I believe i say this question in a test some where, could anyone tell me what the hell b mod (a) is
It's called modulo. "b mod a" is the remainder of b/a.
Example: 8 mod 5 is 3, 6 mod 2 is 0, 7 mod 8 is 7.
if b is 15 and a is 4, then you can write
15 = 4*3 + 1
the thing that you have to plus, namely 1, is b mod a, so 15 mod 4 is 1. When you have to take the biggest integer and multiply with the a, such that you don't get a bigger number than b, then what you have to add is b mod a.
b = a*n+c
then b mod a = c.
where c is the smallest positive number so b = a*n+c is true
you got two integers a and b. Try two write b as a integer n times a that is thry to write
b = a*n
for some integer n. You will see that this is not possible if not a devides b. So what is the next you could try, you vould try
b = a*n + c
where c in an integer. This can always be done, but c is not unique ex.
b=20 a=3 then
20 = 3*7-1 or 20 = 3*6+2 or 20 = 3*4+8
so how could we make this c unique? If we demand that c is positive and that c is the smallest number possible then it is unique, then the only answer would be
so now c is unique, and we call that c for b mod a, pronunced b modulo a.
I think modulo is easier to understand with a practical example. Think af an old analog clock with two hands and 12 numbers, it is modulo 12.
With a clock, you can't go past 12 o' clock, whenever you do, you just start from 0 again. This is excactly what modulo is. So if you have the time 14:00, we all know that it's 2 o' clock, and mathematically it's just 14 mod(12) = 2.
Hope it helps.
Kurt, just apply the division algorithm to a and b.
a/b = q and a remainder r, well, that r is the result of "a mod b"
Example: 25/4 = 6 and the remainder 1. Then, 25 mod 4 = 1
I hope they're not teaching you that New Math I keep hearing about.
Not exactly a new problem...
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