| New Reply |
Problem with mod operation |
Share Thread | Thread Tools |
| Sep4-12, 01:00 AM | #1 |
|
|
Problem with mod operation
Hello
i know that it shows Remaining for example 5 mod 3=2 1 mod 3=1 but if i select negative number what does it do? example: -10 mod 27 =? Thanks |
| Sep4-12, 01:18 AM | #2 |
|
|
5 [itex]\equiv[/itex] 2 (mod 3) and 1 [itex]\equiv[/itex] 1 (mod 3) where the [itex]\equiv[/itex] in this context is pronounced "is congruent to." In other words a mod-n statement returns "true" or "false" when applied to pairs of numbers. The general rule is that a [itex]\equiv[/itex] b (mod n) if the number n divides a - b. Now a lot of people come to mod from programming languages, where mod is not a relation, but is rather an operator, meaning that it returns a single value. That's the usage you've written, so we say 5 mod 3=2 and so forth. But even though 5 [itex]\equiv[/itex] 2 (mod 3), it's also true that 5 [itex]\equiv[/itex] 47 (mod 3), right? Both 47 and 5 give the same remainder when divided by 3. [That's equivalent to the definition I gave earlier; but you should actually convince yourself of that] So if someone asks us what is 5 mod 3, what should the answer be? The convention is that we take the unique number x such that 5 [itex]\equiv[/itex] x (mod 3) and x is greater than or equal to 0, but less than 3. With that background, what is the answer to -10 mod 27 = ? Well, let's find x such that -10 [itex]\equiv[/itex] = x (mod 27), and x is between 0 and 26 inclusive. A moment's thought will convince you that x = 17 is the right answer here. So -10 mod 27 = 17 That's because a) -10 - 17 is divisible by 27; and b) 17 is the unique number with that property that's also between 0 and 26, inclusive. That's a long answer but it's everything you need to know to make sense of this kind of problem. |
| Sep4-12, 01:40 AM | #3 |
|
Recognitions:
|
remainder of dividing a by c is b*. And complement it with Stevel27's answer. * This is a technical point, since we usually choose the remainder to be within a given range, but we can add multiples. |
| Sep4-12, 06:53 AM | #4 |
|
|
Problem with mod operation
Thanks Dear SteveL27 & Bacle2
you are my best teacher that dedicate your time to telling me the right answer. Thanks again |
| Sep4-12, 04:37 PM | #5 |
|
Recognitions:
|
Thanks, but I did only a minimal part.
That's how I like it, Stevel27 does 99%+ of the work and we split the credit in half ;) . |
| Sep4-12, 04:46 PM | #6 |
|
|
|
| New Reply |
| Thread Tools | |
Similar Threads for: Problem with mod operation
|
||||
| Thread | Forum | Replies | ||
| ladder operation problem | Introductory Physics Homework | 6 | ||
| Initial value problem - value returns invalid operation | Calculus & Beyond Homework | 1 | ||
| Set operation | Set Theory, Logic, Probability, Statistics | 1 | ||
| multi-quote button doesn't work | Forum Feedback & Announcements | 2 | ||
| Set operation problem | Set Theory, Logic, Probability, Statistics | 5 | ||