January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 15 Aug 2005
Posts: 766
Location: Singapore

If positive integer x is a multiple of 6 and positive
[#permalink]
Show Tags
09 Oct 2005, 00:53
Question Stats:
63% (01:37) correct 37% (02:10) wrong based on 868 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Cheers, Rahul.




Math Expert
Joined: 02 Sep 2009
Posts: 52386

Re: DS QUESTION
[#permalink]
Show Tags
06 Aug 2010, 08:21
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. Answer: B.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




SVP
Joined: 24 Sep 2005
Posts: 1783

Re: DS  Multiple
[#permalink]
Show Tags
09 Oct 2005, 05:07
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.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4803
Location: Singapore

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



VP
Joined: 07 Apr 2009
Posts: 1208
Concentration: General Management, Strategy
Schools: Duke (Fuqua)  Class of 2012

Re: multiples
[#permalink]
Show Tags
26 Apr 2009, 14:01
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



Manager
Joined: 11 Sep 2009
Posts: 129

Re: multiple of 14
[#permalink]
Show Tags
11 Nov 2009, 23:09
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.



Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1220
Concentration: Strategy, General Management

Re: multiple of 14
[#permalink]
Show Tags
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.
_________________
[ From 470 to 680My Story ] [ My Last Month Before Test ] [ GMAT Prep Analysis Tool ] [ US. Business School Dashboard ] [ Int. Business School Dashboard ]
I Can, I Will
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 19 Nov 2007
Posts: 179

Re: multiple of 14
[#permalink]
Show Tags
Updated on: 12 Nov 2009, 01:28
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 11 Nov 2009, 23:38.
Last edited by jade3 on 12 Nov 2009, 01:28, edited 1 time in total.



Manager
Joined: 11 Sep 2009
Posts: 129

Re: multiple of 14
[#permalink]
Show Tags
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.



Manager
Joined: 13 Aug 2009
Posts: 176

Re: multiple of 14
[#permalink]
Show Tags
12 Nov 2009, 22:59
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
ANSWER: B.



Manager
Joined: 17 Aug 2009
Posts: 175

Re: multiple of 14
[#permalink]
Show Tags
10 Dec 2009, 12:01
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 2SUFFICIENT
Y is a multiple of 25 therefore Y has a 5*5 in it Bingo! We needed a 5. Hence Sufficient
Ans  B



Manager
Joined: 29 Oct 2009
Posts: 166
Concentration: General Management, Sustainability
WE: Consulting (Computer Software)

Re: Multiples question
[#permalink]
Show Tags
23 Feb 2010, 07:18
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.



Manager
Joined: 26 May 2005
Posts: 188

Re: Multiples question
[#permalink]
Show Tags
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
B



Manager
Joined: 10 Feb 2010
Posts: 155

Re: Multiples question
[#permalink]
Show Tags
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 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



Manager
Joined: 13 Dec 2009
Posts: 224

Re: Multiples question
[#permalink]
Show Tags
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 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. Hence, B is the answer
_________________
My debrief: doneanddusted730q49v40



Manager
Joined: 05 Mar 2010
Posts: 172

Re: Multiples question
[#permalink]
Show Tags
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 Am i doing anything wrong here?
_________________
Success is my Destiny



Manager
Joined: 13 Dec 2009
Posts: 224

Re: Multiples question
[#permalink]
Show Tags
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.
_________________
My debrief: doneanddusted730q49v40



Intern
Joined: 17 Dec 2009
Posts: 24

Re: OG DS 82
[#permalink]
Show Tags
01 Apr 2010, 14:21
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.



Manager
Joined: 24 Mar 2010
Posts: 88

Re: OG DS 82
[#permalink]
Show Tags
03 Apr 2010, 14:54
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
_________________
Please do consider giving kudos if you like my posts



Manager
Joined: 07 Jan 2010
Posts: 112
Location: So. CA
WE 1: 2 IT
WE 2: 4 Software Analyst

Re: DS QUESTION
[#permalink]
Show Tags
25 Oct 2010, 21:23
always making number properties look so simple Bunuel!




Re: DS QUESTION &nbs
[#permalink]
25 Oct 2010, 21:23



Go to page
1 2
Next
[ 35 posts ]



