Hlares
  • Hlares
Question: "For which n does the expression 1+2+...+(n-1) congruent to 0(mod n) hold?" If I am reading this correctly, for the expression 0 is the least positive residue of (1+2+...+(n-1))(mod n). However, if that is correct, I am not sure where to go from there. Would n be 0 or 1 then? Or something else entirely?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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freckles
  • freckles
let's look at some examples let's say n is 1... then you have 0 congruent to 0 mod 1 let's say n is 2 then you have 0+1 congruent to 0 mod 2 which is false let's say n is 3 what happens?
Hlares
  • Hlares
Then if n=3, it would be 1+2=3 congruent to 0mod3, which would be true then, aye?
freckles
  • freckles
yes so let's look at the more general thing you have 0+1+2+3+...+(n-1) or 1+2+3+...+(n-1) or 1+(n-1) + 2+(n-2) + 3+(n-3) +..... now I can pair the terms of the sum up if there are an even amount of terms by the way you do know that k+(n-k) is n right? so each of those rows I wrote have sum n and what is n mod n

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freckles
  • freckles
for example n=4 gives 1+2+3 which doesn't have a odd amount of terms you have (1+3)+2 for n=5 you have 1+2+3+4 you have an even amount of terms so we have (1+4)+(2+3) for n=6 you have 1+2+3+4+5 or (1+5)+(2+4)+3 ...
freckles
  • freckles
do you know how to evaluate k+(n-k) mod n?
Hlares
  • Hlares
I do not off the top of my head. I am looking back through my text right now to see if it was mentioned and if so where. As for n(mod n), that would be 0 though, aye?
freckles
  • freckles
right and k+(n-k) is n so k+(n-k) mod n is 0
Hlares
  • Hlares
Ah okay! And if I am understanding pairing up the terms correctly: that is being done to see if the sum is divisible by n, right? For example, for n=6, that would be false because it has an extra 3, aye?
freckles
  • freckles
right and 3 mod 6 isn't 0
freckles
  • freckles
6+6+3=3 but 3 mod 6 isn't 0
freckles
  • freckles
if you have an even amount of terms in your sum you can pair them up and such away you have n+n+n+n+...+n mod n
freckles
  • freckles
I think I also have another way to look at it too after you get this way
Hlares
  • Hlares
Okay, that makes sense! We did not cover pairing them up like that at all, so that's really useful to know, thank you. And if I am noticing the pattern correctly here, the only n's for which the statement hold thus far are for odd integers as the even ones result in false statements, right?
freckles
  • freckles
right
freckles
  • freckles
another way: now \[1+2+3+ \cdots +(n-1)=\frac{n(n-1)}{2} \text{ do you remember this formula }\]
freckles
  • freckles
think back to calculus 1 days
freckles
  • freckles
if that isn't too far back
freckles
  • freckles
\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]
freckles
  • freckles
except we have to n-1
Hlares
  • Hlares
Calculus 1 was awhile ago, but that was covered under a section on mathematical induction in this class recently.
freckles
  • freckles
\[\sum_{i=1}^{n-1}i=\frac{(n-1)n}{2} \text{ or } \frac{n(n-1)}{2}\]
freckles
  • freckles
oh cool
freckles
  • freckles
so we can solve this equation \[\frac{n(n-1)}{2} \equiv 0 \mod n \]
freckles
  • freckles
now if n-1 is even then we can cancel out the two's and we are left with \[\frac{n(2a)}{2} =n a \equiv 0 a \equiv 0 (\mod n)\]
freckles
  • freckles
if n-1 is even then n is odd
freckles
  • freckles
if n-1 is odd we can't cancel out the 2 on the bottom but n would be even and n would get destroyed by that division by 2 if you know what I mean so we wouldn't get 0 when modding it by n
freckles
  • freckles
you can write the fraction like this: \[n \frac{n-1}{2} \equiv 0 \mod n \text{ As long as } \frac{n-1}{2} \in \mathbb{Z} \]
Hlares
  • Hlares
Aye, I do. That makes a lot of sense. Thank you for explaining it so well; your explanation was a lot more understandable than the section of the text relating to this problem.
freckles
  • freckles
np
freckles
  • freckles
you know when (n-1)/2 is an integer right? I kind of already said but just want to be sure you know
Hlares
  • Hlares
Aye, I do! Can only deal with integers with congruences, right?
freckles
  • freckles
well I think you can actually use other numbers but in number theory yeah I think you only use integers that is all I ever seen in number theory n is a positive integer and... for (n-1)/2 to be an integer n-1 would have to be even because that is the only sorta of number to be divisible by 2 and if n-1 is even then n is odd since n and n-1 are consecutive integers
Hlares
  • Hlares
Ah okay. This is all relating to brief stint into elementary number theory for this class, so all we have dealt with with problems is either all integers or positive integers/natural numbers, which is why I presumed that. And okay, that makes sense!
freckles
  • freckles
anyways glad I can help
Hlares
  • Hlares
Thank you very much for the assistance, I greatly appreciate it!

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