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# If positive integer x is a multiple of 6 and positive

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Director
Joined: 15 Aug 2005
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If positive integer x is a multiple of 6 and positive  [#permalink]

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09 Oct 2005, 01:53
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64% (01:37) correct 36% (02:13) wrong based on 814 sessions

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If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105?

(1) x is a multiple of 9.
(2) y is a multiple of 25.

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Cheers, Rahul.
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06 Aug 2010, 09:21
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5
zest4mba wrote:
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ?
(1) x is a multiple of 9.
(2) y is a multiple of 25.

Can someone explain this

$$105=3*5*7$$. Since $$x$$ is a multiple of 6 and $$y$$ is a multiple of 14, then $$xy$$ is a multiple of $$LCM(x,y)=2*3*7$$: we have 3 and 7 as factors of $$xy$$, so in order $$xy$$ to be a multiple of 105 we need missing 5 to be a factor of either $$x$$ or $$y$$.

(1) x is a multiple of 9 --> we don't know whether 5 is a factor of either $$x$$ or $$y$$. Not sufficient.

(2) y is a multiple of 25 --> 5 is a factor of $$y$$, hence $$xy$$ is a multiple of 105. Sufficient.

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26 Apr 2009, 15:01
3
2
B
factors of 105 are 3, 5, 7
14 gets you the 7
6 gets you the 3
so all you need is something to get you the 5

A) doesn't solve anything, you can never get a 5 in the unit place with just 14, 6, and 9
B) gives you the 5 you need
##### General Discussion
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09 Oct 2005, 06:07
rahulraao wrote:
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105?

(1) x is a multiple of 9.
(2) y is a multiple of 25.

xy is a multiple or 105---->at least x or y must be a multiple of 5
(1) is insufficient.
(2) is sufficient.
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09 Oct 2005, 06:17
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1
105 has prime factors: 21 * 5 = 3 * 7 * 5

Using (1), x is also a multiple of 9. Then smallest possible value of X is 18. We can't tell whether xy is a multiple of 105 as we cannot determine if there is a prime factor 5 present in Y.

Using (2), y is a multple of 25. So we have prime factors 7, 5, 2, and 3. So xy is a multiple of 105.

Ans: B
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12 Nov 2009, 00:09
This quesiton can be solved if reduced to prime factors.

Given Information:

Let n represent some integer...

a) x = 6n = (3)(2)n
b) y = 14n = (7)(2)n

We know that therefore: xy = (3)(7)(2)(2)n
Is xy = 105n? = (5)(3)(7)n?

So essentially we need to know that either x or y is ALSO a multiple of 5

Statement 1: x is a multiple of 9.

i.e. x = (3)(3)n. This gives no information whether x is a multiple of 5. Therefore, insufficient.

Statement 2: y is a multiple of 25.

i.e. y = (5)(5)n. Bingo! Sufficient.

Therefore the correct answer is B.
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12 Nov 2009, 00:20
1
x is a multiple of 6==> x has at least two factors, 2 & 3
y is a multiple of14==> y has at least two factors, 2 & 7

xy will be a multiple of 105, if the factors of xy (combined factors of x & y) can be formed together to make 105.

Statement 1: x is a multiple of 9==> In addtion to 2 & 3 , x also has one more factor 3. So x has at least three factors 2,3 & 3. Now if we multiply x & y and hence multiply the possible factors of x & y we will have the following possible factors of xy: 2,3,3,2 & 7. Now we can't form 105 by multiplying any of these numbers. Hence we can say that statement 1 is not sufficient.

Statement 2: y is a multiple of 25>> In addition to 2 & 7, y also has two more factors, 5 & 5 or 5^2. Now if we multiply x & y and hence multiply the possible factors of x & y we will have the following possible group 2,3,7,5 & 5. You can see that we can easily form 105 by multiplying 3 , 7 & 5. Statement 2 is sufficient

This is my way of thinking. Kindly correct me if I am wrong.
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Updated on: 12 Nov 2009, 02:28
Although the answer is correct. The method could be wrong

AKProdigy87 wrote:
a) x = 6n = (3)(2)n
b) y = 14n = (7)(2)n

If x= 6n, you cannot assume that y= 14n; y could also be 14m

AKProdigy87 wrote:
We know that therefore: xy = (3)(7)(2)(2)n

xy need not be a equal to (3)(7)(2)(2)n; xy is a multiple of (3)(7)(2) =42

Originally posted by jade3 on 12 Nov 2009, 00:38.
Last edited by jade3 on 12 Nov 2009, 02:28, edited 1 time in total.
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12 Nov 2009, 01:24
Although the answer is correct. The method is wrong

AKProdigy87 wrote:
a) x = 6n = (3)(2)n
b) y = 14n = (7)(2)n

If x= 6n, you cannot assume that y= 14n; y could also be 14m

AKProdigy87 wrote:
We know that therefore: xy = (3)(7)(2)(2)n

xy need not be a equal to (3)(7)(2)(2)n; xy is a multiple of (3)(7)(2) =42

I used n to represent an integer (any integer)... not as a means of equating that the integer n was the same in both cases. I can see how the confusion could arise though.
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12 Nov 2009, 23:59
1
First we need to figure out what the factors of 105 are:

Factors of 105: 1, 3, 5, 7

So we need figure out if x and y share the factors [1,3,5,7]:

Factors of x (6): 1,2,3,6
Factors of y (14): 1,2,7

Between factors of 6 and 14, we are missing only the factor 5.

