Euler's theorem

Re: Euler's theorem

Wed, 01 Jul 2009 22:20:12 +0000

ah. i see now

Re: Euler's theorem

Wed, 01 Jul 2009 21:49:05 +0000

What I mean is that 7⁴ obviously isn't equal to 1 — it's only equivalent to 1 mod 10. So you can substitute 1 for it, but only when you're working in mod 10.

Re: Euler's theorem

Wed, 01 Jul 2009 04:08:04 +0000

I have never seen modular arithmetic before, but it seems to imply the elementary idea of remainders in division. For example, you can say that (although it is misleading and simply not mathematically proper) 17/4 is 4 with a remainder of 1. Therefore, 17 mod 4 is 1. I do not have any idea if this is entirely correct. Upon research, it seems there is more to this than what I hypothesize.

In mod 10, this is easy. The equivalence is simply the last digit of the integer. 134 = 4 mod 10. Therefore, 7⁴ = 2401, which in mod 10 is 1. This is how this equivalence is made above.

Re: Euler's theorem

Wed, 01 Jul 2009 03:01:55 +0000

what do you mean "occurs in an equivalence"?

and how would your explanation work with other numbers? im horribly inept at substitution for some reason

Re: Euler's theorem

Wed, 01 Jul 2009 02:58:48 +0000

yeah. i always failed at spanish. thank goodness im finished with it though

Re: Euler's theorem

Wed, 01 Jul 2009 01:56:59 +0000

Right, and (X^a)^b = X^(a*b).

I just couldn't understand how 7^4 = 1.

Re: Euler's theorem

Wed, 01 Jul 2009 01:19:09 +0000

@graphmastur (7⁴)⁵⁵ = 7²²⁰ since (x^a)^b = x^ab.

If you multiplied to numbers of the same base, then you add the exponents. x^a*x^b = x^a+b

Re: Euler's theorem

Wed, 01 Jul 2009 01:13:19 +0000

yeah. i know that much, i think. the clock would be 7^4 mod 10, but no further

Re: Euler's theorem

Tue, 30 Jun 2009 23:37:59 +0000

Modular arithmetic would be very useful knowledge, so you should read up on it. …Imagine a clock face with ten hours. Starting at 0:00/10:00, we then move forward 7^4 hours. What hour would that be?

Also, you might remember that the last digit of a number can be acquired with 10fPart(.1X). This immediately follows from 10fPart(X/10), which can act as a replacement for X-10int(X/10) – essentially, we're using X (mod 10).

Re: Euler's theorem

Tue, 30 Jun 2009 22:36:40 +0000

Actually, when you ask someone "what" in spanish, as in what did you say, or what was that, you actually do Cómo.

As for the problem, I have no earthly idea. Power wise, the 7 to the 4th to the 55th, is 7 to the 59th. However, he said that 7 to the 4th is equal to 1, so I don't know.

Re: Euler's theorem

Tue, 30 Jun 2009 22:18:59 +0000

so because 10 has 4 values comprime to it we can substitute 1 for it:?

¿¿¿que???

Re: Euler's theorem

Tue, 30 Jun 2009 15:28:07 +0000

is actually the meat of Euler's theorem, since 4 is φ(10). This means that whenever 7⁴ occurs in an equivalence mod 10, we can substitute 1 for it: if , then (raising both sides to a power), and (multiplying both sides by 7²).

Another way to see it is this: taking things mod 10 is equivalent to finding the last digit of a number. When multiplying two large numbers (e.g. 7⁴*7² = 2401*49), if all we care about is the last digit of the result, we don't need to know what the other digits of the numbers themselves are, so we might as well just multiply 1*9. This is exactly what we're doing here: the last digit of 7⁴ is 1, so we replace 7⁴ by 1 when doing a last-digit calculation.

Euler's theorem

Tue, 30 Jun 2009 14:43:38 +0000

(1)

can someone tell me why this part is true?

(2)

how does the 7^4 come out?