9^2007 devided by 8. find the remainder

Results 1 to 7 of 7

- May 8th 2007, 03:55 AM #1

- Joined
- May 2007
- Posts
- 4

- May 8th 2007, 04:49 AM #2

- May 8th 2007, 05:27 AM #3

- Joined
- May 2007
- Posts
- 4

- May 8th 2007, 05:31 AM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- May 8th 2007, 05:35 AM #5

- Joined
- May 2007
- Posts
- 4

- May 8th 2007, 05:43 AM #6

- Joined
- May 2007
- Posts
- 4

- May 8th 2007, 05:54 AM #7
2003*2004*2005 = (2006 - 3)(2006 - 2)(2006 - 1)

= 2006^3 - 6*2006^2 + 11*2006 - 6

So

2003*2004*2005 + 2006^3 = 2*2006^3 - 6*2006^2 + 11*2006 - 6

Now, similar to the idea from before, let's get the remainders of each of the powers of 2006:

2006/7 leaves a remainder of 4

2006^2/7 leaves a remainder of 4^2 = 16

2006^3/7 leaves a remainder of 4^3 = 64

Now,

16 = 2*7 + 2, so this is a remainder of 2.

64 = 9*7 + 1, so this is a remainder of 1.

Thus

(2003*2004*2005 + 2006^3)/7 = (2*2006^3 - 6*2006^2 + 11*2006 - 6)/7

leaves a remainder of:

2*1 - 6*2 + 11*4 - 6 = 2 - 12 + 44 - 6 = 28

And 28 = 4*7 + 0

So you are correct, this leaves a remainder of 0 when you divide by 7.

-Dan