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UnkleRhaukus
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The sum of any five consecutive integers is divisible by 5 .
Proof:
 2 years ago
 2 years ago
UnkleRhaukus Group Title
The sum of any five consecutive integers is divisible by 5 . Proof:
 2 years ago
 2 years 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. \[\square\]
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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