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

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- myininaya

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

- myininaya

smallest*

- anonymous

Looks like the Chinese remainder theorem.

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## More answers

- myininaya

yay! :)

- myininaya

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

- myininaya

i really like this problem

- anonymous

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

- myininaya

omg polpak i love the RSA

- myininaya

i want to be an expert at number theory

- anonymous

pretty cool. You can work for nsa ;)

- 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

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

- 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

- anonymous

all algorithms are just descrete math stuff ;p

- myininaya

yes thats true

- myininaya

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

- anonymous

About to transfer to 4 year university.

- anonymous

just finished the local CC.

- myininaya

what?!
no way

- myininaya

how are you super smart?

- anonymous

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

- myininaya

i think you are really cool :)

- anonymous

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

- myininaya

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

- myininaya

math instructor?
engineer?
...

- anonymous

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

- myininaya

you can do it
you have the knowledge already for math

- myininaya

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

- myininaya

or some differential equations

- anonymous

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

- myininaya

are you american?

- myininaya

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

- anonymous

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

- myininaya

lol

- anonymous

I study cause it makes things easier

- anonymous

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

- myininaya

yes i agree

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

evaluate the above is the question

- myininaya

or the problem

- myininaya

let me know if you want a hint k?

- anonymous

sorry, had a crisis..

- myininaya

its cool

- anonymous

looks like just u-sub then by parts..

- 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.

- anonymous

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

- anonymous

those things took pages =(

- anonymous

Whoops, should be 2u/3 I think actually

- myininaya

wait i think you can teach me something one sec

- myininaya

##### 1 Attachment

- myininaya

then integration by parts blah blah

- anonymous

some mistakes there

- myininaya

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

- anonymous

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

- 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

oh darn it lol

- 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

- anonymous

I don't square nothin

- anonymous

I think you misunderstand what I did

- myininaya

me too

- myininaya

but that squaring does work also

- anonymous

my way is easier ;p

- anonymous

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

- anonymous

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

- myininaya

oh i see lol

- myininaya

your awesome

- anonymous

you're ;p

- anonymous

ok, I'm off to watch a movie

- myininaya

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

- anonymous

ok =)

- anonymous

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

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