dtweah wrote:There are how many integers n, n > 3, such that (n – 3) divides (n2 – n)?

(a) 1
(b) 2
(c) 3
(d) More than 3 but a finite number
(e) infinitely many

My answer would be D.

My interpretation is [n^2 -n]/(n-3) leaves no remainder for how many numbers ??


I reduced the above equation to [n(n-1)]/(n-3), and then substituted different values,

I saw that 4,5,6,9 satisfied the given condition and thus D.

Please correct if wrong.

IMO D. Did a manual check. I hope there is a smart way of answering this problem.

IMO D

N^2 - N can be written as N(N-1) so we need to find values of N in which N-3 divides N(N-1)

Let's assume that N= X + 3 and therefore N-1 = X + 2 and N - 3 = X

the new equation can be written as

(X+3)(X+2) / X ---> X^2 + 5X + 6 / X . This is true if X is a factor of 6 so X values can be {1,2,3,6} which is a finite set of numbers.

Can an expert check if my solution is correct?

Vemuri wrote:IMO D. Did a manual check. I hope there is a smart way of answering this problem.

Yes there is:

Do the long division:
n2-n/n-3= n + 2 + 6/(n-3). You see that for n>9, you cant' get an integer. So D

dtweah wrote:Do the long division:
n2-n/n-3= n + 2 + 6/(n-3). You see that for n>9, you cant' get an integer. So D

I actually did the long division when first attempting the question. This is what I did:

(n-3)(n+2) + 6 = n^2-n

But, I did not know what to do next to simplyfy & quickly answer the question. So, went about checking the answer manually :roll:

Given that n^2 - n is divisible by n-3

In questions like these try to express n^2 - n as as a combination of
n-3 .

We have n^2 - n = (n-3)(n+2) + 6

Now (n-3)(n+2) + 6 must be divisible by n-3

Since (n-3)(n+2) is divisible by n-3, 6 must be divisible by n-3.

So we can conclude that for n^2-n to be divisible by n-3, 6 must be divisible by n-3.

So n-3 can take the values of all the factors of 6 i.e. 1, 2, 3 and 6.

Hence n = 4, 5, 6 and 9.

Also given that n > 3 and all the above values satisfy this condition.

So n take only 4 values