To add to Jason's i9000 answer:You can speed the process up (which might end up being useful for very large exponents) making use of the binary development of the exponent. I suspect that it'h actually much quicker (in common) to avoid the éxponentiation by squaring, ánd rather search straight for the minimum exponent $t$ like that $5^t 5 (mod 221)$. This will of program rely on size of exponent vérsus modulus, but once you possess that exponent, you just need a one calculation (exponent mod k) and look for.

1. How To Figure Out Mode In Math
2. How To Find The Modulus Of Elasticity
3. How To Find The Mod Folder Macbook

1) ((a mod N) + (-b mod N)) mod N 2) ((a mod N) - (b mod N)) mod N I will call the just-cited algorithms ' methods '. While in 'simpler' modulus calculations (like 11 mod 3 ) N is already lower than a, in examples reported in the defined answering, it also happens that we find a negative first argument and a positive second argument. That’s simple, 1. Divide the two numbers ( eg. 7/3 = 2.333333) 2. Eliminate the decimal part (i.e., make the 2.33333 → 2) ( If there is no decimal part, the MOD value is 0, eg. 6/2 = 3, since there’s nothing after the decimal 6MOD2 = 0) 3. For example -121 / 26 = -4 17/26, thus, mod is -17 which is +9 in mod 26. Alternatively you can add the modulo base to the computation for negative numbers: -121 / 26 + 26 = 21 9/26 (mod is 9). EDIT2: As @simpatico pointed out, this method will not work for numbers that are out of calculator's precision. I really can't get my head around this 'modulo' thing. Can someone show me a general step-by-step procedure on how I would be able to find out the 5 modulo 10, or 10 modulo 5.

Take note it't also therefore definitely better if you need to repeat similar calculations. (You can't in general look for $a^k 1 (mod 221)$ since this only occurs if $a$ and 221 are relatively prime)-Feb 1 '10 at 16:06. 5^55 mod221= ( 5^10. 5^10. 5^10. 5^10.

5^10. 5^5) mod221= ( ( 5^10) mod221. 5^10. 5^10. 5^10.

5^10. 5^5) mod221= ( 77. 5^10. 5^10. 5^10.

5^10. 5^5) mod221= ( ( 77. 5^10) mod221.

5^10. 5^10. 5^10. 5^5) mod221= ( 183.

5^10. 5^10. 5^10. 5^5) mod221= ( ( 183. 5^10) mod221.

5^10. 5^10. 5^5) mod221= ( 168. 5^10. 5^10.

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5^5) mod221= ( ( 168. 5^10) mod 221. 5^10. 5^5) mod221= ( 118. 5^10. 5^5) mod221= ( ( 118.

5^10) mod 221. 5^5) mod221= ( 25.

5^5) mod221= 112. Comes along to mind as an initial point as 221 = 13. 17. So, break this down into 2 components that obtain mixed in the finish, one for mód 13 and one for mod 17. Second, I think there can be some evidence of a^(g-1) = 1 mod p for all nón zero á which also helps decrease your issue as 5^55 becomes 5^3 for the mod 13 case as 13.4=52. If you look under the issue of 'Finite Fields' you may find some good results on how to resolve this.EDIT: The reason I point out the elements is usually that this creates a method to element zero into non-zero components as if you tried something like 13^2.

How To Figure Out Mode In Math

17^4 mod 221, the solution is zero since 13.17=221. A great deal of large amounts aren't going to be leading, though there are usually ways to find Iarge primes as théy are usually utilized a great deal in cryptography and some other places within Math.

How To Find The Modulus Of Elasticity

It'beds calculated exactly like the mód of a optimistic amount. In arithmetic modulo $c$, we look for to convey any $back button$ as $qc+r$, where $ur$ must become a non-négative integer.Why put on't we test it out with an example?Consider $-100$ mod $8 = 4$. This can be because $8 cdot -13 = -104$. The remainder will be $4$.So now allow's get $(37-54)$ mod $5$. It's i9000 equal to $-17$ mod $5 = 3$. Substitute in and perform the computation: Technique $1$ gives $3$, which can be what we wish, and method $2$ gives $-2$, so the right approach is usually method $1$.

