@blackpenredpen the problem is-THERE ARE 6 MEN AND 4 BOYS AND THEY ARE NEEDED TO BE ARRANGED IN A ROW OF 10 VACANT SPACES.FIND THE PROBABLITY THAT ONLY 2 BOYS ARE NEXT TO EACH OTHER(together).
10:16 There is a fast way! Observe the following: 186 should be written as 1 + p + p^2 + ... + p^a Notice after that first +1 term every term has a factor of p So just test the prime factors of 185 (5 and 37) and you're set! In general, if you're looking to represent a factor F in the form 1 + p + p^2 + ... + p^a, the only values you need to test are the prime factors of F-1
I always liked math because of little things like this, it turns illegible information into knowledge, you can see how people from the past viewed math with a level of mysticism.
at a certain level, it feels like the problems are hard not because the problem is innately hard, but just deliberately created hard so fewer people can solve them
Reference is 10:54 for 6x31: weirdly 31 can also be written as 5^0 + 5^1 + 5^2 ... so once you consider 5, you can't consider it again ... I don't understand the part of logic that I am missing ...
An integer number n has the unique decomposition n = p1^a * p2^b * ... * pk^z, with px a different prime every factor. And that the sum of its divisors is (1 + p1 + p1^2 + ... + p1^a) * (1 + p2 + p2^2 + ... + p2^b) * ... * (1 + pk + pk^z + ... + pk^z) As you can see, the largest term of each factor is one of the primes in the decomposition of n to a power Therefore, if 6 gives you the prime 5 to the 1st power and 31 gives you the same prime 5 to the 2nd power, it means that this unique decomposition contains both the factor 5 AND the factor 5^2, which is not possible as all primes must be different.
I was just thinking, must the prime in each bracket be distinct? Cuz for the 6*31 i was thinking that 31 = 1+5+5^2, but since we already used 5 in 6=1+5, idk if its valid.
It's not valid, by definition the prime factorisation he started with assumed that each p is a distinct prime, so the prime used for each bracket must also be distinct
Not sure if anyone will see this but I have devised a formula for a^3-b^3 which I believe is easier to use and can help solve a lot of different types of problems. The original formula: (a-b)(a^2+b^2+ab) My formula: x(x^2+3bx+3b^2) where x = a-b The reason I believe this formula is better is because it doesn’t rely on the variable a, if they ask a problem to find a while giving a-b and a^3-b^3 this could make the solving process easier because when I tried to solve it then traditional way it took a lot of time. This also is useful for finding a^3-b^3 by hand. Let a=2763 and b=2760 3(2763^2+2760^2+2763*2760) vs 3(9+9*2760+2760^2) In the traditional way, you have to find 2763*2763, 2760*2760 and 2763*2760 which would take forever but in my way you just need to find 2760^2 and 2763*3 which is way easier because as I timed it and it took me 5x longer to do it the traditional way. If you have any opinion on this formula and have any changes please let me know in the reply section!
idk how well it could work but to check if a number can be written as 1+x+x^2...x^n we can subtract 1 from the number and fsctorize it.. to narrow down the possibilities, because if we take 186= 1+x+x^2+x^3...x^n then 185 = x+x^2+x^3..x^n, so 185 is always a multiple of x.. so 185 has only 2 factors of 5 and 37.. and 37^2 is way greater than 185, and 5 doesnt work as he said, idk how much easuer this makes it but maybe(?)
Is it possible that one of the factors (like 62, in this example) could theoretically be expressed as the sum of powers of primes in that manner, but in two different ways? In other words, maybe finding (1+2)(1+61) isn't quite finishing off the case of 3·62?
Notably, 31 can come from a prime power of 2^4, but it could also be from a prime power of 5². (But not in this specific case because it occurs beside 6, which would be from a factor of 5¹ and that violates the requirement that the powers are of distinct primes)
Sigh ... what a lovely problem. I just wish it didn't feel impossible to come up with this factorization at 04:15. I have solid degree in math and yet ... feels bad
bprp, I have a math problem that needs to be solved before the end of summer or else bad things will happen to me! The problem is: give an example of f(x), where f(n) = a_n, such that the lim n --> infinity of a_n = L but lim x --> infinity does not equal L. Please help, this is urgent.
ua-cam.com/video/1TBVeuOcY1w/v-deo.html Sir in this video, at 38:00 I can't understand how is it a sum of products of positive numbers gives a negative value?
