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Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:18
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If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n 260 = 52*5 Possible values of n = 52 —> 169 is not a divisor of n n = 52*13 —> 169 is a divisor of n
Insufficient
(2) 1,352 is the least common multiple of 104 and n
1352 = 169*8 = \(13^2\)*8 104 = 13*8 Possible values of n = \(13^2\) —> 169 is a factor of n n = \(13^2\)*2 —> 169 is a factor of n n = \(13^2\)*4 —> 169 is a factor of n n = \(13^2\)*8 —> 169 is a factor of n
If n is a positive integer, is 169 a factor of n?
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Updated on: 24 Jul 2019, 18:33
1
From statement we have that n has to be a multiple of 13^2 if we want to prove that 169 is a factor of n From (1) 52 is the greatest common divisor of 260 and n, we have that:
GCD(260,n)= 52 = 2^2* 13
260 = 2^2*5*13
From this one, we have that n could have 13 or 13^2, or 13^p, so we cannot assure that n is going to be a multiple of 169, so insufficient.
From (2) 1,352 is the least common multiple of 104 and n, we have that
1352 = 2^3 * 13^2, and since 104 = 2^3 * 13, n must have a factor of 13^2 or 169, so sufficient. (B) is our answer.
Originally posted by Mizar18 on 24 Jul 2019, 08:29.
Last edited by Mizar18 on 24 Jul 2019, 18:33, edited 1 time in total.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:32
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B I think.
We are basically looking for n, and checking is 169 or 13^2 can be a factor of n.
1) 52 is the greatest common divisor of 260 and n In this case, n has to be a multiple of 52 or n = 52*x = 2^2*13*x, where x is an integer, except multiples of 5. Because if x is a multiple of 5, n = 260*y then the GCD will be 260. So, x may not be 13 necessarily, so insufficient.
(2) 1,352 is the least common multiple of 104 and n 1352 = 2^3*13^2 104 = 2^3*13 So, n has to be 2^3*13^2 else 1352 will not be a LCM of n and 104. So, n is divisible by 13^2 or 169. Sufficient.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:32
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If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n (2) 1,352 is the least common multiple of 104 and n
169 ; 13^2 #1 52 is the greatest common divisor of 260 and n not necessarily that 52 being divisor of n would also give 169 as its factor #2 1,352 is the least common multiple of 104 and n 1352/104 = 13 so n has to be be a multiple factor of 13 , so 169 would be factor of n as n=169 IMO B
If n is a positive integer, is 169 a factor of n?
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Updated on: 24 Jul 2019, 21:09
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Given that, n is a positive integer. We need to find whether 169 a factor of n? I.e. n = 169*A? = 13*13*A? We need to determine whether n is at least two 13s as its prime factor(s).
(1) 52 is the greatest common divisor of 260 and n Therefore, 52 * X = 13*4*X = 13*2*2*5 = 260 52 * Y = 13*4*Y = 13*2*2*Y = n
We are not sure what value Y can take. It may or not consist of 13 in its factors.
Not Sufficient.
(2) 1,352 is the least common multiple of 104 and n Therefore, 104 * A = 1352 n * B = 1352 => n * B = 2*2*2*13*13
Now, LCM of two numbers will have prime factors of both the numbers raised to their highest power. \(1352 = 2^3 * 13^2\) \(104 = 2^3 * 13\) Thus, n has to be \(n = 2^x * 13^2\), where x <= 3
Hence, n has at least two 13s as its prime factors.
Sufficient
Answer B
Originally posted by Sayon on 24 Jul 2019, 08:34.
Last edited by Sayon on 24 Jul 2019, 21:09, edited 1 time in total.
If n is a positive integer, is 169 a factor of n?
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Updated on: 24 Jul 2019, 09:10
1
Quote:
If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n (2) 1,352 is the least common multiple of 104 and n
As 169 = 13 * 13 n should have\(13^2\) in its prime factorization.
Statement 1: 52 = 4*13 =\(2^2 * 13\) and as its the GCD, n should have\(2^2 * 13\) but that does not justify if it has 13 * 13 as a factor. n can be 169 or 52 hence we get a yes or a no Insufficient.
Statement 2: As least common multiple either of the numbers should have the factors of 1352 1352 = 13 * 13 * 8 But 104 = 13* 4 therefore the extra 13 in the LCM should come from n, hence n has to have a factor of atleast\(13^2\) Hence 169 will be a factor of n Sufficient.
Option B is the answer
Originally posted by techloverforever on 24 Jul 2019, 08:35.
Last edited by techloverforever on 24 Jul 2019, 09:10, edited 1 time in total.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:36
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If n is a positive integer, is 169 a factor of n?
