At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Let's try this the old fashioned long division way..

|dw:1328418115277:dw|

|dw:1328418167975:dw|

|dw:1328418236084:dw|

sorry those are my random thoughts

don't have scrap paper on me

just gotta get the k in terms of n and 7.. hold on

im still thinkin btw haha

i think it's 1, just gotta prove it

okay, i think i got it:
Suppose n=7a+k and n^2=7b+k

subtracting the two equations:
n-n^2 = (7a+k) - (7b+k)
or
n-n^2 = 7a-7b = 7(a-b)

or
n(n-1)=7(a-b). Now this means that either n is divisible by 7 or n-1 is divisible by 7.

We know that n is not divisible by 7 (given), that means n-1 must be divisible by 7

Thus, if n-1 is some multiple of 7, n-1=7*x; or n = 7*x+1

which makes k = 1. Actually let's test this.

8/7 has remainder 1, so does 64/7

Try 30 and then 30^2 and test for equal remainders.

Use congruences
k=1