You already use modulo computation when you look at the clock and e.g. needs to figure out what time it's 3 hours after 11 o'clock, which is 2 o'clock. In math we write that as:
( (11 + 3) : mod : 12 = 2 )
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where 12 is the modulus because we want the time as an integer between 0 and 11 (12 o'clock is in this case denoted by 0). In words we say 11 plus 3 modulo 12 is equal 2. The result of a modulo computation is an integer between 0 and the modulus minus 1. E.g. with the modulus 3 we have that:
- ( 1 : mod : 3 = 1 )
- ( 2 : mod : 3 = 2 )
- ( 3 : mod : 3 = 0 )
- ( 4 : mod : 3 = 1 )
- ( 5 : mod : 3 = 2 )
- ( 6 : mod : 3 = 0 )
- etc.
If we e.g. look at ( 27 : mod : 5 ) then modulo computes the number of times 5 divides 27 and then returns the remainder of the result which is 2 in this case, i.e. ( 27 : mod : 5 = 2 ). But how did we get this result?
First we compute the number of times it's possible to multiply 5 with the number ( x ) such that we get an integer as close as possible to 27 without exceeding it, i.e. we have to find the maximun value of ( x ) such that ( 5 cdot x leq 27 ). In this case we have that ( x = 5 ) because ( 5 cdot 5 = 25 leq 27 ). Then by subtracting 27 with 25 we get the answer ( 27 - 25 = 2).
If the integer is negative e.g. ( -27 : mod : 5 ) we have to do it slightly different and the answer is ( -27 : mod : 5 = 3 ). In this case the integer ( x ) is negative and should be the closest integer that exceed -27, i.e. we have to find the minimum value of ( -x ) such that ( 5 cdot -x geq -27 ). Now we have that ( -x = -6 ) because ( 5 cdot -6 = -30 geq -27 ). Then by subtracting -27 with -30 we get the answer ( -27 - (-30) = -27 + 30 = 3).
It's important that ( x ) or ( -x ) is an integer such as ( -14, 3, 17 ) etc. and NOT a fraction or float such as ( frac{1}{4}, frac{-3}{7}, 2.5, 5.1 ) etc.
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If two integers ( a ) and ( b ) modulo the same modulus ( c ) returns the same remainder ( r ), then we say that ( a ) and ( b ) are congruent modulo ( c ). I.e. if ( a : mod : c = r ) and ( b : mod : c = r ) then ( a equiv b : (mod : c) ). Also, notice that if the modulus ( c ) is greater than the integer ( a ), i.e. ( c > a ), the result will always be equal ( a : mod : c = a ).