Question

The sum of any five consecutive integers is divisible by 5 .
Proof:

Answer 1

Proof:
\[\text{If \(x\) is the smallest of the integers,}\]\[\text{the sum of 5 consecutive integers can be written as }\]\[x+(x+1)+(x+2)+(x+3)+(x+4)\]\[\qquad\qquad=5x+(1+2+3+4)\]\[\qquad\qquad=5x+10\]\[\qquad\qquad=5(x+2)\]\[\text{If this is divisible by 5, then }\]\[5|5(x+2)\]\[\text{ is a true statement.}\]\[\text{This is true when \((x+2)\) is an integer which is always }\]\[\text{ true because integers are closed under addition.}\]

This proves the sum of any five consecutive integers is divisible by 5.
\[\square\]

Answer 2

is this good?

Answer 3

Works for me, but I'm pretty lenient when it comes to rigour.

Answer 4

a + (a+1) + (a+2) + (a+3) + (a+4) = 5a + 10 QED

Answer 5

you dont think i need so many words?

Answer 6

That's right...and x+2 is neither here nor there....

Answer 7

5x +10 is already divisible by 5

Answer 8

Yeah, I do like the extra step of showing that 5x+10 = 5(x+2), then it's obvious that the number is a multiple of five, and hence is divisible by 5.

Answer 9

The extra step is superfluous (really really superfluous)

Answer 10

n1=1
n2=2
n3=3
n4=4
n5=5

n1 +n2 +n3 +n4 +n5 =n15 /5 is true

n=k suppose is true

k1 +k2 +k3 +k4 +k5 =k15 /5

than for k=k+1

(k1 +1) +(k2 +1) +(k3 +1) +(k4 +1) +(k5 +1) =
= (k1 +k2 +k3 +k4 +k5) +5=
=5+5+5+5=
=20/5

Q:E:D:

Answer 11

oppinions ???

Answer 12

(a-2) + (a-1) + a + (a+1) + (a+2) = 5a QED

Answer 13

i like your induction @jhonyy9,
but it is long

Answer 14

so yes thank you but i think that is very very more ,,mathematically"
thematically

yes???

Answer 15

my method is a flow of implication of truth

Answer 16

yes but here on openstudy have told me that a proof need being more mathematically,understandably and not for the last step more understandably for sombody inclusive who not know mathematics just a ,,little"

this was for my,, Goldbach's and Collatz"s conjecture proofs"