If n is a positive integer greater than 1, what is the smallest positi

My bad, mistook factors for multiples.

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.


Kudos for a correct solution.

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.


So its A.

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

Ans = A

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MANHATTAN GMAT OFFICIAL SOLUTION:

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:

\(\frac{n+1}{100}=p\)
\(n+1=100p\)
\(n=100p-1\)
\(n=(10\sqrt{p}+1)(10\sqrt{p}-1)\)

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.

The correct answer is A.

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