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Divisibility by 3 in Three Consecutive Numbers

Date: 10/07/2002 at 08:21:55
From: JoeyLynn Paquette
Subject: Divisibility of 3 in 3 consecutive numbers

I know that every combination of 3 consecutive natural numbers 
inlcudes at least one even number and one multiple of 3. But with any 
combination of consecutive natural numbers, why is one integer 
divisible by three and why is ONLY one number divisible by 3?

This question is for my algebraic structures for the elementary 
teacher course. I appreciate the help!

Thank you!  
JoeyLynn Paquette


Date: 10/07/2002 at 12:38:27
From: Doctor Peterson
Subject: Re: Divisibility of 3 in 3 consecutive numbers

Hi, JoeyLynn.

Sometimes the obvious is hard to put into words. For an intuitive 
explanation, you might just think about how you count by threes:

    1 2 3 4 5 6 7 8 9 10 ...
        ^     ^     ^

Every third number is a multiple of 3. Now if I take any group of 
three consecutive numbers, it's like putting a stick of length 3 along 
this line:

    1 2 3 4 5 6 7 8 9 10 ...
        ^     ^     ^
            =====

No matter where I put it, it will have to touch one multiple of 3, 
because it's too long to fit between two multiples, and too short to 
span the distance between them so it can cover more than one.

How can we make this more formal? Suppose the first of our three 
consecutive numbers is N, so that they are N, N+1, and N+2. Now, our 
number N must leave a remainder of either 0, 1, or 2 when divided by 
3, so it can be written as either 3k, 3k+1, or 3k+2 for some k (the 
quotient). So we have three cases:

    N = 3k, N+1 = 3k+1, N+2 = 3k+2    ==>  only N is a multiple of 3
    N = 3k+1, N+1 = 3k+2, N+2 = 3k+3  ==>  only N+2 is a multiple of 3
    N = 3k+2, N+1 = 3k+3, N+2 = 3k+4  ==>  only N+1 is a multiple of 3

So there is always exactly one multiple of 3 among them.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 10/07/2002 at 12:57:24
From: JoeyLynn Paquette
Subject: Divisibility of 3 in consecutive numbers

Is there are way to solve this problem using the qualities of prime 
numbers?

Thanks again!
JoeyLynn Paquette


Date: 10/07/2002 at 13:07:05
From: Doctor Peterson
Subject: Re: Divisibility of 3 in consecutive numbers

Hi, JoeyLynn.

I don't see that prime numbers are really relevant here; the same 
fact would be true if we replaced 3 with any number, prime or not:

    among any N consecutive natural numbers,
    exactly one will be a multiple of N.

But the fact that 3 is prime makes this fact itself more useful than 
the equivalent fact involving a composite number. For example, you 
can use it to show that the product of any three consecutive numbers 
is a multiple of 6, because our theorem applied to N=2 and N=3 (both 
prime) implies that the product is a multiple of both 2 and 3, and 
therefore of 6. If we had used, say, 2 and 4, then divisibility by 2 
and 4 would not imply divisibility by 8.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Number Theory
Middle School Division
Middle School Factoring Numbers

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