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/
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