How can we be sure that there are not Two different primes that could lead to 31 for example? I mean could it be possible that 31 = 1+p^1+...+p^i = 1 + q^1 + ... + q^j ?
@jojijoestar4762 Thank you... In fact we should consider like this all the possible divisors (if there is no other big impossibility). I think the video is a bit to quick on this topic.
take a look in the playlists section on his channel, and find calculus 1. i recommend watching them in order, there are a lot but many of them are just revisers that you can skip.
Hey bprp, can you try finding the sides of a triangle with a given perimeter and area? I've been wondering about some trick to the sol. using herons formula but it didnt work out. Can you please try it too?
Same area and perimeter unfortunately don’t guarantee you get a unique triangle. Consider triangle A with sides 17, 25, 28 and triangle B with sides 20, 21, 29. They both have area 210 and perimeter 210 but they are not congruent.
sir, could u pls solve integration of sec³x via D,I METHOD🙂 I AM BADLY STUCK IN THIS PROBLEM (USING D,I METHOD) also pls make one more video to solve integration of inverse fuction via F,I method, log.inverse, algebra.inverse, trigo.inverse etc. i hope u will make video for your this student love from india❤️🇮🇳
Hey man i've been watching a lot of your integral videos, and i must say you are an absolute math genius, and i really like your writing a lot. I'm currently doing my calculus 2 course and bumped into this integral: sqrt(4x^2 -1). I searched the entire internet and couldn't find anyone that showed a way to solve it. I solved it on my own but it took a while and the answer is like over 2 pages long. I would be intrigued to see how you do it.
Vaiya, Please integrate x^2√(25-x^2) with respect to x without doing trigonometric substitution. Please Please Please Please Please🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏 Best wishes from Bangladesh
@@muntasirstudent1746 √(25-x^2) this can be integrate without trigonometric substitution using simple by parts formula. So, I thik x^2√(25-x^2) also can be integrate without trigonometric substitution. But I am not sure
@@mojiburrahman6608 First Integration by parts will be a common way and even if u apply that u will need to use trig sub there also so yeah in a way or the other trig sub is needed
Hey bprp! I had an great idea for a video : for u to solve some of jee main/advanced math questions! They r one of the toughest exams ever, and it could make a great video with very deep math questions!
I'm just here to vent a little. I hate irrational numbers. I think they are a result of a faulty mathematical language. For instance.... a circle's circumference is complete. We can see a circle and see that it is complete. But pi never ends and so therefore there shouldn't be any completely enclosed circles. I've been trying to break/eliminate pi for about a year now and I'm driving myself crazy. I've tried using squares and triangles, I've tried improving on Achimedes method to calculate the circumference to no avail. It's these god damned angles. There seems to be an infinite amount of them!!! Ahhhhhhhhhhhhhhhhhhhhjhhhhh!!!!! 😢
hey bprp! i have a habit of screwing around on desmos, and ended up chasing this one odd number -- approximately 1.6937.. -- which resulted from trying to find x in the equation "x^y = x^(y-1) + 1" and then seeing the point at which that curve intersects with the line x = y. plugging into WFA gave no exact answer, only an approximation, and i havent found this number listed in any database (including OEIS, wolframalpha search, and ebyte.it). i suspect there's a usage of productlog to be seen here but i cant figure it out for the life of me, and im wondering if youd be interested in making a video on it? should i contact you elsewhere?
0:22 I blame the autocorrect for not catching my typo 😆
Fair
Hey, try this next time: solve for x in sin(x) = 1, cos(x) = i. It turns out that sin(theta) = e^((theta - x) i), I think.
if i give you a problem would you solve it and make a video on it
@blackpenredpen the problem is-THERE ARE 6 MEN AND 4 BOYS AND THEY ARE NEEDED TO BE ARRANGED IN A ROW OF 10 VACANT SPACES.FIND THE PROBABLITY THAT ONLY 2 BOYS ARE NEXT TO EACH OTHER(together).
pls tell the solution@picup30296
A (slightly) easier way to find if
186 = 1 + p + p² + ... + pⁿ
is possible, is by noticing that p mist divide 186-1 = 5 * 37, so p is either 5 or 37.
