Question

Julie claims that:
"The sum of any five consecutive positive integers is always divisible by five." Peter claims that: "the sum of any six consecutive integers is always divisible by six." Are they correct? justify your answer

Answer 1

Let's test julie's claim first
sum of five consecutive positive integers being divisible by 5
the no.s will be of the form
n, n+1, n+2, n+3, n+4 \(\text{n is a positive integer}\)
Let's add them, let the sum be S
\[S=n+n+1+n+2+n+3+n+4\]
We get
\[S=5n+1+2+3+4=>S=5n+10\]
let's divide this by 5
\[\frac{s}{5}=\frac{5n+10}{5}=n+2\]
so this is true

Answer 2

let the first number be x
then the others are x+1, x+2, x+3, x+4
sum of them= x+x+1+x+2+x+3+x+4=5x+10=5(x+2) which is divisible by 5
therefore, it is correct.

For peter's statment,
use the same method to check if it is correct.

Answer 3

let take 1,2,3,4,5,6 as consecutive no and check 1+2+3+4+5 is divisible by five

Answer 4

@Lachlan1996 do you get this?

Answer 5

hold on one sec @ash2326 , could you explain what your doing please?

Answer 6

why are you adding on a second lot of integers to the first bit?

Answer 7

would it just be such that 5(n) where n is a positive integer, wouldnt this just mean that because it is 5n it divisible by 5, and thus the rule holds true?

Answer 8

Julie claims that: "The sum of any five consecutive positive integers is always divisible by five."
so I took 5 consecutive positive integers, they will be of the form n, n+1, n+2, n+3 and n+4
here n can be any positve integer
say if it's n=2
then the no.s are 2, 3, 4, 5, 6
Do you get this part?

Answer 9

ah yes, youve just got a recursive sequence running as your integers

Answer 10

Next I'd summed them and then I divide the sum by 5
the quotient, We get is also a positive integer
so Julie's claim is true

Answer 11

@Lachlan1996 do you get the idea?

Answer 12

uhh somewhat, this was just a random question in the misc section of my textbook. we have never been taught proofs, so i wasnt sure of how to approach it.

Answer 13

so basically i just let a variable, ie S = any positive integer then sum them.

Answer 14

yeah

Answer 15

okie doke, i see the idea now. Cheers for that mate.

Answer 16

Sorry, im a little slow picking up on things i havent been taught. Doing integration at the minute but came across this question and it stumped me... Anyway cheers for the help mate. Have a good night, and keep up the good work :)

Answer 17

yeah, Would you try the same for the second claim?

Answer 18

no problem, we are here to help.
You're welcome:)

Answer 19

Okie doke, so answering the second

Answer 20

actually no sorry, stuffed that up

Answer 21

excuse my idiocy

Answer 22

you get 6n + 15, which is not divisible by 6 so it is not true

Answer 23

Right.
@sai1234 's method is also good since it proves Peter's idea can't be always true.

Answer 24

okie doke, thanks very much mate.

Answer 25

@Lachlan1996 sorry I was afk
second is not correct, you are right:)