The computation after 8:00 can be replaced by a much simpler one. Since N = 0 (mod 5^5), we can write N = (5^5)n for some integer n; inserting this in N = 5^5 (mod 2^5) we obtain (5^5)n = 5^5 (mod 2^5). Since gcd (5^5,2^5) = 1, we can divide both sides by 5^5, obtaining n = 1 (mod 2^5), or n = 1 + (2^5)m for some integer m. Therefore, N = (5^5)n = (5^5) [1 + (2^5)m] = 5^5 + (10^5)m = 5^5 (mod 10^5) = 3125 (mod 10^5).
Yep. You can easily prove (e.g. by induction) that the cycle will continue, and 5^5^5^5 mod 8 is easily determined as well (those values just alternate between 1 and 5).
@@angelmendez-rivera351 that's different though and much less explicit we sont use or think mod x or something like that..so my point still stands..so why dont you not talk out of your ass for once ..
I have a somewhat easier solution: I put the numbers 5^0, 5^1,5^2,5^3 ,... mod 100,000 in a row and what I realised is, that there is a cycle of length 8 starting at n^5: 3125 (n^5) 15625 78125 90625 53125 65625 28125 40625 3125 (n^13) Thus I only need to know what 5^5^5^5 mod 8 is. But for 5^k there is a cycle of length 2 mod 8 , that is 5 mod 8 for odd k and 1 for even k. As 5^5^5 is clearly odd, we have 5^5^5^5 = 5 mod 8. And that tells us, that 5^5^5^5^5 = 3125 mod 100,000.
6:22 Let’s maybe go ahead and do that 12:52 That’s a good place to stop Homework. Number theory. Putnam 1963. Find all integers n for which x^2 - x + n divides x^13 + x + 90.
When you are getting ready to find the values of n at 6:33, I gasped. That's the same combination on my luggage! What a coincidence, that's amazing! I'd better change it.
2:20 What?!?! You meant (5^5) ^ (5^5^5) because that is 5 raised to 5*(5^5^5) and that is different from N. You just made a common mistake when it comes to towers of powers, which is thinking that a^b^c^d is EQUAL to (a^b)^(c^d), which is not. The tower is divisible by 5^5 because it can be written as 5^(5+n) with n = non-negative integer, and any number in this format has no remainder when divided by 5^5, and in fact the quotient is 5^n.
Typically, exponentiation is right-associative, e.g. n^n^n = n^(n^n), n^n^n^n = n^(n^(n^n)), n^n^n^n^n = n^(n^(n^(n^n))), etc. This is the best convention, since the left-associative interpretation can be reduced: (n^n)^n = n^(n^2), ((n^n)^n)^n = (n^(n^2))^n = n^(n^3)), etc. Look up "tetration" for more on these power towers!
@@samsonblack Yep. Think about it: a left assoc power towers would be pretty useless. Like you said, a left assoc exponentiation would give 5^5^5^5^5 = 5^(5^4) which has "only" 437 digits. The right assoc power tower 5^5^5 already has 2185 digits. Clearly, right assoc is the way to go if you want to reach those absurdly large numbers. :D
@@jamirimaj6880 For sure not. 5^^4 already has many, many more digits than 10k. On the order of 10^1000 something digits, so clearly 5^^5 has even more than that. And by by m^^n I meant double Knuth's up-arrow notation, ie. tetration. 5^^5 = 5^(5^(5^(5^5)))
Quick solution. Observations: 5^5=03725 and the last five digits of 5^13 are also 03725. Therefore, the last five digits of 5^n is cyclical by the period of 8 when n>=5. Then question boils down to find the value of (5^5^5^5-5)(mod 8). Easy to see it is 0 using Euler's theorem. Hence, the last five digits of 5^5^5^5^5 is 03725.
Honestly, when I saw the title I was excited how Prof. Penn wanted to find the 10,000th digit of that number. And then I got a n interesting video how the find the fifth digit from the right. :-) Thank you for this interesting insight into number theory.
Maybe I missed that step, but why is the 10000th digit equal to the 5th digit from the right? (Edit: Ok, language barrier ^^ there is a difference between ten-thousands and ten-thousandth ...)
