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Mathematics 78 Online
myininaya (myininaya):

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?

myininaya (myininaya):

or i mean what is the smaller possible value for n?

myininaya (myininaya):

smallest*

OpenStudy (anonymous):

Looks like the Chinese remainder theorem.

myininaya (myininaya):

yay! :)

myininaya (myininaya):

lol very good yes i used euclidean's algorithm which is mixed up with that Chinese stuff

myininaya (myininaya):

i really like this problem

OpenStudy (anonymous):

yeah I did something similar a few weeks ago to implement by hand RSA encryption

myininaya (myininaya):

omg polpak i love the RSA

myininaya (myininaya):

i want to be an expert at number theory

OpenStudy (anonymous):

pretty cool. You can work for nsa ;)

myininaya (myininaya):

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

myininaya (myininaya):

I also looked at the secure hash algorithm 1 which is really just discrete math but still fun :)

myininaya (myininaya):

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

OpenStudy (anonymous):

all algorithms are just descrete math stuff ;p

myininaya (myininaya):

yes thats true

myininaya (myininaya):

what education level are you if you don't mind me asking?

OpenStudy (anonymous):

About to transfer to 4 year university.

OpenStudy (anonymous):

just finished the local CC.

myininaya (myininaya):

what?! no way

myininaya (myininaya):

how are you super smart?

OpenStudy (anonymous):

I'm old. Some smarts come with age. I also study a lot (comes with being old)

myininaya (myininaya):

i think you are really cool :)

OpenStudy (anonymous):

Well thanks. I'm 33 and started college a few years ago when my son was born.

myininaya (myininaya):

are you going to be a math person? i mean you really already are, but do something with math?

myininaya (myininaya):

math instructor? engineer? ...

OpenStudy (anonymous):

Shooting for a masters in computer science, though I'm thinking I'll probably do math too. Maybe two masters! =)

myininaya (myininaya):

you can do it you have the knowledge already for math

myininaya (myininaya):

you probably know how to solve differential equations already i think you done one here before dont really remember

myininaya (myininaya):

or some differential equations

OpenStudy (anonymous):

I've done 1 class in diff eq's and 1 class in linear algebra thus far

myininaya (myininaya):

are you american?

myininaya (myininaya):

americans are lazy. you can't be american because you did that self study stuff

OpenStudy (anonymous):

lol, yes I'm american. and yes, I'm lazy. most smart people are ;)

myininaya (myininaya):

lol

OpenStudy (anonymous):

I study cause it makes things easier

OpenStudy (anonymous):

it's a lot more work to not know how to do the homework and struggle through it for hours and hours

myininaya (myininaya):

yes i agree

myininaya (myininaya):

i do have a another challenge question for you if you are willing to accept?

OpenStudy (anonymous):

oh sure, though if it takes as long as euclid's algorithm I may have to hold off a while ;)

myininaya (myininaya):

\[\int\limits_{}^{} e^{\sqrt{3x+9}} dx\]

myininaya (myininaya):

evaluate the above is the question

myininaya (myininaya):

or the problem

myininaya (myininaya):

let me know if you want a hint k?

OpenStudy (anonymous):

sorry, had a crisis..

myininaya (myininaya):

its cool

OpenStudy (anonymous):

looks like just u-sub then by parts..

OpenStudy (anonymous):

\[u = \sqrt{3x + 9} \implies du = \frac{3}{u}dx \implies dx = \frac{u}{3}du\] \[\implies \int e^{\sqrt{3x+9}} dx = \frac{1}{3}\int ue^udu\] Then go with by-parts.

OpenStudy (anonymous):

not nearly as bad as some of the triple integrals I had last year in calc 3.

OpenStudy (anonymous):

those things took pages =(

OpenStudy (anonymous):

Whoops, should be 2u/3 I think actually

myininaya (myininaya):

wait i think you can teach me something one sec

myininaya (myininaya):

myininaya (myininaya):

then integration by parts blah blah

OpenStudy (anonymous):

some mistakes there

myininaya (myininaya):

the only mistake i see is du=3/[2(3x+9)] dx but i really didn't use this simplification

OpenStudy (anonymous):

It should be \[\frac{2}{3}\int ue^udu\]

OpenStudy (anonymous):

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

myininaya (myininaya):

oh darn it lol

myininaya (myininaya):

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

OpenStudy (anonymous):

I don't square nothin

OpenStudy (anonymous):

I think you misunderstand what I did

myininaya (myininaya):

me too

myininaya (myininaya):

but that squaring does work also

OpenStudy (anonymous):

my way is easier ;p

OpenStudy (anonymous):

Ok, Let \(u = \sqrt{3x + 9}\) \[\implies du = (\frac{1}{2\sqrt{3x+9}} * 3 )dx\] \[ = \frac{3}{2u}dx\] \[\implies dx = \frac{2u}{3}du\]

OpenStudy (anonymous):

Then go back to the integral and sub that for dx, and the exponent of e just becomes u also.

myininaya (myininaya):

oh i see lol

myininaya (myininaya):

your awesome

OpenStudy (anonymous):

you're ;p

OpenStudy (anonymous):

ok, I'm off to watch a movie

myininaya (myininaya):

k later peace i will see if i can come up with more challenging problems next time these sucked obviously

OpenStudy (anonymous):

ok =)

OpenStudy (anonymous):

Mod[11, 3] = 2, Mod[11, 5] = 1

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