Show that the sum of 11 consecutive integers is always d

[Guest]

a) Show that the sum of 11 consecutive integers is always divisible by 11.

b) Show that the sum of 12 consecutive integers is never divisible by 12.

[Guest]

To answer the first question:
The sum of any 11 consecutive integers can be written as:
[tex]k + (k+1) + \cdots (k+10)[/tex]
A little algebra, and the identity
[tex]\sum_{k=1}^n k = \frac{n(n+1)}2[/tex]
yields
[tex]k + (k+1) + \cdots + (k+10) = 11k + (1 + 2 + \cdots +10) = 11(k+5)[/tex]

[TomHE]

To answer the first question:
The sum of any 11 consecutive integers can be written as:
[tex]k + (k+1) + \cdots (k+10)[/tex]
A little algebra, and the identity
[tex]\sum_{k=1}^n k = \frac{n(n+1)}2[/tex]
yields
[tex]k + (k+1) + \cdots + (k+10) = 11k + (1 + 2 + \cdots +10) = 11(k+5)[/tex]

The second question can be answered using the same technique.
Any 12 consecutive integers can be written as:
[tex]k + (k+1) + \cdots (k+11) = 12(k + 5) + 6[/tex]
which is not divisible by 12.