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GMATPrep Integers

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by Hate Standardized Tests » Thu Nov 13, 2008 11:01 am
You'd think that integers is the easiest math topic, but I'm absolutely stumped as to why the correct answer is right. Maybe I'm interpreting the question wrong?

Is the integer x divisible by 6?
(1) x + 3 is divisible by 3
(2) x + 3 is an odd number.

Answer choices:
(a) Statement 1 alone is sufficient, but Statement 2 alone is not sufficient.
(b) Statement 2 alone is sufficient, but Statement 1 alone is not sufficient.
(c) Either statements are NOT sufficient, but BOTH together are sufficient.
(d) Each statement ALONE is sufficient.
(e) Neither statement Alone is sufficient.


Any one have any insights?[spoiler] Correct Answer: c [/spoiler]
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by cramya » Thu Nov 13, 2008 11:16 am
Hate Standardized Tests,

Stmt I)

x+3 is divisible by 3

x+3/3 = someintger
x = 3(someinteger) + 3

For eg: someintger = 1 Then x is divisible by 6
someintger = 2 Then x is not divisible by 6

INSUFF

Stmt II

x+3 is an odd number

x+3 = odd

From this we know x is even (since even+odd = odd)
x=4 not divisble by 6
x=6 divisible by 6

INSUFF

Stmt I and II together

x+3 is diviisble by 3
x+3 is odd

For both of these to be true x has to be a mutiple of 6

Eg:0 3 6 9 12 15 18 21

x could be 0,6, 12,18 etc...
Last edited by cramya on Thu Nov 13, 2008 11:27 am, edited 2 times in total.
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Re: GMATPrep Integers

by logitech » Thu Nov 13, 2008 11:19 am
Is the integer x divisible by 6?

(1) x + 3 is divisible by 3

x+3=3,6,9,12,15,18, etc...

x=0,3,6,9,12,15, etc

If you pay attention:

6 and 12 are divisible by 6 and they are even, so x+3= ODD

INSUF

(2) x + 3 is an odd number.



3,5,7,9,11,13, etc

so x=0,2,4,6,8,

1+2 together we know that X is divisible by 6

Hence, C
LGTCH
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by srisl11 » Thu Nov 13, 2008 12:00 pm
Is x divisible by 6?

We can rephrase that as is x an even multiple of 3?

1)
since 3 is divisible by 3 , x should be divisible by 3 for x+3 to be divisible by 3
So Statement 1 tells us that x is divisible by 3
But we don't know whether x is even or odd

2)
For x+3 to be odd , x must be even
Statement 2 tells us that x is even
but we dont know whether x is a multiple of 3

Statement 1 & 2 tells us that x is a even multiple of 3
=> x is divisible by 6

So C is the Answer
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by Tryingmybest » Thu Nov 13, 2008 12:01 pm
If you dont want to play with pluggin numbers. This will help.

Statement 1
X+3 is divisible by 3
=> 3(X+3) + 0 where 0 is remainder

This is not sufficient
Statement 2:

X+3 is odd => 2K+1 format

Not enuf

Combine both
3(2K+1+3) +0
6k+12
which is divisible by 6 as 12/6 leaves raminder 0
So choice C :D
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by sushanta57021 » Thu Nov 13, 2008 12:14 pm
what if X= 0 ; is 0 divisible by 6?
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by Tryingmybest » Thu Nov 13, 2008 12:55 pm
3(2K+1+3) +0
6k+12

here k can be negative or positive or zero , no matter what it is ,the remainder will be 0 when divided by 6

Also,0 is divisible by any number other than 0.
Hope this helps. :D
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by Hate Standardized Tests » Tue Nov 25, 2008 1:45 pm
Thanks everyone who gave their thoughts.

I talked with a friend and here's her way of solving it.

Statement (1):
We know that x can be 0, 3, 6, 9, 12, 15, 18, 21 … because 0 + 3 = 3 and 3 is divisible by 3; 3 + 3 = 6 and 6 is divisible by 3; and so on. However, we can’t say that 0, 3, 6, 9, 12, 15, 18, 21 … is each divisible by 6. So Statement (1) alone is NOT sufficient.