Ah that’s so nice! I didn’t think of that! Thanks.
I did the same
@@blackpenredpen additionally, we probably want to consider the geometric sum formula.
Hey blackpenredpen, I have a proof of an infinite series. docs.google.com/document/d/1exnCCsNHCqhKH1BaFi9XU6uqR15K5tH-sO95Xy-fR9Q/edit?usp=sharing
hello
I like that whenever I'm lacking and not studying, bprp comes with a new video and I feel like to do math again
10:16
There is a fast way!
Observe the following:
186 should be written as 1 + p + p^2 + ... + p^a
Notice after that first +1 term every term has a factor of p
So just test the prime factors of 185 (5 and 37) and you're set!
In general, if you're looking to represent a factor F in the form 1 + p + p^2 + ... + p^a, the only values you need to test are the prime factors of F-1
I always liked math because of little things like this, it turns illegible information into knowledge, you can see how people from the past viewed math with a level of mysticism.
Some are blackpilled, others are redpilled. We are happy to be blackandredpenned.
i’m permablackpilled 😔
Try this next
Oxford-Level Counting Problem: ua-cam.com/video/9WKHZB76BN0/v-deo.html
at a certain level, it feels like the problems are hard not because the problem is innately hard, but just deliberately created hard so fewer people can solve them
your way of solving is very interesting and worth watching ,
thanks bprp!!!
Reference is 10:54 for 6x31: weirdly 31 can also be written as 5^0 + 5^1 + 5^2 ... so once you consider 5, you can't consider it again ... I don't understand the part of logic that I am missing ...
The other number is 8191, which is a repunit in bases 2 and 90. The Goormaghtigh conjecture is that there are no others.
An integer number n has the unique decomposition n = p1^a * p2^b * ... * pk^z, with px a different prime every factor.
And that the sum of its divisors is (1 + p1 + p1^2 + ... + p1^a) * (1 + p2 + p2^2 + ... + p2^b) * ... * (1 + pk + pk^z + ... + pk^z)
As you can see, the largest term of each factor is one of the primes in the decomposition of n to a power
Therefore, if 6 gives you the prime 5 to the 1st power and 31 gives you the same prime 5 to the 2nd power, it means that this unique decomposition contains both the factor 5 AND the factor 5^2, which is not possible as all primes must be different.
Last time i was this early he didn't even add a +c
I proctored for this competition! Didn't really get to think about the problems much though.
Nice to hear from you again Fematica! Hope all is well!
what does proctoring mean?
@@NoNameAtAll2 I led the students into their room, gave them their tests, and made sure they didn’t cheat.
For the 186 case, notice that after subtracting one the RHS divides the prime, so 185=5*37 must divide it. Neither 5 nor 37 works.
I was just thinking, must the prime in each bracket be distinct? Cuz for the 6*31 i was thinking that 31 = 1+5+5^2, but since we already used 5 in 6=1+5, idk if its valid.
It's not valid, by definition the prime factorisation he started with assumed that each p is a distinct prime, so the prime used for each bracket must also be distinct
@@aadhavan7127 ok thank you for clarifying.
Not sure if anyone will see this but I have devised a formula for a^3-b^3 which I believe is easier to use and can help solve a lot of different types of problems.
The original formula: (a-b)(a^2+b^2+ab)
My formula: x(x^2+3bx+3b^2) where x = a-b
The reason I believe this formula is better is because it doesn’t rely on the variable a, if they ask a problem to find a while giving a-b and a^3-b^3 this could make the solving process easier because when I tried to solve it then traditional way it took a lot of time.
This also is useful for finding a^3-b^3 by hand.
Let a=2763 and b=2760
3(2763^2+2760^2+2763*2760) vs 3(9+9*2760+2760^2)
In the traditional way, you have to find 2763*2763, 2760*2760 and 2763*2760 which would take forever but in my way you just need to find 2760^2 and 2763*3 which is way easier because as I timed it and it took me 5x longer to do it the traditional way.
