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If n is a positive integer, is 169 a factor of n?

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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 24 Jul 2019, 08:00
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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

Sufficient

IMO Option B

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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post Updated on: 25 Jul 2019, 09:56
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


Hence B

1. 52= 13*4
We have no idea if n has one more 13 as factor. Not sufficient

2. 1352=13*13*8
n definitely has 13*13
Therefore, 169 is a factor of n

Hence B
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Originally posted by kitipriyanka on 24 Jul 2019, 08:19.
Last edited by kitipriyanka on 25 Jul 2019, 09:56, edited 2 times in total.
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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post Updated on: 24 Jul 2019, 09:14
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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


n=Z+ is 169 or 13ˆ2 factor of n?

(1) 52 is the greatest common divisor of 260 and n:
GCF(260,n)=52=13*4… n=13*4*some multiple or 13*4*13*some multiple… insufficient;

(2) 1,352 is the least common multiple of 104 and n:
LCM(104,n)=1352=13*104=13*2*52=13*13*8,… since 104=13*8, then, n=13ˆ2*some multiple, sufficient;

Answer (B).

Originally posted by exc4libur on 24 Jul 2019, 08:23.
Last edited by exc4libur on 24 Jul 2019, 09:14, edited 1 time in total.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 24 Jul 2019, 08:23
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(1) 52 is the greatest common divisor of 260 and n
not sufficient

(2) 1,352 is the least common multiple of 104 and n
divide number by 104 and 169 you can tell the number is divisible by 169
sufficient
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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post Updated on: 24 Jul 2019, 18:33
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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.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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
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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.
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If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post Updated on: 24 Jul 2019, 09:10
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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.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.

The answer is B.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 24 Jul 2019, 08:42
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Given: n is an integer.
To determine: n is divisible by 169 or 13^2

(1) HCF ( 2^2 x 13 x 5, n) is 2^2 x 13
n can be 2^2 x 13 or 2^2 x 13^2. So insufficient.

(2) LCM ( 2^3 x 13, n) is 2^3 x 13^2
n has to have 13^2 as a factor. Sufficient.

B is correct.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.

Sufficient.


IMO B
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.

Answer: B
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 24 Jul 2019, 08:53
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Statement 1:
(1) 52 is the greatest common divisor of 260 and n

260 = 52X5
n=52(A) ==> A is positive integer

if A = 3 then 169 is not a factor of n
if A=13 then 169 is not factor of n.
no Conclusive answer from Statement 1 so it is insufficient.

Statement 2:
(2) 1,352 is the least common multiple of 104 and n

1352 = 169X8
104 = 13X8
For LCM we need 169
So n==>169K where k is positive integer.

Hence, B is the answer.
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 24 Jul 2019, 09:02
(1) 52 is the greatest common divisor of 260 and n
Says that 52 is a factor of n

(2) 1,352 is the least common multiple of 104 and n
1352=104x13
Says that 13 is a factor of n

But we still don't know if 169 is a factor of n
Answer: E
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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

So the answer is B
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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.

IMO B
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Re: If n is a positive integer, is 169 a factor of n?  [#permalink]

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New post 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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Re: If n is a positive integer, is 169 a factor of n?   [#permalink] 24 Jul 2019, 09:29

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