Problem solving

do you're own homework brainlet

14

easy

>>8963675
Obviously, so do it

>>8963679
it's so easy that it's not worth typing it out, are you really having problem with this?

>>8963681
Not op, starting easy i presume

>>8963666
Do your homework

This is a bmo question actually, im intrigued if any of you brainlets can actually do it. From 2001 i believe

N must be divisible by 10, so there's not too many things to check. Can do the rest by brute force / repeated squaring to check what you get from big powers of 10 mod 17

>>8963666
I would compute all ~90 values of 10^N-N. I'm guessing it's hard to divide numbers that big so I'd make a lookup table on multiples of 170. How did I do? This might be overly complicated.

Op here, nobody has the answer yet

This would have been a non calculator paper for teenagers, comon /sci/ really??

>>8963882
I told you how to do it you fucking piggot. do the work yourself
>>8963694

>>8963666
20, 39, 58, 77, 96

for N from 10 to 99
if ((10^N - N) mod 170 is equal to 0)
print N;
end
end

>>8963914
this but add the digits of 10^N-N instead, takes like 2 lines more

>>8963914
>>8963924
good luck computing 10^N - N

>>8963901
This is correct.

#include <stdio.h>

void main(void)
{
int n,nines,tens,ones;

for( n=10; n<=99; n++ )
{
nines = n-2;
tens = (100 - n) / 10;
ones = (100 - n) % 10;

if( (9*nines+tens+ones) % 170 == 0 )
printf("n %d is game\n", n);
}
}

>>8963950
That's a 100 digits number at most, a microwave with a big number maths library will have no problem with it

>>8963691
What is bmo?
Wants more.

>>8963666
well I am a phd student in theoretical computer science, I have a strong mathematical background but I can only solve problems that I find interesting

people ask me all the time, how the hell did you end up doing CS when you hated math in high school

well my friends, I hated math in high school because math in high school is shit with no real meaning.

when I started my PhD, I was like holy fucking shit math can actually be used to solve cool problems. That's when I started diving in, I learned everything there is to learn about math in computer science, and now I am sitting in my office, finished my second paper that I am about to submit in a couple of hours (deadline is in 3 hours)

I just wanted to get this out of my chest, it's off topic, but I don't care, I am tired as shit

>>8964138
You find hanging out here, NOT solving what you call uninteresting problems more interesting than actually solving these problems.

I'm a dimwit noob and even I could solve that. You should have been able to do it on a minute.

>>8964126
It's a bigger number than there's atoms in the universe. It's basically impossible to do calculations with numbers that large.

>>8963901

I got this too.

Write it as 10^N-100 + (100-N). That's
"(N-2) repeated 9's" * 100 + (100-N). Let M = 100-N = 10 m + n, where 0<=m,n<=9. You get
the sum of the digits mod 170 = 0 requires
32 - 89 m - 8 n = 0
so m is even, m = 2 k
32 - 8 k - 8 n = 0
and you just calculate n when k=0...4.

>>8964129
British mathematical olympiad, for those autistic kids that are insane at math

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>>8964157
Bingo!
Next one, little harder

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OP here, this one should take 5 mins also..

>>8964172
Dude, am I invisible >>8964119

>>8964119
Saw your computing stuff and skipped bud

>>8964174
cute if you can guess which year paper this was from too, i chuckled

>>8964174
its 1997

#include <stdio.h>
#include <math.h>

int r(int n)
{
int rn=0;
rn += n % 10 * 1000;
rn += n / 10 % 10 * 100;
rn += n / 100 % 10 * 10;
rn += n / 1000 % 10 * 1;

return rn;
}

void main(void)
{

int n, rn;

for( n=1001; n<9999; n++ )
{
if( n % 10 == 0 )
continue;
rn = r(n);

if( 4*n+3 == rn )
printf("n %d rn %d\n", n, rn );
}
}

>>8964194
told you this one was easy, why the code tho i dont get it

>>8964198
How else would one solve this?

>>8964218
i mean the notation

>>8964218
wait are you computing the answer with brute force?

>>8964225
This one, yes.

>>8963666
Not going to solve it because I'm at my job right now so it feels bad to take out my notebook and start solving this but this is how you would do it:

First characterize the numbers you are actually working with. Note that you will always have a sum of the form 9 + 9 + 9 + ... + k + i where k ranges from 0 to 9 and it is determined by the second digit of your two-digit integer, while i ranges from 0 to 9 and it is determined by the first digit of your two-digit number.

