modular arithmetic - How to calculate a Modulo? - Mathematics Stack Exchange
Oakley Mod 3 vs Mod 5 Hemlets
elementary number theory - How do you prove that $ n^5$ is congruent to $ n$ mod 10? - Mathematics Stack Exchange
why is n ≡ n^5 (mod 10)?
How much is 17 mod 3?
17 mod 3 equals 2 since dividing 17 by 3 gives a quotient of 5 and a remainder of 2. The remainder is the result of the modulus operation. In simpler terms, 17 mod 3 = 2.
What is modulo (mod) operation?
What is the difference between mod and remainder?
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When you see "modulo", especially if you are using a calculator, think of it as the remainder term when you do division.
Examples:
The result of 10 modulo 5 is 0 because the remainder of 10 / 5 is 0.
The result of 7 modulo 5 is 2 because the remainder of 7 / 5 is 2.
The reason your calculator says 113 modulo 120 = 113 is because 113 < 120, so it isn't doing any division.
More generally, the idea is that two numbers are congruent if they are the same modulo a given number (or modulus)
For example, as above, 
where 
is our modulus.
Another issue is that of inverses, which is where the confusion of 
comes in.
We say that 
and 
are inverses modulo 
, if 
, and we might write 
.
For example $17\cdot 113 = 1921 = 120\cdot 16 +1 \equiv 1 \mod 120
17^{-1} = 113$ modulo 
.
There are ways to calculate it, modulo is remainder counting basically.
because 
because 
and so on, so if you want to calculate for example 
you can do this, that is want to get 
, take 73 and continue subtracting 7 until you no longer can. 
, 
etc until we get 
which gives us that 
in it's simplest form (any of the results along the way can technically be a).
As for what 
the question is simply what times 17 gives remainder 1 when divided by 120?
$113\cdot 17 = 1921 = 120\cdot 16+1$
Hello,
What is the difference between the Oakley Mod 3 and Mod 5. To me, they both look great. So is it worth paying the extra £50 for the mod 5? Are these helmets both compatible with all Oakley goggles?
Side note: this would be for my second proper skiing trip to France and I can confidently ski blues and reds, is it actually worth it buying a helmet or should I just rent?
Thanks
Here's a direct proof: note that $$n^5 - n = n(n^4-1) = n(n-1)(n+1)(n^2+1)$$
You need to prove that this number is divisible by $2$ and $5$.
For divisibility by $2$, note that either $n$ or $n+1$ must be even.
For divisibility by $5$, you should prove that if neither $n$ nor $n \pm 1$ is divisible by $5$ (i.e. if $n \equiv 2\ \text{or}\ 3 \bmod 5$) then $n^2+1$ is divisible by $5$.
Hint: In order to use Fermat's Little Theorem, you need the modulus to be prime which $10$ is not. However, $10 = 2\times 5$. Use Fermat's Little Theorem for each prime and see if you can combine these results to arrive at your claim.
I was going over some modular arithmetic, and tried to answer why n is always congruent to n^5 (mod 10).
Sure, there are not that many digits anyway, so for some whole number x, it's bound to happen where n ≡ n^x (mod 10) for all integers n. But x=5 seems very low if you think about this as pure chance.
| unit digit | n2 (mod 10) | n3 (mod 10) | n4 (mod 10) | n5 (mod 10) | period |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 4 | 8 | 6 | 2 | 4 |
| 3 | 9 | 7 | 1 | 3 | 4 |
| 4 | 6 | 4 | 6 | 4 | 2 |
| 5 | 5 | 5 | 5 | 5 | 1 |
| 6 | 6 | 6 | 6 | 6 | 1 |
| 7 | 9 | 3 | 1 | 7 | 4 |
| 8 | 4 | 2 | 6 | 8 | 4 |
| 9 | 1 | 9 | 1 | 9 | 2 |
All periods are 1,2,4, so I see why it'll come back to itself at 5th power for any values of n. I mean, it works bc it works, but is there any behind-the-scene explanation to this? If I didn't make the table, and just try to answer it with my intuition and gut feeling, I would probably say the common period will be like idk 36 or something that's definitely greater than 5.