "POSITIVE integer", not just "integer"?

Hi Ron,
Yes, I can confirm that the problem says "integer", not "positive integer". It came up in GMAT Prep1.

[mschwrtz]

Notice that the apparently tricky case 0 isn't really a tricky case at all. It gives the same answer as every other legal value of x.

Well, that might be a bit strong. The way that 0 could trick you here is if it occurred to you to test 0, but you didn't know its properties.

You're smart to consider that 0 works differently than other numbers in lots of circumstances, but Ron is right (as usual) that the GMAT won't generally exploit special properties of 0 in questions like this one.

Students will sometimes ask "But isn't 0 divisible by __?" I will answer "Yeah, but that doesn't change the answer."

[k.mayank]

The correct Answer is (C)

First of all X is an integer.Integer can be positive,0 and negative numbers.

Also it is important to note "A number is even if it is a multiple of two, and is odd otherwise".
Therefore, a negative number can be even or odd.Because definition of odd and even suggests just a number.No where mentioned positive number.Therefore a negative number can be odd or even depending in its divisibility by 2.If it is leaving a remainder 0 when divided by 2 then it is even otherwise odd.

Now let us discuss the solution ,

CASE [1]
Consider x =0
Statement(1) : 0+3 = 3 => divisible by 3.Here,x=0
Therefore,
x=0 is also divisible by 6 leaving remainder as 0.

Statement(2) : 0+3 is an odd number..Here,x=0
Therefore,
x=0 is also divisible by 6 leaving remainder as 0.

CASE[2]
Consider x= -3
Statement(1) : -3+3 = 0 => divisible by 3.Here,x=-3
Therefore,
x=-3 is not divisible by 6 leaving remainder as 3.Always remember remainder by definition are non negative in nature.For example when -26 is divided by 6 we say remainder is 4 (not -2)

Statement(2) : -3+3= 0 is not an odd number.0 is neither odd nor even.Here,x=-3
Therefore,
x=-3 is not divisible by 6.


CASE[3]
Consider x= -6
Statement(1) : -6+3 = -3 => divisible by 3.Here,x=-6
Therefore,
x=-6 is divisible by 6.
Statement(2) : -6+3= -3 is an odd number.
Therefore,
x=-6 is divisible by 6.


CASE[4]
Consider x= 3
Statement(1) : 3+3 = 6 => divisible by 3.Here,x=3
Therefore,
x=3 is not divisible by 6.
Statement(2) : 3+3= 6 is an even number.

Therefore,using Statement(1) we ca not clearly say that whether integer X will be divisible by 6 or not.
But the moment we say x+3 is odd then we can say yes X will be divisible by 6.

[jlucero]

mayank,

Your method here works fine, but in this particular problem, you don't actually need to use negative numbers. The same theory would be true if you chose 3, 6, 9, 12... etc. Also, the logic behind the question makes the problem much easier:

even + even = even
div by 3 + div by 3 = div by 3
div by 7 + div by 7 = div by 7

Think of it this way, if you had a bunch of $20 bills in your pocket and picked up some more $20 bills, you couldn't have $64 in your pocket, you'd need some multiple of 20. Statement 1 tells us x is divisible by 3, Statement 2 tells us x is divisible by 2. Together, x must be divisible by 6.