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- UnkleRhaukus

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

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- UnkleRhaukus

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

- katieb

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- UnkleRhaukus

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.
\[\square\]

- UnkleRhaukus

is this good?

- anonymous

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

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- anonymous

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

- UnkleRhaukus

|dw:1351169356249:dw|

- UnkleRhaukus

you dont think i need so many words?

- anonymous

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

- anonymous

5x +10 is already divisible by 5

- anonymous

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.

- anonymous

The extra step is superfluous (really really superfluous)

- jhonyy9

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:

- jhonyy9

oppinions ???

- anonymous

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

- UnkleRhaukus

i like your induction @jhonyy9,
but it is long

- jhonyy9

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

- UnkleRhaukus

my method is a flow of implication of truth

- jhonyy9

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"

- jhonyy9

sorry on the second understandably i wann writing acceptably
sorry

- anonymous

@UnkleRhaukus This is sufficient:-
Let x, x+-1, x+-2 represent 5 consecutive integers
Their sum is 5x for all x in Z (if x = 0 then x/5 is zero)
Hence the sum of 5 consecutive integers is always divisible by 5

- UnkleRhaukus

That is real good @estudier !

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