Polpak's Challenge Question Round 1 Let n be a positive integer. If we divide n by 3, the remainder is 2 If we divide n by 5, the remainder is 1 what is n?
or i mean what is the smaller possible value for n?
smallest*
Looks like the Chinese remainder theorem.
yay! :)
lol very good yes i used euclidean's algorithm which is mixed up with that Chinese stuff
i really like this problem
yeah I did something similar a few weeks ago to implement by hand RSA encryption
omg polpak i love the RSA
i want to be an expert at number theory
pretty cool. You can work for nsa ;)
my instructor may me look at the RSA because I said I was interested in number theory and i was kindof aftraid because i thought i was going to have to know some computer stuff, but all it really was was number theory i looked at the DSA which is fun too but not as fun as the RSA
I also looked at the secure hash algorithm 1 which is really just discrete math but still fun :)
i'm one of those people who get's the joke: There are 10 kinds of people in this world; those who know binary and those who don't
all algorithms are just descrete math stuff ;p
yes thats true
what education level are you if you don't mind me asking?
About to transfer to 4 year university.
just finished the local CC.
what?! no way
how are you super smart?
I'm old. Some smarts come with age. I also study a lot (comes with being old)
i think you are really cool :)
Well thanks. I'm 33 and started college a few years ago when my son was born.
are you going to be a math person? i mean you really already are, but do something with math?
math instructor? engineer? ...
Shooting for a masters in computer science, though I'm thinking I'll probably do math too. Maybe two masters! =)
you can do it you have the knowledge already for math
you probably know how to solve differential equations already i think you done one here before dont really remember
or some differential equations
I've done 1 class in diff eq's and 1 class in linear algebra thus far
are you american?
americans are lazy. you can't be american because you did that self study stuff
lol, yes I'm american. and yes, I'm lazy. most smart people are ;)
lol
I study cause it makes things easier
it's a lot more work to not know how to do the homework and struggle through it for hours and hours
yes i agree
i do have a another challenge question for you if you are willing to accept?
oh sure, though if it takes as long as euclid's algorithm I may have to hold off a while ;)
\[\int\limits_{}^{} e^{\sqrt{3x+9}} dx\]
evaluate the above is the question
or the problem
let me know if you want a hint k?
sorry, had a crisis..
its cool
looks like just u-sub then by parts..
\[u = \sqrt{3x + 9} \implies du = \frac{3}{u}dx \implies dx = \frac{u}{3}du\] \[\implies \int e^{\sqrt{3x+9}} dx = \frac{1}{3}\int ue^udu\] Then go with by-parts.
not nearly as bad as some of the triple integrals I had last year in calc 3.
those things took pages =(
Whoops, should be 2u/3 I think actually
wait i think you can teach me something one sec
then integration by parts blah blah
some mistakes there
the only mistake i see is du=3/[2(3x+9)] dx but i really didn't use this simplification
It should be \[\frac{2}{3}\int ue^udu\]
What you have on the right side, second line is correct. but recall that sqrt(3x + 9) is u, so you just have u in the denominator
oh darn it lol
you way is much more easier instead of multiply by 1 you just freaking squared both sides of u=sqrt(3x+9) and got u^2=3x+9 so 2 du=3dx so dx=2du/3 you are freaking brillant lol not really i just didn't think of that i mean you are smart though
I don't square nothin
I think you misunderstand what I did
me too
but that squaring does work also
my way is easier ;p
Ok, Let \(u = \sqrt{3x + 9}\) \[\implies du = (\frac{1}{2\sqrt{3x+9}} * 3 )dx\] \[ = \frac{3}{2u}dx\] \[\implies dx = \frac{2u}{3}du\]
Then go back to the integral and sub that for dx, and the exponent of e just becomes u also.
oh i see lol
your awesome
you're ;p
ok, I'm off to watch a movie
k later peace i will see if i can come up with more challenging problems next time these sucked obviously
ok =)
Mod[11, 3] = 2, Mod[11, 5] = 1
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