Alternative Ways To Find The Remainder of A/B?

[iScience]

I guess this would be a Number Theory question. Short of actually going through the division process, is there another way to find the decimal remainder of an arbitrary set of integers { A , B }

$$\frac{A}{B} , A > B$$

 

[mathwonk]

I would say no. it may seem as if there is, for example when finding the remainder of division by 3 you just add up the digits. but that ignores the fact that in doing that you are really doing the division just not recording the answer. i.e. the remainder of 457 on division by 3 is 4+5+7 = 16, hence going on it is 1+6 = 7, or 7-6 = 1. but you are really "casting out 3's" which is division by 3. e.g. the first step of replacing 457 by 16, has cast out 4x33 of them from the 400, plus 5x3 of them from the 50, or 147 of the 3's, leaving 16 or five more 3's, so the answer to the division has actually been computed as 152 with remainder 1, but this answer has been ignored.

similarly to compute any remainder one could "cast out B's" but you are really dividing by B. e.g. the remainder of 457 on division by 19, could be successively computed by casting out 190 + 190 or 380, i.e. twenty 19's, leaving 77, then casting out three more or 57, leaving 20, then one more 19, leaving remainder 1. but if you look back at that process you will see the division is there, and the answer is 10+10+3+1 = 24 with remainder 1.

 

[iScience]

I would say no. it may seem as if there is, for example when finding the remainder of division by 3 you just add up the digits. but that ignores the fact that in doing that you are really doing the division just not recording the answer. i.e. the remainder of 457 on division by 3 is 4+5+7 = 16, hence going on it is 1+6 = 7, or 7-6 = 1. but you are really "casting out 3's" which is division by 3. e.g. the first step of replacing 457 by 16, has cast out 4x33 of them from the 400, plus 5x3 of them from the 50, or 147 of the 3's, leaving 16 or five more 3's, so the answer to the division has actually been computed as 152 with remainder 1, but this answer has been ignored.

similarly to compute any remainder one could "cast out B's" but you are really dividing by B. e.g. the remainder of 457 on division by 19, could be successively computed by casting out 190 + 190 or 380, i.e. twenty 19's, leaving 77, then casting out three more or 57, leaving 20, then one more 19, leaving remainder 1. but if you look back at that process you will see the division is there, and the answer is 10+10+3+1 = 24 with remainder 1.

Where were you getting the double tens, and the single three from? Were the double tens systematically determined or just randomly? what if the dividend was 371, what would the double tens then be?

 

[mathwonk]

190 is ten 19's so 380 = 190 +190 is 10+10 nineteens.

if i divide 457 by 371 i just cast out the one copy of 371 by subtracting, and get 29+57 = 86 as remainder, and dividend 1. you just count how many copies of the divisor you have cast out, i.e. subtracted out.

 

[iScience]

190 is ten 19's so 380 = 190 +190 is 10+10 nineteens.

if i divide 457 by 371 i just cast out the one copy of 371 by subtracting, and get 29+57 = 86 as remainder, and dividend 1. you just count how many copies of the divisor you have cast out, i.e. subtracted out.

Are you just multiplying it by ten because it's quick and easy to do?

I meant 371 / 19. But I think I get the flow. But just for my understanding what would you do if it was 155 / 19?

 

[mathwonk]

yes. the point is to =cast out, i.e. subtract, copies of the divisor, so to make it somehow easier than actual full division, you just do ones that are easy.

like the trick for dividing by 3, is to know in advance that 100 = 3x33 + 1, and 1000 = 3x333 + 1, etc... so the easy ones here are 9,99,999,9999,...155/19 maybe i would throw out 190, which is 35 too much, so i get -35, then add in 38 getting 3. does that work? let's see 10-2 = 8 and, 8x19 = 152, yep, that's off by 3.