After you have characterized them, factorize 170= 2 times 5 times 17

And from here you can start chopping numbers. First leave out the numbers that won't be divisible by two. This should be easy because you simply have to calculate them mod 2, which is completely determined by i.

Then scratch out the numbers not divisible by 5, which is also easy because you only need to study the numbers mod 5 and again, you only need i to know what the number is mod 5.

From that you should have chopped off most of the non-answers while having done very little effort.

And the thing that follows is now studying the numbers mod 17, which should be harder but I conjecture that when you reach this step, you will be left with very few options so you could even now just brute force a solution by doing a bunch of calculations. That said, there is obviously a better way but to figure that out you need to first find out which possible solutions are left.

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Guys I know how to get famous in math.

Actually proving/disproving things is hard.

Just create a conjecture that's not easy to test. Boom. You've just kept mathematicians busy for the next 30 years and everyone using your name. Call it "The (Your Name) Conjecture"

>>8964398
>Just create a conjecture that's not easy to test. Boom. You've just kept mathematicians busy for the next 30 years and everyone using your name. Call it "The (Your Name) Conjecture"

>Daily reminder that many mathematicians, including famous ones like Gauss, mocked Fermat for even bothering to conjecture something he had not even a clue about how to prove

>Daily reminder that the only reason people even remember Fermat is because he actually proved some interesting things here and there

>Daily reminder that it is Wiles Theorem and if you call it by anything else you are a literal brainletto.

>>8964229
im looking for an intuitive solution, its for 11-16 year old kidds without a calculator, or computer.

>>8964229
>>8964610
e.g you know N<2500 because R(N) is 4 digits, therefore you know the first digit is 1 or two. howerver it cant be two because R(N) is odd(4n+3) etc.

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How many of you are as smart as you think you are?

>>8965653
Where did you get this problem from? Just tell me if you came up with it or not.

>>8965658
Didnt come up with it.

>>8964174
Solved this but it took a good 25 minutes or so

t. high school brainlet

answer is 1997

>>8965653
Fuck imma do it... stand back brainlets!

Take the limit as a,b,c --> 0

I can make the expression be arbitrarily close to zero. Including less than 1.

Boom.

>>8963666
got this fairly easily

main = do
let s = map (\(k, x) -> (k, 10^x - x)) (zip [10..99] [10..99])
s' = map (\(k, x) -> (k, digitSum x)) s
print $ map fst $ filter (\(_, x) -> mod x 170 == 0) s'

digitSum :: Integer -> Integer
digitSum 0 = 0
digitSum n = (mod n 10) + digitSum (div n 10)

it returns:
[20,39,58,77,96]

think it's right, not sure

>>8964174
>>8964194
get cucked C brainlets

digitReverse :: Integer -> Integer
digitReverse = read . reverse . show

main = print $ filter (\x -> digitReverse x == 4*x + 3) [1000..9999]

>>8963666
Try a real world problem/scenario and maybe i'll give a shit. Not a math problem.

>>8967314
does this board do [code]tags[/code]

doubt it

>>8963784
Try reading the question again

>>8963784
>I'm guessing it's hard to divide numbers that big
It's not. Most programing languages have built-in ways to handle arithmetic using arbitrarily long numbers. A simple computer program could provide an answer
https://en.wikipedia.org/wiki/Arbitrary-precision_arithmetic

Here is the code in python

def sum_digits(n):
res = 0
for dig in n:
res += int(dig)
return res

result_list = []
for ii in range(10,100):
s = str(pow(10, ii) - ii)
if ( 0 == sum_digits(s) % 170):
result_list.append(ii)

print( result_list )

>>8967317
brainlet engineer detected

>>8967386
How is "arbitrarily long numbers" a brainlet thing?
https://en.wikipedia.org/wiki/Arbitrarily_large

>>8967360
Most? Really?

>>8967399
You're subtracting a 2 digit number, N, from 10^N. You're going to get N-2 9's followed by (100-N). Were you sick the day they taught borrowing in kindergarten?

>>8965653
Jest use Cauchy schwarz or the more general inequality i forgot the name
Too lazy to type everything out

>>8967786
youre actually correct, cauchy is a good start. holders is probably the easiest in my eyes

the compsci faggots really are brainlets. problems designed for autistic teenagers to do in their heads, and these adults need a computer to do it for them.

>>8967727
So I could either perform two calculations
>determine N^2
>subtract N from the answer

or I can perform 6 calculations
>determine N-2
>make a string containing that many 9's
>determine 100-N
>convert that to a string
>concatenate the two strings
>convert the new string into a number

Keep trying to solve all your problems using kindergarten logic.