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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.

Re: multiple of 14 [#permalink]
11 Nov 2009, 23:20

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

Hence answer is "B".

This is my way of thinking. Kindly correct me if I am wrong. _________________

Re: multiple of 14 [#permalink]
12 Nov 2009, 00:24

jade3 wrote:

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.

Re: Multiples question [#permalink]
23 Feb 2010, 07:56

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

Re: Multiples question [#permalink]
24 Feb 2010, 09:12

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

Re: Multiples question [#permalink]
12 Mar 2010, 01:15

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

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.

Re: Multiples question [#permalink]
22 Mar 2010, 02:55

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

Re: Multiples question [#permalink]
22 Mar 2010, 03:04

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. _________________

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.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...