This is an easy question, All you need to check is if N is greater or equal to 169.
Let's see the statements.
(1) 52 is the greatest common divisor of 260 and n This is insufficient as N could be 52 or 104 or 260 etc.
(2) 1,352 is the least common multiple of 104 and n This is the correct choice as the number is greater than in 1352 or equal to 1352. Hence solves our purpose.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:46
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If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n \(52=2^2*13\) \(260=2^2*5*13\) This statement only gives us that n has no factor 5. n could have different values, including \(2^2\) and \(13\) as factors. Thus, insufficient.
(2) 1,352 is the least common multiple of 104 and n
\(1352=2^3*13^2\) \(104=2^3*13\)
This statement tells us that n definitely has \(13^2\) as a factor, also it could or could not have \(2\), or \(2^2\) or even \(2^3\) as factors, but this is optional.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:46
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Quote:
If n is a positive integer, is 169 a factor of n?
This is Data Sufficiency (DS) question, and we are told that n>0. We need to find out if 169 is a factor of n or, in other words, if n is divided by 169. If we write down 169 as the product of its prime factors, we get \(169 = 13^2\) Let us analyze each statement separately and find out if we can come to any conclusion.
Statement 1: (1) 52 is the greatest common divisor of 260 and n Let us write down the number 52 as multiplication of its prime factors. In this case, \(52 = 2^2 * 13\) As it can be seen from above, the numbers 52 and 169 have only one common factor which is 13. From this information, we cannot conclude whether number 169 is a factor of n or not. Insufficient.
Statement 2: (2) 1,352 is the least common multiple of 104 and n Just as in the previous case, let us factorize the number 1352 which is LCM for 104 and n: \(1352 = 2^3 * 13^2\) And \(104 = 2^3 * 13\) As LCM is the least number that can be divided by both n and 104, and it will have all prime factors of n and 104, let us find out what are distinct prime factors which are included in 1352, but not in 104: this is 13. As the number 104 already has 13 in it, it is obvious that number n is multiple of \(13^2 = 169\) since \(13^2\) is a factor of 1352. Thus, the number 169 is a factor of n. Sufficient. Statement 2 alone is sufficient.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 08:49
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The answer is B.
For 169 to be a factor of n, then n needs to have the prime factors of 13^2 within.
(1) 260 prime factors are 2^2 x 5^1 x 13^1 while 52 prime factors are 2^2 x 13^1. from this, we can tell that n has at least one power of 13, but we cannot tell if it has 13^2. hence, insufficient.
(2) 104 prime factors are 2^3 x 13^1, while 1,352 prime factors are 2^4 x 13^2. Since 104 only has one power of 13, we know the 13^2 comes from n. Hence, this is sufficient.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 09:05
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If n is a positive integer, is 169 a factor of n? Given: n > 0 & integer n = 169*m ? where m is an integer from 1 to any m => n = 13^2*m?
(1) 52 is the greatest common divisor of 260 and n --> insufficient: 260 = 52*5, so GCD of 52*5 and n = 52, so n = 52*p=13*4*p, but p can or can't be multiple of 13 (2) 1,352 is the least common multiple of 104 and n --> sufficient: 104 = 13*2^3, so LCM of 13*2^3 & n = 1,352 = 13^2*2^3, so n must be multiple of 13^2
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 09:25
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If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n - Insufficient - 52 factorized is 13*2^2 and 260 factorized is 13*5*2^2. Hence N factorized might contain 13^2 or just 13. If n contains 13^2 then 169 will be a factor of n, if not then 169 won't be a factor. Example n can be = 676 <52*12> or 52 itself. when n=676 then 169 is a factor and when n=52 169 is not a factor (2) 1,352 is the least common multiple of 104 and n - Sufficient - 1352<factorized> =13^2*2^3 and 104<factorized>=13*2^3. So for LCM to contain 13^2, n must contain 13^2. Hence 169 must be a factor of n.
Re: If n is a positive integer, is 169 a factor of n?
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24 Jul 2019, 09:29
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If n is a positive integer, is 169 a factor of n?
(1) 52 is the greatest common divisor of 260 and n 52=2*2*13 n=\(2^2*13k\) But 169=\(13^2\) and n is not necessarily a multiple of 169 or 169 is not necessarily a factor of n NOT SUFFICIENT
(2) 1,352 is the least common multiple of 104 and n 1352=13*104 = \(2^3*13^2\) 104=2*2*2*13 = \(2^3*13\) Since 1352 is least common multiple of 104 and n =>\(13^2\) is a factor of n => 169 is a factor of n SUFFICIENT
IMO B
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