SOLUTION: 9. a) How many different numbers, including zero, can there be as remainders if 19 is divided into 2? Am I supposed to know how to find this, or do I need to divide it out? If t

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Question 198684This question is from textbook Elementary Algebra
: 9. a) How many different numbers, including zero, can there be as remainders if 19 is divided into 2?
Am I supposed to know how to find this, or do I need to divide it out? If there is a way to find it without dividing it out, can you show me how?
Thanks.
This question is from textbook Elementary Algebra

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Well let's divide some numbers into 2 to see what possible remainders we get

0: 0/2 = 0 remainder 0
1: 1/2 = 0 remainder 1
2: 2/2 = 1 remainder 0
3: 3/2 = 1 remainder 1
4: 4/2 = 2 remainder 0
5: 5/2 = 2 remainder 1
6: 6/2 = 3 remainder 0

etc...

Starting to see the pattern? It turns out that if the dividend (ie the numerator) is even, then the remainder is 0. Otherwise, if the dividend is odd, then the remainder is 1. So the only possible choices of remainders are 0 and 1. These are the ONLY two possible choices.

So because 19 is odd, this means that the remainder of dividing 19 into 2 gets you a remainder of 1.


So if you did the above work in a previous exercise, then you wouldn't have to do any dividing, but it helps see what's going on.

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