If you have any opinion on this formula and have any changes please let me know in the reply section!
For the 186 case, notice that after subtracting one the RHS divides the prime, so it must be a divisor of 185=5*37. Neither of these work.
idk how well it could work but to check if a number can be written as 1+x+x^2...x^n we can subtract 1 from the number and fsctorize it.. to narrow down the possibilities, because if we take 186= 1+x+x^2+x^3...x^n then 185 = x+x^2+x^3..x^n, so 185 is always a multiple of x.. so 185 has only 2 factors of 5 and 37.. and 37^2 is way greater than 185, and 5 doesnt work as he said, idk how much easuer this makes it but maybe(?)
For 10:30 you could use the formula for a geometric series (1-q^n+1)/(1-q) equal 186 and solve for q to see if there is a prime.
In the beginning why did he stop at 6, ie how do you decide when to stop looking for divisors
Is it possible that one of the factors (like 62, in this example) could theoretically be expressed as the sum of powers of primes in that manner, but in two different ways? In other words, maybe finding (1+2)(1+61) isn't quite finishing off the case of 3·62?
Notably, 31 can come from a prime power of 2^4, but it could also be from a prime power of 5². (But not in this specific case because it occurs beside 6, which would be from a factor of 5¹ and that violates the requirement that the powers are of distinct primes)
Hello, can I communicate with you? I need to solve a differential equation that has been highly controversial by senior professors
plz also make a vid about the integral of 1/(x^5 + 1)
Sigh ... what a lovely problem. I just wish it didn't feel impossible to come up with this factorization at 04:15. I have solid degree in math and yet ... feels bad
Wow this way is so fast to solve
bprp, I have a math problem that needs to be solved before the end of summer or else bad things will happen to me!
The problem is: give an example of f(x), where f(n) = a_n, such that the lim n --> infinity of a_n = L but lim x --> infinity does not equal L.
Please help, this is urgent.
Take f(x)=sin(π x).
@@FedericoRulli oh my gosh, thank you sir!
Make an introduction to contoured integration. Love your vids ❤
please solve this integral (sinx/sinhx )dx
How would one compute the indefinite integral
ʃ(e^(sin(x))) dx
ua-cam.com/video/1TBVeuOcY1w/v-deo.html
Sir in this video, at 38:00 I can't understand how is it a sum of products of positive numbers gives a negative value?
Hello, how are you? You seem really nice
@@thenew3dworldfan Hi, I'm good, I hope you too, glad to hear this from you, and it's clear that you're very kind too :)
How can we be sure that there are not Two different primes that could lead to 31 for example? I mean could it be possible that 31 = 1+p^1+...+p^i = 1 + q^1 + ... + q^j ?
@jojijoestar4762 Thank you... In fact we should consider like this all the possible divisors (if there is no other big impossibility). I think the video is a bit to quick on this topic.
6x31 =(1+5)(1+5+5²)
They should be different numbers.
Otherwise it will be 5¹•5² = 5³
And sum of whose divisors is 156
Do i^sqrt(i) next
Can u teach me calculus? (From the beginning) I really love your videos.
take a look in the playlists section on his channel, and find calculus 1.
i recommend watching them in order, there are a lot but many of them are just revisers that you can skip.
can you find tan(x/3) while also trying to find tan(arctan(x)/3)
Go back go back!
So cool. 🎉
[789under root x-1 divided by 100-x]+ [196/x⁵+66] this whole expression divided by 100-x=197/98 then find the value of x
Спасибо большое!
Hey bprp, can you try finding the sides of a triangle with a given perimeter and area? I've been wondering about some trick to the sol. using herons formula but it didnt work out. Can you please try it too?
Same area and perimeter unfortunately don’t guarantee you get a unique triangle. Consider triangle A with sides 17, 25, 28 and triangle B with sides 20, 21, 29. They both have area 210 and perimeter 210 but they are not congruent.
