RonPurewal wrote:
i've noticed that you're all just taking ONE case for each remainder of n/5, and blindly trusting that \(n^{2}\)/5 (for statement 1) and 2n/5 (for statement 2) will always have the same result for ALL cases of that remainder.
more specifically:
for the case in which n/5 gives remainder = 1... you're just examining n = 6, and blindly trusting that the same result will obtain for n = 11, 16, 21, 26, etc.
if you understand the actual theory behind why this is so — AND/OR if you've worked out the algebra to prove it (which i'll do in the answer key i'm going to post in 24-36 hours) — THEN you can reliably generalize from these single cases.
otherwise that's a huge risk (...1 data point doesn't constitute a "pattern"!), and you should mitigate that risk by testing a couple of different n's for each possible remainder of n/5.
for calculating remainder of n/5, there can be only 5 possible values for r = 0,1,2,3,4 . Since we are given that n is not divisible by 5 , there are only cases r=1,2,3,4
Lets say n = 5 + x ,
1. Remainder [5+x] > Remainder [(5+x)^2]
Remainder [x] > Remainder [x^2]
Since rest of the terms are divisible by 5 and hence remainder = 0 for them.
Now since we are checking for remainder for 5. x can have only 5 unique values, because all other values beyond x=5 can again be reduced to 1,2,3,4 and 5.
This is the reason only checking for 6,7,8,9,10 is sufficient
2. Similarly for second statement, it boils down to
Remainder [2x] < Remainder [x]