# exponetial remainder problem

• Sep 27th 2006, 05:48 PM
ceasar_19134
exponetial remainder problem
What is the remainder when 7^348+25^605 is divided by 8?

The regular calculater errors-out
• Sep 27th 2006, 06:11 PM
topsquark
Quote:

Originally Posted by ceasar_19134
What is the remainder when 7^348+25^605 is divided by 8?

The regular calculater errors-out

7 = -1 (mod 8) so 7^348 = (-1)^348 (mod 8) = 1 (mod 8)

25 = 1 (mod 8) so 25^605 = (1)^605 (mod 8) = 1 (mod 8)

Thus
7^348+25^605 = 1 + 1 (mod 8) = 2 (mod 8)

Thus you will have a remainder of 2.

-Dan
• Sep 27th 2006, 06:13 PM
dan
Quote:

Originally Posted by ceasar_19134
What is the remainder when 7^348+25^605 is divided by 8?

The regular calculater errors-out

do you mean [(7^348) + (25^605)]/8 ?

dan
• Sep 27th 2006, 06:20 PM
dan
using win XP system calc i got ((7^348) + (25^605)) mod 8 = 2

but i cant prove that to your teacher ^_^

dan
• Sep 27th 2006, 06:27 PM
ceasar_19134
dan
yeah
how do you solve it
• Sep 27th 2006, 06:36 PM
dan
Quote:

Originally Posted by ceasar_19134
dan
yeah
how do you solve it

use topsquarks method... or a win XP calc.....^_^

see yea,

dan
• Sep 28th 2006, 04:25 PM
ceasar_19134
Quote:

Originally Posted by topsquark
7 = -1 (mod 8) so 7^348 = (-1)^348 (mod 8) = 1 (mod 8)

25 = 1 (mod 8) so 25^605 = (1)^605 (mod 8) = 1 (mod 8)

Thus
7^348+25^605 = 1 + 1 (mod 8) = 2 (mod 8)

Thus you will have a remainder of 2.

-Dan

Im still not sure on how 7=-1 and 25=1, could someone explain?
• Sep 28th 2006, 04:40 PM
dan
Quote:

Originally Posted by ceasar_19134
Im still not sure on how 7=-1 and 25=1, could someone explain?

7 does not = -1 what topsquark said was
Quote:

7=-1(mod8)
same with the other
• Sep 28th 2006, 05:32 PM
topsquark
Quote:

Originally Posted by ceasar_19134
Im still not sure on how 7=-1 and 25=1, could someone explain?

My apologies. I had figured if you had been given this problem that you would have been taught this. Honestly I really don't know any other (good) way to do your problem.

Let me start with 25 = 1 (mod 8) first, it's easier.

This is shorthand for saying that the remainder when we divide 25 by 8 is 1. Another example would be 5 = 2 (mod 3). When you divide 5 by 3 you get a remainder of 2.

Now, when you divide 7 by 8 you get a remainder of, well, 7. So 7 = 7 (mod 8). Where do I get the -1 from? Well, we're playing around with something called an "equivalence relation." In actuality I shouldn't be using the "=" symbol, it should have 3 bars across, not 2. But since LaTeX is still down I'm using it for shorthand.

Anyway, this -1 business. We can show that a remainder of 7 (mod 8) is equivalent to a remainder -1 (mod 8). Really it isn't that hard. Let's look at 25 = 1 (mod 8) again. When we do 25/8 we pick the largest integer "k" such that 8k < 25, then we get the remainder by r = 25 - 8k. In this case k = 3. BUT what if we took k = 4 and found r = 25 - 8k? Now r = -7. So 25 = 1 (mod 8) = -7 (mod 8).

This is all a way of leading you into the more general definition of this "modular" number system. When we say 25 = r (mod 8) what we are saying is that r = 25 - 8k, where k is some integer. This defines a set that r can be equal to. Any member of this set is said to be "equivalent a modulo 8."

Typically we choose positive numbers for r, but to make things easier we occasionally also pick negative numbers, as I did to solve the problem in this thread.

This really isn't the best introduction to the subject. (I'm not a bad writer, but neither am I a good one! :) ) But it isn't really that hard, it just takes a little getting used to. You might find this link to be helpful. The first section is probably all that you need, though if you know the algebra the rest of it is decently informative.

Hope it helps!

-Dan
• Nov 16th 2006, 08:36 AM
TriKri
Is it 25 = 1 (mod 8) and not $25 \equiv 1\ (mod\ 8)$? Or what's the difference?
• Nov 16th 2006, 08:57 AM
CaptainBlack
Quote:

Originally Posted by TriKri
Is it 25 = 1 (mod 8) and not $25 \equiv 1\ (mod\ 8)$? Or what's the difference?

There is no difference.

RonL
• Nov 16th 2006, 09:28 AM
TriKri
I'm a noob... -_-;;

But is there some other difference? (I don't want to start a new thread for this though)
• Nov 16th 2006, 12:32 PM
CaptainBlack
Quote:

Originally Posted by TriKri
I'm a noob... -_-;;

But is there some other difference? (I don't want to start a new thread for this though)

Only that you can type "=" in plain ASCII, but a special symbol for
conguency is redundant as its the (mod 8) qualified identifies what the
symbol is doing. Also the $\equiv$ symbol has other uses as
well.

RonL
• Nov 16th 2006, 07:34 PM
Soroban
Hello, ceasar_19134!

Exactly what course does this question come from?
If you aren't familiar with modular arithmetic, it's not a fair question.

Quote:

What is the remainder when $7^{348}+25^{605}$ is divided by 8?

For reference, let $N \,=\,7^{348} + 24^{605}$

Any multiple of 1000 is divisible by 8.
Hence, we are concerned with only the last three digits of $N.$

We will indicate " $7^6$ ends in $649$" with: $7^6 \rightarrow 649$.

[1] We find that $25^n \rightarrow 625$ for $n \geq 2.$

[2] We find that:
. . . . . . $\begin{array}{ccccc}7^4 \rightarrow 401 \\ 7^8 \rightarrow 801 \\ 7^{12} \rightarrow 201 \\ 7^{16} \rightarrow 601 \\ 7^{20} \rightarrow 001\end{array}$

Hence: . $7^{348} \:=\:\left(7^{340}\right)\left(7^8\right) \;=\;\left(7^{20}\right)^{17}\left(7^8\right) \rightarrow (001)^{17}(801) \rightarrow 801$

Then: . $N \;\rightarrow \;625 + 801 \rightarrow 426$

And: . $426 \div 8 \:=\:53,\;\boxed{\text{remainder }2}$