## UnkleRhaukus Group Title The sum of any five consecutive integers is divisible by 5 . Proof: one year ago one year ago

1. 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. $\square$

2. UnkleRhaukus

is this good?

3. CliffSedge

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

4. estudier

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

5. UnkleRhaukus

|dw:1351169356249:dw|

6. UnkleRhaukus

you dont think i need so many words?

7. estudier

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

8. estudier

5x +10 is already divisible by 5

9. CliffSedge

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.

10. estudier

The extra step is superfluous (really really superfluous)

11. 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:

12. jhonyy9

oppinions ???

13. estudier

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

14. UnkleRhaukus

i like your induction @jhonyy9, but it is long

15. jhonyy9

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

16. UnkleRhaukus

my method is a flow of implication of truth

17. 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"

18. jhonyy9

sorry on the second understandably i wann writing acceptably sorry

19. estudier

@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

20. UnkleRhaukus

That is real good @estudier !