Additional answers have dealt with the instant question, therefore I'd like to tackle a philosophical one.I believe that the way you're thinking about of 'mod' is definitely a bit deceptive. You appear to end up being considering of 'mod' as an agent: therefore that '13 mod 8' is certainly another way to create the quantity '5'.

This is usually the method that modulo workers often work in development dialects: in Python you can create '13% 8' and get back again the quantity 5.Mathematically, though, I think it will be much better to think of 'mod 8' as an adverb enhancing '=': when we state '5 = 13 (mod 8)' we are really saying '5 is definitely similar to 13, if you believe of equality as operating modulo 8'. When you believe of 'mod' this way, it doesn't really make sense to request about the appearance '((a mod In) + (-c mod In)) mod In': it's not also actually an expression, under this interpretation.I'meters not attempting to say that you are incorrect for thinking of 'mod' as an procedure, because the procedure of 'consuming a residue mod $michael$' will be a helpful operation. However, I think it is certainly also useful to maintain the additional meaning of 'mod' in brain.(After writing this reply I observe that the issue was published even more than a season ago. Nicely, maybe someone else will find this useful.).

A modulo of a quantity is should become in the range of 0 and (divisor-1).Illustrations:.The worth of 7 mod 3, we discover that the quotient is 2 and the rest can be 1. The modulus is certainly properly within the range of 0 and 2 (divisor -1). Outcome is 1.The value of -7 mod 3, we observe that the quotient is usually -2, and the remainder is certainly -1. But (-1) will be not really in the range 0 and 2 (divisor -1). Therefore we make a minor adjustment, we add the rest to the divisór (-1 + 3 ).

We get the modulus ás 2, which is in the variety specified. Outcome is certainly 2. The worth of 7 mod -3, we notice that the quotient is usually -2 and the remainder is certainly 1, which is certainly not in the range of -2 and 0.

How To Find The Mod Folder Macbook

Therefore we create a slight adjustment, we include the remainder to the divisór (1 + -3). We get the modulus ás -2, which is usually in the variety specified.

Outcome is usually -2. The worth of -7 mod -3, we notice that the quotient can be 2 and the remainder can be -1, which is in the variety of -2 and 0. Result is -1. I needed to try out and change a mistake I believe Newb can be making. World of warships sound mods.

Possibly I simply misinterpreted, but basing ón what he himseIf wrote (' Take −100 mod 8=4. This is certainly because 8⋅−13=−104. The remainder is 4.' And ' qc+ur=x'), I get that we should constantly ignore down to the optimum element of N ( N can end up being found in the subsequent formulations) lower than á. 1) ((a mod N) + (-t mod In)) mod N2) ((a mod N) - (c mod In)) mod NI will call the just-cited algorithms ' strategies'.While in 'simpler' modulus calculations (like 11 mod 3) N is already lower than a, in good examples reported in the defined answering, it furthermore happens that we find a unfavorable first disagreement and a positive second case.For the instance of (37-54)mod5 it'beds been mentioned that 3 should result with the 1st method (which will be the appropriate result as nicely) and -2 should end result from the program of the second technique. Modulo operations with unfavorable values like as these show up to me really unclear, but I tried to follow what Newb stated, locating myself that it seemed proper.

But if I use the principle I mentioned previously, 3 should actually appear from both methods. 1) ((a mod In) + (-c mod In)) mod N((37 mod 5) + (-54 mod 5)) mod 5 = ((2) + (1)) mod 5 = 3% 5 = 32) ((a mod N) - (b mod D)) mod N((37 mod 5) - (54 mod 5)) mod 5 = ((2) - (4)) mod 5 = -2 mod 5 = 3Checking with a calculator before publishing I find my computations specific, but I'michael still open up to all points of view.Sadly I do not possess enough reputation to remark under Newb'h own answer, I imagine it might have got been much better. Also, British is not my main language, so, please, reveal any technical (or non) mistakes.