An integer n is chosen at random from a set of first 100 positive inte

Question

An integer n is chosen at random from a set of first  100  positive integers. What is the probability that  (n-1)(n+1)  is divisible by  8 ?

(A)  0.49 
(B)  0.50 
(C)  0.60 
(D)  0.61 
(E)  0.25

IanStewart · Accepted Answer

If n is even, n-1 and n+1 are consecutive odd numbers, and (n-1)(n+1) won't even be divisible by 2, let alone by 8. If n is odd, n-1 and n+1 are consecutive even numbers, and since you always have one multiple of 4 among any two consecutive even numbers, (n-1)(n+1) is the product of a multiple of 4 and a multiple of 2, and will definitely always be divisible by 8 if n is odd. Since exactly half of the possible values of n here are odd, the answer is 1/2.

Vetrick · Answer

When we take any odd number as n,  (n+1) * (n-1)  is divisible by 8.

1 can also be included for (n+1) (n-1) = 0 in this case since 0 is divisible by any number including 8.

Since there are 50 odd numbers between 1 to 100.

0.5 is the probability and B is the answer.

ermohit86 · Answer

Vetrick 
I can find only 3 numbers - 5,7,9 which satisfies the given condition (n-1)(n+1) -> n^2 - 1 should be divisible by 8.
Can you please elaborate if I misinterpret the question?

PyjamaScientist · Answer

ermohit86 : If you take n = 11. You will get (11-1)(11+1) = 10*12 = 2*5*2*2*3 =  8 *15. Similarly, if you take n = 13, (13-1)(13+1) = 12*14 = 4*3*2*7 =  8 *21.
So, for all odd values of 'n' you will get a product of (n-1)(n+1) a multiple of 8.

Vetrick · Answer

ermohit86

It's nice that you have simplified the (n-1) (n+1) to n^2-1. But even if you take that simplified formula, you will get answers which are divisible by 8 for all odd numbers. 3^2-1 = 9-1 =8 ; 91^2 - 1 = 8281 - 1 = 8280 (when divided by 8 you get a quotient of 1035 and a remainder of 0). Try it for a few more numbers.

ermohit86 · Answer

Vetrick   PyjamaScientist 
Hi, I know my mistake now. I misinterpret the question as the value of n^2-1 is given between 1 to 100 but actually its value of n. Thanks for your help.

ermohit86 · Answer

I want to share another method of solving such questions:

Given:
The range of n is 1 to 100.
(n-1)(n+1) should be divisible by 8.

Solution: Make 12 set of 8 numbers: Set1: 1,2,3,4,5,6,7,8 Set2: 9,10,11,12,13,14,15,16 .... Set13: 97,98,99,100
You don't have to write all the sets.

Using 1-2 sets and you will see the pattern that each set has 4 numbers, for which n^2-1 is divisible by 8.
Set 1 numbers are (1,3,5,7), for set 2 number are (9,11,13,15).
So for 12 sets, we got 48 numbers. The last set (set 13) has 2 numbers (97,99) which also satisfies the given condition so. Overall we have 50 numbers. 
So B is the answer.

Note: This method is applicable for various other combinations as well like 100 numbers and n(n-1) divisible by 6 etc.

Abir77 · Answer

Interesting !
I have thought about it too. But made set1 : 1,2,3,4,5,6,7,8,9,10

Can you elaborate how did you come up with set that is 1 to 8 and then on and on ?

Vishal120 · Answer

When we are taking odd number then above condition is not divisable by 8

so we need to take even numbers 
also

there is total 50 number between 1 to 100 so its probability will be

0.50

Posted from my mobile device

ermohit86 · Answer

For this particular question, making a group of 10 numbers should work as well. But as I mentioned GMAT may give different conditions e.g. in another question it is asked if n(n+1) is divisible by 6 with n from 1 to 100. 
For this equation n(n+1)
set1 as (1,2,..10) ... we got 5 numbers that satisfy given condition
set2 as (11,12,..20) we got 6 numbers that satisfy given condition. Hence pattern break.
But if we make sets of 6 numbers:
set1 as (1,2,..6) ... we got 4 numbers that satisfy given condition
set2 as (7,8,..12) ... we got 4 numbers that satisfy given condition. Pattern remains.

So to generalize the process of solving such questions it is best practice to form a set of n numbers where n is given in the question stem as a condition. 
Hope this clarifies.

liammoyn · Answer

The question says that n is between 1 and 100, not that (n-1)(n+1) is between 1 and 100. Therefore n=11 is a valid solution along with every other odd number less than 100.

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