# Divisible by 49

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• Aug 25th 2006, 10:00 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
wouldn't it imply that n+1=k+1???

Are you familiar with induction?
If not another excellent thing for you to learn.
(But first you would have to answer the questions I gave you before we can procede to a new topic).
• Sep 6th 2006, 04:18 AM
malaygoel
a TRY
I followed TPH's advice and see what I got.
Let $S_n=n^2 + n +2$.
I divide the set of n in two parts:
One is of the form 7k-4 and second the rest.
For the first form, you can directly check by substituting the general form 7a-4 for n.
Now the second case.
We could see that $S_1, S_2,S_4,S_5,S_6,S_7$ also satisfy the given condition.
Now, $S_{7+k}=S_k + 7(7+2k+1)$
If you substitute the value of k=1,2,4,5,6,7 you get that $S_8.....S_{14}$ are not divisible by 7.
Continuing this , now you can substitute the value k=8,9,11,12,13,14 and you get the non-divisibility of next six(out of seven, one you have alredy proved) by 7.
In this way you can complete the proof by mathamatical induction.

Bye.
• Sep 6th 2006, 03:03 PM
Quick
Welcome Back
Hello Malay!
I thought I wouldn't hear from you for another 6-9 months?
• Sep 8th 2006, 02:27 AM
malaygoel
Quote:

Originally Posted by Quick
Hello Malay!
I thought I wouldn't hear from you for another 6-9 months?

Even I was thinking the same.
It is fortunate that there is a computer centre in the college where I can aceess you mhf anytime. Whenever I find time from college studies(which I barely get), I try to visit MHF. Although, by mid october, I will have internet in the room.

Keep Smiling
Malay
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