please help me

a.h.m.a.d · #1 $Old$ May 11th, 2009, 02:03 PM

please help me

if A= 1^5 + 2^5 + 3^5 + .......+ (101)^5

find the remains after we divide A by 7

TheEmptySet · #2 $Old$ May 11th, 2009, 02:45 PM

Quote:

Originally Posted by a.h.m.a.d $View Post$

please help me

if A= 1^5 + 2^5 + 3^5 + .......+ (101)^5

find the remains after we divide A by 7

Note that by fermat's little thereom that

$a^{6} \equiv 1 \mod (7)$
If we multply by sides by $a^{-1}$ we get

$a^5 \equiv a^{-1} \mod (7)$

We can do this becuase we know that every element has an inverse mod 7.

8=1 mod 7
9=2 mod 7
ect....

see if this helps

Soroban · #3 $Old$ May 11th, 2009, 02:59 PM

Hello, a.h.m.a.d!

Are we allowed to use Modulo Arithmetic?

Quote:

If $A\:=\: 1^5 + 2^5 + 3^5 + \hdots + 101^5$

find the remainder after we divide $A$ by 7.

We find that:

. . $\begin{array}{cccc}1^5 &\equiv& 1^5 & \text{(mod 7)} \\ 2^5 &\equiv& 2^5 & "\\ 3^5 &\equiv& 3^5 & "\\ 4^5 &\equiv& (\text{-}3)^5 & "\\ 5^5 &\equiv& (\text{-}2)^5 & "\\ 6^5 &\equiv& (\text{-}1)^5 & "\\ 7^5 &\equiv& 0^5 & " \\ \hline \text{Total} &\equiv & 0 & \text{(mod 7)} \end{array}$

The sum of the first seven 5th powers has a remainder of 0.

We find that the same is true for the next seven 5th powers:
. . $8^5 + 9^5 + \hdots + 14^5 \:\equiv\:0\text{ (mod 7)}$
and the next seven 5th powers . . . and the next . . . and so on.

So we have: . $A \;=\;\underbrace{\left(1^5+2^5+\hdots+7^5\right)}_{\equiv\; 0} + \hdots + \underbrace{\left(92^5 + 93^5 + \hdots + 98^5\right)}_{\equiv\; 0} + 99^5 + 100^5 + 101^5$

Hence, we have: . $\begin{array}{ccccccccc}99^5 &\equiv& 1^5 &=& 1 \\ 100^5 &\equiv& 2^5 &=& 32 \\ 101^5 &\equiv& 3^5 &=& 243 \\ \hline \text{Total:} & & & & 276 \end{array}$

And: . $276 \:\equiv\:3\text{ (mod 7)}$

. . Therefore, the remainder is 3.

a.h.m.a.d · #4 $Old$ May 12th, 2009, 05:06 PM

thanks soroban

Michel Matos · #5 $Old$ November 4th, 2009, 10:57 PM

I was looking for this prove in all of brazilian sites about number theory. Your explanation is just amazing. Thanks man.