Exploring Remainders in Division by 3 for Consecutive Positive Integers

[AznBoi]

When you divide any positive integer by 3, the remainder must be less than or equal to 2.

Does this hold true for all integers from 2-9?? I've tried it with 10 but it doesn't seem to work, nor 1 because anything divided by 1 is itself; 0 has the same issue.

Please give your insight about this phenomenon.. Did is use phenomenon correctly?? xD

 

[D H]

When you divide any positive integer by 3, the remainder must be less than or equal to 2.

Does this hold true for all integers from 2-9??

Does what hold true - that the remainder is less than or equal to 2, or that the remainder is less than the divisor?

The former is not true. The latter is true for all positive integers. (It is inherent to the definition of remainder.) Division by zero is undefined, so the remainder resulting from dividing by zero is undefined.

 

[Werg22]

When you divide any positive integer by 3, the remainder must be less than or equal to 2.

Does this hold true for all integers from 2-9?? I've tried it with 10 but it doesn't seem to work, nor 1 because anything divided by 1 is itself; 0 has the same issue.

Please give your insight about this phenomenon.. Did is use phenomenon correctly?? xD

Look: any number, N, can be written as 3m + K, where 3m is closest inferior multiple of 3. For example 16 = 3*5 + 1, or 26 = 3*8 + 2. Of course, N / 3 = m + K/3. Notice that K is in fact the remainder. K cannot be 3, since the result of N/3 would be m + 1, which means that N is a multiple of 3. Thus for a number that is not a multiple of 3, the remainder is either 1 or 2. You can easily prove a general rule R < n, for any integer n by which you divide, with the same concept.

 

[AznBoi]

Ooops, my bad. I meant like: The largest remainder of an integer 2-9 is 1 less than itself.

Example: The integer 5 , 29/5=4 , which makes 4 the largest remainder for the integer 5.

Does this hold true for integers 2-9 only?

 

[AznBoi]

I'm still confused.. Actually I think it works for any postivie integer. For the integer 10, the largest remainder is 9. 19, 119, 199, etc.

Is there another equation that relates to this? N=3m+k confused me, and I think that you wrote it for the integer 3??

 

[d_leet]

I'm still confused.. Actually I think it works for any postivie integer. For the integer 10, the largest remainder is 9. 19, 119, 199, etc.

Is there another equation that relates to this? N=3m+k confused me, and I think that you wrote it for the integer 3??

This is basically the division algorithm which you can read more about here http://en.wikipedia.org/wiki/Division_algorithm

 

[cristo]

Ooops, my bad. I meant like: The largest remainder of an integer 2-9 is 1 less than itself.

Example: The integer 5 , 29/5=4 , which makes 4 the largest remainder for the integer 5.

Does this hold true for integers 2-9 only?

What does this equation mean, since 29/5 is certainly not equal to 4! (edit: i think you mean that the remainder when dividing 29 by 5 is 4).

If you read Werg22's post below, you will see the case for n=3. This holds true for any integer n.

As an example, consider n=12. Now, suppose that 12 divides m such that the remainder is 13. Let's write m=12*q+13. We immediately see that this can also be written m=12*(q+1)+1 (thus giving a remainder of 1).

So, yes, when dividing by an integer n, the largest remainder possible is n-1.

 

[AznBoi]

I see now. I've tried it out myself with the equation. Here is the SAT question btw which caused me to think about this: Which of the following could be the remainders when four consecutive positive integers are each divided by 3? Answer: 0,1,2,0 because starting from a multiple of 3 and adding 3 gives you another multiple of three. and the integers in between the multiples cannot exceed 3. Thanks everyone!