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

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

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.

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.

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

Hi Bunuel, are there any more questions like this one? It's a subject I need practice on. Thanks!

Hi tgubbay1,

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.

Also, you may try your hand at this question which tests similar concepts if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html

Regards
Harsh

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.

Also, you may try your hand at this question which tests similar concepts if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html

Regards
Harsh[/quote]

Hi Celestial09,

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.

Regards
Harsh

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

Detailed Solution:

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.

To know more in detail about the likely mistakes in LCM-GCD question, read the article 3 mistakes in LCM-GCD questions

Regards
Harsh

Target question: Is xy a multiple of 105?

Important stuff:
First, If N is a multiple of k, then N is divisible by k.

Second, a lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7


Since 105 = (3)(5)(7), then we can rewrite the target question as . . .
Rephrased target question: Is there a 3, a 5 and a 7 hiding in the prime factorization of xy?

Given: x is a multiple of 6
In other words, x = (2)(3)(other possible prime numbers)

Given: y is a multiple of 14
In other words,y = (2)(7)(other possible prime numbers)

Combine both of the above to see that xy = (2)(2)(3)(7)(other possible prime numbers)

So, the given information tells us that we ALREADY have a 3 and a 7 hiding in the prime factorization of xy. The only piece missing is the 5.

So, we can rephrase our target question one last time. . .

Rephrased target question: Is there a 5 hiding in the prime factorization of xy?

Now we can check the statements.

Statement 1: x is a multiple of 9.
Since 9 = (3)(3), all this tells us is that there are two 3's hiding in the prime factorization of xy.
So, there may or may not be a 5 hiding in the prime factorization of xy.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y is a multiple of 25.
Since 25 = (5)(5), this tells us is that there is definitely a 5 hiding in the prime factorization of xy.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

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

105 = 3*5*7

x = 6k; where k is an integer
y = 14m; where m is an integer
xy = 6k * 14m = 84km = 2^2*3*7km; where km is an integer
xy is a multiple of 3*7. If it also multiple of 5, then xy will be a multiple of 105.

(1) x is a multiple of 9.
NOT SUFFICIENT

(2) y is a multiple of 25.
xy is a multiple of 3*7*5 = 105
SUFFICIENT

IMO B

Hi Bunuel, can you help me with compilation of such questions??
I need more practice in this topic

5. Divisibility/Multiples/Factors

• Theory
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  Tips and hints: Divisibility/Multiples/Factors
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  THE MOST EFFICIENT WAY TO STUDY LEAST COMMON MULTIPLES ON THE GMAT
  
  • Questions
    The E-GMAT Primes Trio - 3 Ques on No. of factors & Prime factors
    DS Questions
    PS Questions

For other topics:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

I understand the solution for the question, but how it is possible to have an integer which is multiple of both 14 and 25? According to the main statement y is a multiple of 14, then according to the 2nd statement y is also a multiple of 25. How such an integer is possible?

MiraliYahya, a number can be a multiple of many different numbers.

"y is a multiple of 14" simply means that "14 is a factor of y", meaning "y includes 2 and 7 as prime factors (and may or may not also include other factors)".

Similarly, "y is a multiple of 25" means "y includes \(5^2\) as prime factors."

Both of these statements can be true at the same time.

The least common multiple of 14 and 25 is 2*7*\(5^2\) = 350. So, y could be anything in this arithmetic sequence: 350, 700, 1050, etc.

Let us modify the question,

If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 420 ?
(1) x is a multiple of 9.
(2) y is a multiple of 25.


What will be the answer?