Remainder question?

Question

As n goes through the positive integers (i.e. n=1,n=2,...) How many different values can the remainder take when n^2 is divided by n+4?

a) 1
b) 4
c) 8
d) 9
e) infinitely many

Answer 1

The answer is "d". There are only 9 different remainders you can get. Here's why.

First, find the first few remainders. For n^2 < n+4, the remainder is just going to be n^2 itself. So n=1 gives a remainder of 1 and n=2 gives a remainder of 4. Then you can show manually that n=3 gives a remainder of 2, and n=4 gives a remainder of 0.

Now consider the remainders for n>4. Using long division, you can show that n^2/(n+4) = (n-4) + 16/(n+4). So for n>4, the (n-4) term is going to be positive, and the remainder is just going to be the same remainder you get if you divide 16 by (n+4). But notice that if 16 < (n+4), then 16/(n+4) is just going to be 0 with a remainder of 16. So for n > 12, the remainder is ALWAYS going to be 16.

Now we just have to find the remainders for 4 < n < 13. Using the 16/(n+4) short cut, we just have to find the remainders of 16 divided by 9, 10, 11, 12, 13, 14, 15, and 16. 16/9 gives a remainder of 7, so the other remainders are just going to decrease: 7, 6, 5, 4, 3, 2, 1, 0. The other remainders we found were 1, 4, 2, 0, 16. So eliminating the repeats, that makes 9 different values for remainders.

Answer 2

n^2/(n+4)
=(n-4)*(n+4) +16. The remainder is always 16 if n+4>16 n>12
If n<=12 there are 8 different values of the remainder
So in total there are nine values of the remainder

Answer 3

c, but even tho this question is in the intemediate maths challenge, and i'm 22 i don't know why!!

Answer 4

e.