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Re: xy multiple of 105 [#permalink]
05 Mar 2011, 01:19
write this as follows:
x=6*a y=14*b
where a, b integers
is xy is a multiple of 105? (or 35*3 or 3*7*5) ?
combine x*y=6*a*14*b= 2*3*2*7*a*b so to asnwer the question we must have at least a multiple of 3*5*7=105 out of which 3*7 exist . So the question reduces to whether either x or y have a 5 in itself or not.
1) not sufficient, we have been provided with data that x is a multipple of 9, but we do not have any additional information about y it may be or may not be a multiple of 5. we don't know exactly.
2) sufficient, becuase we are provided with the information that y is a multiple of 5 and we are provided with info that x and y are multiples of 6 and 14, it means that we have all sufficnet information to answer the question.
To check : replace the y with any other number y=14*25*g , so g is any integer number becuase y is a multiple of both 14 and 25. and x is a multiple of 6 , so if combined you will have x*y=6*a*14*25*g=2100*a*g - this will produce a number which is a multiuple of 105. (2100 is divisible by 105). _________________
Re: If positive integer x is a multiple of 6 and positive [#permalink]
12 Mar 2015, 17:28
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Re: If positive integer x is a multiple of 6 and positive [#permalink]
17 Apr 2015, 09:54
Bunuel wrote:
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.
Hi Bunuel, are there any more questions like this one? It's a subject I need practice on. Thanks!
Re: If positive integer x is a multiple of 6 and positive [#permalink]
23 Apr 2015, 23:20
Expert's post
tgubbay1 wrote:
Bunuel wrote:
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.
Hi Bunuel, are there any more questions like this one? It's a subject I need practice on. Thanks!
Re: If positive integer x is a multiple of 6 and positive [#permalink]
24 Apr 2015, 03:52
Hi, I the answer of given question IS B? approach: we need to have 5 in the y to make x as its factor too statement 1 ) y could be 18/ 54/ or 90... which involves 5 in one case and not in others. hence insufficient statement 2 ) z has unit digit of non zero, but its even. so this removes the probability of having z as 5. hence y has to be 5 or any number with unit digit as zero. this makes y as a factor of 5. this will result into integer. sufficient.
kindly share the answer and the approach. if the answer is correct then it calls for a kudos. thanks
Why don't you give an attempt on this practice question:
Q: \(x, y\) are positive integers such that \(x\) is the lowest number which has 15 and 20 as its factor. Is \(x\) a factor of \(y\)?
(1) \(\frac{y}{2 *3^2}\) is an integer.
(2) \(\frac{y * z}{5}\) is an integer where \(z\) is a positive even integer whose units digit is not 0.
Let me know your approach and analysis for this practice question.
Re: If positive integer x is a multiple of 6 and positive [#permalink]
24 Apr 2015, 04:25
Expert's post
Celestial09 wrote:
Hi, I the answer of given question IS B? approach: we need to have 5 in the y to make x as its factor too statement 1 ) y could be 18/ 54/ or 90... which involves 5 in one case and not in others. hence insufficient statement 2 ) z has unit digit of non zero, but its even. so this removes the probability of having z as 5. hence y has to be 5 or any number with unit digit as zero. this makes y as a factor of 5. this will result into integer. sufficient.
kindly share the answer and the approach. if the answer is correct then it calls for a kudos. thanks
Let me ask you a question. How did you infer that \(y\) needs to have only 5 as its factor to make \(x\) as its factor. Or to rephrase my question, what do you mean by "\(x\) is the lowest number which has 15 and 20 as its factor". Why don't you give it an another try.
Re: If positive integer x is a multiple of 6 and positive [#permalink]
27 Apr 2015, 05:39
Hi, It should be C. But still I would like to have an explanation as still I on my way to hard ds questions. Though I guess it is a milder version of it but lacking in explanation. Thanks Celestial
EgmatQuantExpert wrote:
Celestial09 wrote:
Hi, I the answer of given question IS B? approach: we need to have 5 in the y to make x as its factor too statement 1 ) y could be 18/ 54/ or 90... which involves 5 in one case and not in others. hence insufficient statement 2 ) z has unit digit of non zero, but its even. so this removes the probability of having z as 5. hence y has to be 5 or any number with unit digit as zero. this makes y as a factor of 5. this will result into integer. sufficient.
kindly share the answer and the approach. if the answer is correct then it calls for a kudos. thanks
Let me ask you a question. How did you infer that \(y\) needs to have only 5 as its factor to make \(x\) as its factor. Or to rephrase my question, what do you mean by "\(x\) is the lowest number which has 15 and 20 as its factor". Why don't you give it an another try.
Step-I: Given Info We are given two positive integers \(x\) and \(y\) such that \(x\) is the lowest number that has 15 & 20 as its factors. We are asked to find if \(x\) is a factor of \(y\).
Step-II: Interpreting the Question Statement We are asked to find if \(x\) is a factor of \(y\). For \(x\) to be a factor of \(y\), \(y\) should have the prime factors of \(x\) in their respective powers as its factors.
We are given that \(x\) is the lowest number that has 15 & 20 as its factors. We know that the lowest number which has two numbers as its factor is the LCM of those numbers. Prime factorizing 15 and 20 would give us
\(15 = 3^1 * 5^1\) and \(20 = 2^2 * 5^1\). We know that for calculating LCM we take the highest powers of prime factors. So, LCM (15, 20) = \(2^2 * 3^1 * 5^1\)
\(x=2^2*3^1 * 5^1\) . For \(x\) to be a factor of \(y\), \(y\) should have \(2^2, 3^1\) and \(5^1\)as its factors. Let’s see if the statements provide us sufficient information about the prime factors of \(y\).
Step-III: Analyze Statement-I independently Statement- I tells us that \(\frac{y}{2*3^2}\) is an integer i.e. \(y\) has \(2\), \(3^2\) as its factors. For \(x\) to be a factor of \(y\), \(y\) should have \(2^2, 3^1\) and \(5^1\) as its factors. We don’t know about the other prime factors of \(y\).
So, statement-I is not sufficient to answer the question.
Step-IV: Analyze Statement-II independently Statement-II tells us that \(\frac{y*z}{5}\) is an integer where \(z\) is a positive even integer whose units digit is not 0. Since, \(z\) is a positive even integer whose units digit is not 0, we can say that \(z\) is not a multiple of 5. Hence if \(\frac{y*z}{5}\) is an integer with \(z\) not being a multiple of 5, that would mean that \(y\) is a multiple of 5.
But we don’t know if \(y\) has \(2^2, 3^1\) as its factors.
So, statement-II is not sufficient to answer the question
Step-V: Combining Statements I & II Combining statement- I & II, we can see that \(y\) has \(2,3^2,5\) as its factors. For a moment, combining both the statements seem sufficient to answer the question. But if we look back at step-II, we can see that for \(x\) to be a factor of \(y\), \(y\) should have \(2^2, 3^1\) and \(5^1\) as its factor. The power of the prime factor\(2\) is not sufficient to tell us if \(x\) is a factor of \(y\). Hence, we can’t say if \(x\) is a factor of \(y\).
So, combining both the statements is also not sufficient to answer the question.
Answer: (E)
Key takeaways For most of the LCM-GCD questions, prime factorization is the key to the solution.
If positive integer x is a multiple of 6 and positive [#permalink]
01 May 2015, 09:56
tgubbay1 wrote:
Bunuel wrote:
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.
Hi Bunuel, are there any more questions like this one? It's a subject I need practice on. Thanks!
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