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

Question

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 ?

Bunuel · Accepted Answer

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.

Does this make sense?

bluesquare · Answer

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 .

AbhiJ · Answer

My bad, mistook factors for multiples.

hsbinfy · Answer

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.

Vaibhav21 · Answer

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

Bunuel · Answer

\

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.

architkap · Answer

Have a doubt with 1)

just testing values, n can be 99, 399, 899 etc. If n=99 then 3^2*11 so smallest difference is 2. If n = 399 then smallest difference is 4 (3*7*19). If n = 899 then smallest difference is 2 again (29*31) so how can we come to a conclusion here?

If  Bunuel   KarishmaB  yall can help me out here

egmat · Answer

Hi architkap,

Your setup is right: statement (1) does force n to be  99 ,  399 ,  899  and so on. The slip is the smallest difference you found for  399 .

Recheck n = 399.  You factored it as  3  ×  7  ×  19  and got  4 . But list  all  the factors, not just the primes:

- Factors of  399 :  1, 3, 7, 19, 21, 57, 133, 399 
-  1  and  3  are both factors and differ by exactly  2 . So do  19  and  21 .

So 399 gives  2 , not 4. The 4 came from comparing prime factors while skipping the fact that  1 is always a factor , and any n divisible by 3 automatically has 1 and 3 sitting 2 apart.

All three of your cases now agree:  99 ,  399  and  899  each give  2 .

Why it is always 2.  Statement (1) gives n = ( 10 k -  1 )( 10 k +  1 ).

- Those two factors are  exactly 2 apart , so the answer can never be more than  2 .
- n =  100 k^2 -  1  is  odd , so it has no even factors. A difference of  1  needs one even and one odd factor, impossible here, so the answer can never be less than  2 .

Squeezed between "at most 2" and "at least 2," it is exactly  2  for every allowed n. Same answer every time, so statement (1) is  sufficient .

The takeaway: when hunting the smallest gap between factors, include  1  and the composite factors, not just the primes.

Answer: A

KarishmaB · Answer

As per statement 1,  \frac{\sqrt{n+1}}{10}  is a positive integer so the square root should give a clean multiple of 10. Hence n+1 must be 100*(a perfect square) i.e. 100 or 400 or 900 etc.
So n = 99 or 399 or 899 etc. It will be 1 less than a multiple of 100 so it will definitely be an odd number.

99 has 9 and 11 as factors so the minimum difference between two factors in case of 99 is 2. 
Now the question is whether the difference can be 1 or 0. We know that till now it can be 2.

Since all possible values of n are odd and none of them can be a perfect square (since n+1 is a perfect square), so n cannot have two factors which are the same and it cannot have two consecutive numbers (odd and even) as factors.

That is why the minimum possible difference in case of any value of n is 2. Some values of n may have a minimum difference of more than 2 between their factors, but the minimum possible value is 2.

Hence statement 1 is sufficient.

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

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