Statement (2):
We know that x can be 0, 2, 4, 6, 8 … because 0 + 3 = 3 and 3 is an odd number; 2 + 3 = 5 and 5 is an odd number; and so on. However, we can’t say that 0, 2, 4, 6, 8 … is each divisible by 6. So Statement (2) alone is NOT sufficient.

Combine both Statement (1) and (2):
Now go back to the series of x under Statement (1), which are 0, 3, 6, 9, 12, 15, 18, 21 … and see which one of them when add up with 3 can be an odd number. Let’s take 0 + 3 = 3, yes 3 is an odd number; 3 + 3 = 6, no 6 is not an odd number; 6 + 3 = 9, yes 9 is an odd number; and so on.
Now we have a new series for x, which are 0, 6, 12, 18 … Each of this is divisible by 6, so both Statements combined together ARE sufficient.
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Re: GMATPrep Integers

by Stuart@KaplanGMAT » Tue Nov 25, 2008 1:53 pm
Hate Standardized Tests wrote:You'd think that integers is the easiest math topic, but I'm absolutely stumped as to why the correct answer is right. Maybe I'm interpreting the question wrong?

Is the integer x divisible by 6?
(1) x + 3 is divisible by 3
(2) x + 3 is an odd number.

Answer choices:
(a) Statement 1 alone is sufficient, but Statement 2 alone is not sufficient.
(b) Statement 2 alone is sufficient, but Statement 1 alone is not sufficient.
(c) Either statements are NOT sufficient, but BOTH together are sufficient.
(d) Each statement ALONE is sufficient.
(e) Neither statement Alone is sufficient.

Simplifying the statements often simplifies life!

Q: is x a multiple of 6?

(1) x + 3 is a multiple of 3. Well, since 3 is a multiple of 3 all by itself, statement (1) really says "x is a multiple of 3".

If x is a mult of 3, it may or may not be a mult of 6: insuff.

(2) x + 3 is odd. Well, if x + 3 is odd, then x is even.

If x is even, it may or may not be a mult of 6: insuff.

Together:

(1) x is a mult of 3
(2) x is even; therefore, x is a mult of 2

if x is a mult of 2 AND a mult of 3, it's definitely a mult of 6 (which is just 2*3): sufficient, choose (C).
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by siddhans » Wed Jul 20, 2011 3:12 pm
Hate Standardized Tests wrote:Thanks everyone who gave their thoughts.

I talked with a friend and here's her way of solving it.

Statement (1):
We know that x can be 0, 3, 6, 9, 12, 15, 18, 21 � because 0 + 3 = 3 and 3 is divisible by 3; 3 + 3 = 6 and 6 is divisible by 3; and so on. However, we can�t say that 0, 3, 6, 9, 12, 15, 18, 21 � is each divisible by 6. So Statement (1) alone is NOT sufficient.

Statement (2):
We know that x can be 0, 2, 4, 6, 8 � because 0 + 3 = 3 and 3 is an odd number; 2 + 3 = 5 and 5 is an odd number; and so on. However, we can�t say that 0, 2, 4, 6, 8 � is each divisible by 6. So Statement (2) alone is NOT sufficient.

Combine both Statement (1) and (2):
Now go back to the series of x under Statement (1), which are 0, 3, 6, 9, 12, 15, 18, 21 � and see which one of them when add up with 3 can be an odd number. Let�s take 0 + 3 = 3, yes 3 is an odd number; 3 + 3 = 6, no 6 is not an odd number; 6 + 3 = 9, yes 9 is an odd number; and so on.
Now we have a new series for x, which are 0, 6, 12, 18 � Each of this is divisible by 6, so both Statements combined together ARE sufficient.
I dont understand the new series...why is it 0,6,12...Combined x + 3 has to be odd and not even ...
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