CTK Insights

27 Mar

Divisibility of N(N+1)(2N+1) by 6

Jeffrey Samuelson twitted a question to which I am happy to provide an answer, or two.

Question

Find an integer N such that N(N+1)(2N+1) is NOT divisible by 6. If no such integer exists, prove why.

Since the question is about the properties of a product modulo 6, it seems proper to investigate the dependency of the product on the residues modulo 6. Let's form a table

    N     N+1     2N+1  
  0(6) 1 1
  1 2(2) 3(3)
  2(2) 3(3) 5
  3(3) 4(2) 1
  4(2) 5 3(3)
  5 6(6) 5

The first number in a cell is the remainder of division by 6 of the number in the caption of a column. The number in parentheses - if present - shows a factor of that number. Clearely, in every row, the product of factors is at least 6. So that, for every possible value of N modulo 6, the product N(N+1)(2N+1) is divisible by 6.

An effort of putting together a table could have been spared with an observation that, for every integer N,

N(N+1)(N+2) is divisible by 6

In words, a product of any three consecutive integers is divisible by 6. This is a somewhat simpler variant of the problem at hand. Why a variant? Are not problems rather different? Yes and no. Consider the difference

N(N+1)(2N+1) - N(N+1)(N+2) = N(N+1)(N-1) = (N-1)N(N+1).

Or, separating N(N+1)(2N+1),

N(N+1)(2N+1) = N(N+1)(N-1) = (N-1)N(N+1) + N(N+1)(N+2),

meaning that N(N+1)(2N+1) is the sum of two products of three consecutive integers, each of which is divisible by 6.

Finally, N(N+1)(2N+1)/6 is the sum of the first N squares: 1² + 2² + ... + N² (see, Jeff's derivation. This is called a combinatorial argument. Since N(N+1)(2N+1)/6 is thus an integer, 6 is bound to be a factor of N(N+1)(2N+1).

4 Responses to “Divisibility of N(N+1)(2N+1) by 6”

  1. 1
    Jeff Samuelson Says:

    Thanks for the insights into the problem! I do appreciate it. Your site is a fantastic mathematics resource!

  2. 2
    admin Says:

    Jeff, you questions are a real inspiration. Thinking of a problem, turning it in your head after getting a solution - looking back - is the way to learn how to do mathematics. You are on the right track ...

  3. 3
    Vladimir Nikolin Says:

    Term is divisible by 6 if divisible by 2 and 3.
    Term is divisible by 2 (N or N+1 are even).
    Term is divisible by 3 by putting N=3k, N=3k+1, N=3k-1.

  4. 4
    admin Says:

    Yes, this will also work.

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