Statement 1:
Factors of x (9): 1,3,3
NOT SUFFICIENT

Statement 2:
Factors of y (25): 1,5,5
SUFFICIENT

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10 Dec 2009, 13:01
1
REMEMBER THAT ANY NUMBER WILL BE A MULTIPLE OF ANOTHER NUMBER IF IT SHARES THE SAME PRIME FACTORS

IF x is a multiple of 6 then it will have 2 and 3 as prime factors
If y is a multiple of 14 then it will have 2 and 7 as multiples

XY has prime factors 2,3 and 7
The number 105 has prime factors, 3, 5 and 7

As 5 is the only missing prime factor between X and Y, So we need a 5 somewhere between X or Y to make it a multiple of 105

Statement 1 -----INSUFFICIENT

X is a multiple of 9 and therefore has an additional 3 as prime factor. Not needed for our answer

Statement 2-----SUFFICIENT

Y is a multiple of 25 therefore Y has a 5*5 in it
Bingo! We needed a 5. Hence Sufficient

Ans - B
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23 Feb 2010, 08:18
X is a multiple of 6 => X=6a
Y is a multiple of 14 => Y=14b

Now XY = 6*14*a*b = 3*2*2*7*a*b

The question asks if XY is multiple of 105.
105 = 5*21 = 5*3*7

When you see XY, we have 3 and 7 already present in the multiple. The only other thing required is 5.

So either X or Y should be multiple of 5.

Second option shows Y is multiple of 25 which in turn is multiple of 5. So this is sufficient for the solution.

Hope it helps.
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23 Feb 2010, 08:56
If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105?
x = 2 * 3 * a
y = 2 * 7 * b
xy = 3 * 5 * 7 * c????
where a,b,c etc are integers
(1) x is a multiple of 9
x = 3*3 * d .. so from the above equations x can be written as x=2*3*3 * e
the product of xy can be written as xy = 2*3*3*7 * f
if f is a multiple of 5 then xy will be multiple of 105 else no
Insufficient
(2) y is a multiple of 25
y = 5*5*g ... so from the above equations y can be written as y=2*7*5*5 *h
the product of xy can be written as xy = 2*3*5*5*7 * i = 3*5*7 * 2*5*i = 105 * 2*5*i
xy is a multiple of 105
Sufficient

B
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24 Feb 2010, 10:12
nickk wrote:
It seems I'm quite weak with multiples and factors, even after reading the explanation I still don't get it

If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105?
(1) x is a multiple of 9
(2) y is a multiple of 25

OA:

Let x be 6K and y be 14L

x=2x3xK
y=2x7xL

105=3x5x7

xy=(2^2)(3)(7)(K)(L)

xy has 3,7 but is missing 5. Any opttion that provides this value will be the answer.

So, option B
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12 Mar 2010, 02:15
nickk wrote:
It seems I'm quite weak with multiples and factors, even after reading the explanation I still don't get it

If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105?
(1) x is a multiple of 9
(2) y is a multiple of 25

OA:

if x is multiple of 6 and y is multiple of 14 xy is definitely multiple of 2*3*7 (as this is the LCM of 6 and 14)
now 105 comprises 5*3*7
so xy has to have 5*3*7 as its factor in order to be the multiple of 105
stmt1 says x is a multiple of 9 dat means xy is multiple of 2*3*3*7 for sure but this does not prove if it is multiple of 105 or not
stmt2 says y is a multiple of 25 so xy will definitely be multiple of 2*3*7*5*5 and this LCM contains 105 in it which proves that xy is a multiple of 105.

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22 Mar 2010, 03:55
Friends i dont quite understand the point here

if X is a multiple of 6 and 9 both as per statment 1, then X should be 18, 36, 72 and so on....... in any case xy is not a multiple of 105. Hence, sufficient

Am i doing anything wrong here?
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22 Mar 2010, 04:04
hardnstrong wrote:
Friends i dont quite understand the point here

if X is a multiple of 6 and 9 both as per statment 1, then X should be 18, 36, 72 and so on....... in any case xy is not a multiple of 105. Hence, sufficient

Am i doing anything wrong here?

X is a multiple of 6 and y is a multiple of 14.
but X or y can be multiple of other numbers also like 4,5,11 etc.
xy is definitely a multiple of 2*3*7 = 42 (Taking the LCM)
but if we need to find if xy is a multiple of 18 we have to be sure that xy contains 5*3*7 at least.
Since xy does contain 3,7, so it can be multiple of 105 only if it also has 5 as a multiple too.
Only second statement states that y is also a multiple of 25. which means
XY will be multiple of LCM(6, 14, 25) = 2*3*7*5*5 = 1050
since XY is multiple of 1050 its definitely a multiple of 105.
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01 Apr 2010, 15:21
XY can be expressed as (3)(5)(7)r, as the prime breakdown, for the number to be a multiple of 105

From the stem x will already have the numbers (2)(3) to be a multiple of 6
From the stem y will already have the number (7)(2) to be a multiple of 14

So, all we really need is for one of them to have the number 5

A) Prime breakdown is (3)(3) - that doesn't help since we already have a 3 in 6

B) Prime breakdown is (5)(5) - that's the number we need

So B

Hope this helps.
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03 Apr 2010, 15:54
Or get the LCM of 6, 14 and the numbers in the options and it should be divisible by 105, in this case its 1050, so its B
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25 Oct 2010, 22:23
always making number properties look so simple Bunuel!
Re: DS QUESTION   [#permalink] 25 Oct 2010, 22:23

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