Quick questions about modular arithmetic

[johnnyICON]

99999^{99} + 1
As 99999 \equiv24(mod \25)
Can I say then:
99999^{99} + 1 \equiv24^{99} + 1(mod \25), or is it
99999^{99} + 1 \equiv24^{99}(mod \25) + 1,
or are these two the same things?

 

[shmoe]

99999^{99} + 1 \equiv24^{99} + 1(mod \25)

This one is standard. People would probably understand the second, but the usual format is to write the mod at the end.

 

[johnnyICON]

oh ok, i was just wondering if the two were the same. I was uncertain if by moving the +1 before the mod5 it would change the meaning. Awesome, well then that makes this question a lot easier now. Thanks :D

 

[johnnyICON]

Another quick question.

Are the following two congruencies the same?
1. 24^{99} + 1 \equiv 0 \mod \25
2. 24^{99} + 1 \mod \25 \equiv 0 \mod \25

I am trying to show that a number is divisible by 25, and I found that 25 can be written in terms expressed in equation #1. And I found that number that I am trying to divide by 25 can be expressed in terms expressed in equation #2.

I thought I finished the proof but now that I am looking at it, I am unsure about this one thing.

 

[Zurtex]

When you write 25 don't write it \25 or the 5 just shows up, write it normally, e.g:

24^{99} + 1 \equiv 0 \mod 25

and:

24^{99} + 1 \mod 25 \equiv 0 \mod 25

There two statements are the same, I think you miss the point though. Something like:

24^{99} + 1 \mod 37 \equiv 0 \mod 25

Woule make not really make that much sense, so amoung other reasons there is no reasons to write the mod twice.

 

[johnnyICON]

Okay. Because I was trying to show that equation #2 and #1 are equivalent.

So they are right? LOL sorry, I'm just very uncertain about myself.

 

[funkstar]

Okay. Because I was trying to show that equation #2 and #1 are equivalent.

Well, people tend to view congruence defined as a ternary relation:

x \equiv y \mod n \overset{def}{\Longleftrightarrow} n|x-y.

Sometimes one omits the modulo part, but it is still understood that we're dealing with modulo arithmetic by using the equivalence sign \equiv, instead of an equality sign. Another way of stating a congruence x \equiv y \mod n is by saying that x and y belong to the same residue class (look this up on mathworld.wolfram.com). That is

x \equiv y \mod n \Leftrightarrow [x]_n = [y]_n

So, your equation 2. states (with the missing 2 from 25) that

[[25^{99}+1]_{25}]_{25} = [0]_{25}

 

[ramsey2879]

Remember Matt Grime's post in your other thread. You are trying to show that
24^99 +1 = 0 mod 25.
but 24^99 +1 = (-1)^25 +1 = -1+1 = 0 mod 25, since 24=-1 mod 25 and since -1 raised to an odd power is -1. Q.E.D.

 

[johnnyICON]

Yea, that's exactly how I did it Ramsey :D Sorry, I should of concluded this thread by mentioning that. Thanks for another helpful response though. I appreciate it.