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If n is a positive integer greater than 1 [#permalink]

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15 Jul 2013, 08:40

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If n is a positive integer greater than 1, what is the smallest positive difference between two different factors of n?

(1) \(\frac{\sqrt{n+1}}{10}\) is a positive integer.

(2) n is a multiple of both 11 and 9.

My question is: how would you rephrase the STEM?

a) Since the smallest positive difference between any two different factors of n is n. Would you rephrase the question as: If n can be any one of 2,3, 4..., what is the value of n ?

or

b) Would you go one step ahead and say what is the smallest possible value of n ?

Re: If n is a positive integer greater than 1 [#permalink]

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15 Jul 2013, 09:06

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Expert's post

AbhiJ wrote:

If n is a positive integer greater than 1, what is the smallest positive difference between two different factors of n?

Suppose this is a question stem of a DS problem, how would you rephrase this statement ?

a.) Since the smallest positive difference between any two different factors of n is n. Would you rephrase the question as: If n can be any one of 2,3, 4..., what is the value of n ?

or

b.) Would you go one step ahead and say what is the smallest possible value of n ?

Your conclusion in (a) is not correct.

Generally the smallest positive difference between two different factors of some integers is greater than or equal to 1. For example:

If n=2, then the smallest positive difference between two different factors of n is 1 --> 2-1=1. If n=3, then the smallest positive difference between two different factors of n is 2 --> 3-1=2. If n=4, then the smallest positive difference between two different factors of n is 1 --> 2-1=1. If n=5, then the smallest positive difference between two different factors of n is 4 --> 5-1=4. If n=6, then the smallest positive difference between two different factors of n is 1 --> 2-1=1. If n=7, then the smallest positive difference between two different factors of n is 6 --> 7-1=6. ... If n=9, then the smallest positive difference between two different factors of n is 2 --> 3-1=2. ... If n=15, then the smallest positive difference between two different factors of n is 2 --> 3-1=2. ...

Basically if n=even, then the smallest positive difference between two different factors of n is always 1. If n is odd, then the smallest positive difference between two different factors of n is even, so greater than 1.

Re: If n is a positive integer greater than 1, what is the smallest positi [#permalink]

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01 May 2015, 07:42

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A it is.

1. n can be 99 ,9999,999999 and so on..coz then only it will make a perfect square and result in an interger after division by 10.

so 99=11 * 9 9999= 11 * 9 * 101 999999=11 * 9 * 10101(which is further simplified to 3*7*13*37)

but we are asked the smallest....so 11 and 9 will always appear..sufficient i.e. 2 the smallest diff cant be 1 coz there need to be 2 and 3 as factors...and n is would be never divisible by2.

2.insuff

n can be multiple multiple of 8 or not. either would get diff results.

If n is a positive integer greater than 1, what is the smallest positi [#permalink]

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01 May 2015, 08:41

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I say the answer is A.

(I) indicates that there is some positive integer \(X\) such that \(\frac{\sqrt{n+1}}{10}=X\) --> \(\sqrt{n+1}=10X\) --> \(n+1 = 100X^2\) --> \(n = 100X^2 - 1\) --> \(n = (10X)^2-1^2\) --> \(n = (10X-1)(10X+1)\), and so the smallest difference between any two distinct factors of \(n\) will always be at least \(2\).

Additionally, \(n\) will always be odd because \(10X\) will be even, so \(10X-1\) will be \(odd\), and \(odd*odd = odd\). Because \(n\) is \(odd\), the smallest difference cannot be \(1\), which would imply \(even*odd\).

Therefore, the smallest difference is \(2\), and so (I) is sufficient.

(II) is not sufficient because \(n\) could be \(9*10*11=990\).

Re: If n is a positive integer greater than 1, what is the smallest positi [#permalink]

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02 May 2015, 14:23

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St 1 root(n+1)/10 is a pos int(1,2,3,4.....) only possible when numerator is a multiple of 10(10,20,30) and to get 10 as numerator n=99;20 as numerator n=399; 30 as numerator n=899 and if you carefully notice all the values of n have 2 factors common i.e 3 and 1 hence the min diff = 2 Satisfied St 2 N is a multiple of both 11 and 9 so? if n= 11x9x2 factors = 1,2,3,11 smallest diff = 1 AND if N=11X9 factors 1,3,11 samllest diff = 2 THUS NS

Factors are integers, by definition, so the smallest possible difference between any two factors has to be at least 1. For example, if n = 2, then the number has factors 1 and 2, and the smallest positive difference between those factors is 1. If, on the other hand, n =3, then the number has factors 1 and 3, and the smallest positive difference between those two factors is 2.

On this problem, statement 2 is (arguably) easier, so you might choose to start there.

(2) NOT SUFFICIENT. If n is a multiple of both 11 and 9, then it could be 99. In this case, the factors would be 1, 9, 11, and 99, and the smallest difference between two factors would be 2. On the other hand, n could be 198, with factors 1 and 2 (among others). In this case, the smallest difference is only 1.

(1) SUFFICIENT. What can this strange expression indicate about the value of n? We’re going to need to dig into number theory a bit here.

If that whole expression represents a positive integer, then squaring it would represent a perfect square of an integer:

\(\frac{n+1}{100} = perfect \ square\)

Use the variable p to represent the perfect square, just to make this easier to write:

Remember that p is a perfect square, so the square root of p is still an integer. This last equation means that one factor of n is \(10\sqrt{p}+1\) and another factor of n is \(10\sqrt{p}-1\) (where \(10\sqrt{p}\) is an integer). These two factors, then, are really “an integer + 1” and “that same integer – 1.” In other words, these two integers are 2 units apart.

But is that the smallest possible distance between two factors? Here’s the best (and trickiest) part. Remember this stage of the equation simplification above?

n=100p-1

That step means: n equals an even number minus 1. In other words, n is odd!

The only way that two factors can be a distance of just 1 unit apart is when one of those factors is even and one of those factors is odd. If n itself is odd, though, then it cannot have any even factors.

Because n is odd, it isn’t possible for two of the factors to be just 1 unit apart. Therefore, the smallest possible distance between two factors is indeed 2.

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