Here's another interesting quesiton. (1 Viewer)

[roadrage75]

here's an interesting question. - if you know how to approach it , it is very easy, but if not, it is very difficult.

i'm interested to see how you guys go about it? and also to see whether the way i did it is popular or unusual.

anyway, here's the question.

how many positive integers n are there such that n+3 divides n^2 +7 without a remainder.

 

[Mark576]

three?

 

[pinknails500]

see this is why i did gneral maths pppppppppppehh its probably reallyy easy but i have no clue what your asking

 

[Mark576]

Spoiler

(n^2 + 7)/(n + 3) = n – 3 + 16/(n + 3); for the expression for the remainder to be integral, n = 1, 5 or 13.

 

[roadrage75]

very well done mark, that's indeed the correct answer

 

[Mark576]

Sweet!

 

[GaDaMIt]

You guys are entertaining.

 

[hurikai]

Yup, I did it that way too

 

[tommykins]

Fucking awesome, I have no idea what he did to obtain the answer..

I understand that (n^2 + 7)/(n+3) bit but have no idea what he did to proceed.

 

[undalay]

(n^2 + 7)/(n + 3) = n – 3 + 16/(n + 3);

For it to divide perfectly i.e. no remainder;
(n^2 + 7)/(n + 3) must be an integer.

but (n^2 + 7)/(n + 3) = n – 3 + 16/(n + 3);

thus n – 3 + 16/(n + 3); must be an integer.

thus 16/(n + 3); must be an integer

Then you find n which will make that fraction an integer.

 

[tommykins]

oh of course =\ ta.