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

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
Proof: \[\text{If \(x\) is the smallest of the integers,}\]\[\text{the sum of 5 consecutive integers can be written as }\]\[x+(x+1)+(x+2)+(x+3)+(x+4)\]\[\qquad\qquad=5x+(1+2+3+4)\]\[\qquad\qquad=5x+10\]\[\qquad\qquad=5(x+2)\]\[\text{If this is divisible by 5, then }\]\[55(x+2)\]\[\text{ is a true statement.}\]\[\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. \[\square\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
is this good?
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Works for me, but I'm pretty lenient when it comes to rigour.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
a + (a+1) + (a+2) + (a+3) + (a+4) = 5a + 10 QED
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1351169356249:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
you dont think i need so many words?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
That's right...and x+2 is neither here nor there....
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
5x +10 is already divisible by 5
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
The extra step is superfluous (really really superfluous)
 one year ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
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:
 one year ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
oppinions ???
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
(a2) + (a1) + a + (a+1) + (a+2) = 5a QED
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i like your induction @jhonyy9, but it is long
 one year ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so yes thank you but i think that is very very more ,,mathematically" thematically yes???
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
my method is a flow of implication of truth
 one year ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
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"
 one year ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
sorry on the second understandably i wann writing acceptably sorry
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
@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
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
That is real good @estudier !
 one year ago
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