I want someone to help me in maths i have exams really soon 😢 have some olompiad questions
x^(4)-x+1=0 find x^(10) can you help with that pls. I did not find any solution in the internet.
sir, could u pls solve integration of sec³x via D,I METHOD🙂
I AM BADLY STUCK IN THIS PROBLEM (USING D,I METHOD)
also
pls make one more video to solve integration of inverse fuction via F,I method, log.inverse, algebra.inverse, trigo.inverse etc.
i hope u will make video for your this student
love from india❤️🇮🇳
Here is a problem that I'm curious about: I tetrated to i
Why are you uploading at 2am?
Bud it's near evening here.
@@artophile7777 Who are you?
@@donwald3436 I'm his uncle's nephew's brother's grandmother's sister's aunt's cousin's son.
@@donwald3436 videos can be scheduled to go public at a specific time lmao
@@artophile7777
"What's that make us?"
"Absolutely nothing. Which is what you are about to become."
Number Theory problems are the most interesting Math problems, aren't they?
Hey man i've been watching a lot of your integral videos, and i must say you are an absolute math genius, and i really like your writing a lot.
I'm currently doing my calculus 2 course and bumped into this integral: sqrt(4x^2 -1).
I searched the entire internet and couldn't find anyone that showed a way to solve it.
I solved it on my own but it took a while and the answer is like over 2 pages long.
I would be intrigued to see how you do it.
I would do trig sub: let x = 1/2sec(theta) and go from there
Number theory
i cant even understand the question lol intuitively
fisrt
import math
# Function to calculate the sum of divisors of a number
def sum_of_divisors(n):
result = 0
i = 1
while i
I don't think you'd have Python during the competition...
I always like the solution in R a little better
which(sapply(1:1000, function(n) { divs = 1:n; sum(divs[which(n %% divs == 0)]) }) == 186)
[1] 80 122
@@erikkonstas haha I know thought it was fun though to see if a script would be able to do the same thing
Vaiya, Please integrate x^2√(25-x^2) with respect to x without doing trigonometric substitution.
Please Please Please Please Please🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Best wishes from Bangladesh
Vaiya I am also a Bangladeshi But it can't be solved without trig sub
@@muntasirstudent1746 √(25-x^2) this can be integrate without trigonometric substitution using simple by parts formula. So, I thik x^2√(25-x^2) also can be integrate without trigonometric substitution. But I am not sure
@@mojiburrahman6608 First Integration by parts will be a common way and even if u apply that u will need to use trig sub there also so yeah in a way or the other trig sub is needed
Hey bprp! I had an great idea for a video : for u to solve some of jee main/advanced math questions! They r one of the toughest exams ever, and it could make a great video with very deep math questions!
please no sound effects. no dings.
Were they a bit too loud?
@@blackpenredpenIt wasn't too loud, it was fine
Find f(x) ,f(x)=0 when x from -infinity to 1 and f(x)=1 when x from >1 to infinity without forier series i dare you friend
☺️
I'm just here to vent a little.
I hate irrational numbers. I think they are a result of a faulty mathematical language. For instance.... a circle's circumference is complete. We can see a circle and see that it is complete. But pi never ends and so therefore there shouldn't be any completely enclosed circles.
I've been trying to break/eliminate pi for about a year now and I'm driving myself crazy. I've tried using squares and triangles, I've tried improving on Achimedes method to calculate the circumference to no avail.
It's these god damned angles. There seems to be an infinite amount of them!!! Ahhhhhhhhhhhhhhhhhhhhjhhhhh!!!!! 😢
I could add up the prime factors of 40 within about 5 seconds😂
No one can understand competition MATH i wanted to take the class but then i said ur not ready yet)
hey bprp! i have a habit of screwing around on desmos, and ended up chasing this one odd number -- approximately 1.6937.. -- which resulted from trying to find x in the equation "x^y = x^(y-1) + 1" and then seeing the point at which that curve intersects with the line x = y. plugging into WFA gave no exact answer, only an approximation, and i havent found this number listed in any database (including OEIS, wolframalpha search, and ebyte.it). i suspect there's a usage of productlog to be seen here but i cant figure it out for the life of me, and im wondering if youd be interested in making a video on it? should i contact you elsewhere?
Find f(x) ,f(x)=0 when x from -infinity to 1 and f(x)=1 when x from >1 to infinity without forier series i dare you friend