You think youre a problem solver? Solve this.
Problem solving
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do you're own homework brainlet
14
easy
Obviously, so do it
it's so easy that it's not worth typing it out, are you really having problem with this?
Not op, starting easy i presume
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
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 Veeky Forums really??
I told you how to do it you fucking piggot. do the work yourself
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
this but add the digits of 10^N-N instead, takes like 2 lines more
good luck computing 10^N - N
This is correct.
#include
void main(void)
{
int n,nines,tens,ones;
for( n=10; n
That's a 100 digits number at most, a microwave with a big number maths library will have no problem with it
What is bmo?
Wants more.
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
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.
It's a bigger number than there's atoms in the universe. It's basically impossible to do calculations with numbers that large.
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
British mathematical olympiad, for those autistic kids that are insane at math
Bingo!
Next one, little harder
OP here, this one should take 5 mins also..
Dude, am I invisible
Saw your computing stuff and skipped bud
cute if you can guess which year paper this was from too, i chuckled
its 1997
#include
#include
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
told you this one was easy, why the code tho i dont get it
How else would one solve this?
i mean the notation
wait are you computing the answer with brute force?
This one, yes.
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.
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"
>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.
im looking for an intuitive solution, its for 11-16 year old kidds without a calculator, or computer.
e.g you know N
How many of you are as smart as you think you are?
Where did you get this problem from? Just tell me if you came up with it or not.
Didnt come up with it.
Solved this but it took a good 25 minutes or so
t. high school brainlet
answer is 1997
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.
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
get cucked C brainlets
digitReverse :: Integer -> Integer
digitReverse = read . reverse . show
main = print $ filter (\x -> digitReverse x == 4*x + 3) [1000..9999]
Try a real world problem/scenario and maybe i'll give a shit. Not a math problem.
does this board do [code]tags[/code]
doubt it
Try reading the question again
>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
en.wikipedia.org
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 )
brainlet engineer detected
How is "arbitrarily long numbers" a brainlet thing?
en.wikipedia.org
Most? Really?
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?
Jest use Cauchy schwarz or the more general inequality i forgot the name
Too lazy to type everything out
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.
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.