1. ## help meeee

9^2007 devided by 8. find the remainder

2. Originally Posted by bananalion
9^2007 devided by 8. find the remainder
Are you doing modular mathematics?

9 = 1 (mod 8)

So
9^n = 1^n (mod 8) = 1 (mod 8) for integer n

Thus 9^{2007} = 1 (mod 8) so 9^{2007}/8 has a remainder of 1.

-Dan

3. no......the question is only on the algebra.....u think its possible?

4. Originally Posted by bananalion
no......the question is only on the algebra.....u think its possible?
Just look at the pattern then.

9 leaves remainder 1 upon division by 8.

9^2 = 81 leaves remainder 1 upon division by 8.

9^3 =729 leaves reaminder 1 upon division by 8.

It seems that 9 to any exponenets leaves the same remainder.

5. thank you very much

2003x2004x2005+2006^3 devided by 7

but not using calculator

6. do you think the remainder is 0?

7. Originally Posted by bananalion
thank you very much

2003x2004x2005+2006^3 devided by 7

but not using calculator
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