It was indicated in the book that the 10000th digit is also equal from the 5th digit to the right. Maybe he should've started by saying how many digits the number has.
I was thinking a lot about this problem but I got nowhere. Turns out I misunderstood the problem. I thought we were looking for the Nth ordinal digit, N=10000, not the 10000 radix valued digit. There is a huge difference between the two... So glad I finally unpaused so I could discover my mistake.
I solved this problem based on observation. First and foremost, observe the first few powers of 5 mod 100000. (5^1) mod 100000 = 5 (5^2) mod 100000 = 25 (5^3) mod 100000 = 125 (5^4) mod 100000 = 625 (5^5) mod 100000 = 03125 (5^6) mod 100000 = 15625 (5^7) mod 100000 = 78125 (5^8) mod 100000 = 90625 (5^9) mod 100000 = 53125 (5^10) mod 100000 = 65625 (5^11) mod 100000 = 28125 (5^12) mod 100000 = 40625 (5^13) mod 100000 = 03125 We can observe that the last 5 digits of the 5th power and 13th power are the same. Hence, there will be a cycle of 8 where the last 5 digits of the powers remains the same. Therefore, it is enough to determine what the power 5^5^5^5 evaluates to mod 8. Assume that k=5^5^5 (5^5^5^5) mod 8 = (5^k) mod 8 = ((4+1)^k) mod 8 = (k*4 + 1) mod 8 {Using the binomial theorem} We know that k is odd. Replacing k=2*m+1, we get (5^5^5^5) mod 8 = ((2*m+1)*4 + 1) mod 8 = 5 Therefore the last 5 digits of 5^(5^5^5^5) are the same as that of 5^5, i.e., 03125.
I have a problem, mau you can help me to solve it? If we have a triangle ABC, and point D inside the triangle, then can we find the maximum length of AD+BD+CD mathematically? Thank you..
4:40 He wants to reduce a^b modulo c. Why can he do this by reducing b modulo phi(c) ? I don't see the connection to a^phi(n) congruent to1 modulo n. Would be nice if someone can explain this to me. OK, I think I got it: a^(phi(n)+d) = a^phi(n) * a^d = 1 * a^d = a^d (modulo n), therefore the result is not changed when reducing the exponent modulo phi(n). I think Michael should have explained this in more detail. Also, his reasoning at 2:30 is not really correct, but for sure the conclusion is. But for sure still great videos where there is always a lot to learn from!
We could use the fact that the last 5 digits of 5^(k+8) are the same as of the number 5^k, because 5^(k+8)-5^k is divisible by 10^5 (for k>=5). And now we have to figure out what is 5^5^5^5 congruent to mod 8. It is true that 5 is congruent to 5^5 mod 8 thus 5^(5) is congruent to 5^(5^5) mod 8 and similarly 5^(5^(5)) is songruent to 5^(5^(5^5)) mod 8. This means that 5^5^5^5 is congruent to 5 mod 8. And thus the last five digits of number 5^5^5^5^5 are the same as of the number 5^(13)=1220703125. And from here we can see that the ten-thousands digit is 0.
But presumably you need a computer to work out the ten-thousands digit of 5^13, in which case why not just solve the original problem directly on a computer? No offence intended, but seems a bit of a cheat in the context of a math competition!
It's not that hard to do. I have a program in python that calculates the last 500 digits of G64 in a couple of seconds. I got the idea when I solved problem 188 at project euler.
Hello professor michael , can you plz do some more AIME number theory / combinatorics problems ? they are really great problems and if u make a video about those , it will be the best thing ever :)
I started with a completely wrong interpretation of the problem. I thought it was 10,000th digit, versus the 10000's digit. Is there a reasonable way to approach this harder problem?
Pardon; the first question is how to know that "the fifth digit from the right" is the same as "the ten-thousands digit"? I think you may need a refresher on place-value arithmetic....
@Paul O'Reilly Oh I see, I get it now. But beginners would be very confused that the 10,000 digit of the number is the 5th number to the right. Maybe he should've cleared this up at the beginning.
At the end we are essentially searching for a multiple of 5^5 that is congruent to 21 modulo 32. Since 5^5=(25)^2*5=(-7)^2*5=49*5=17*5=85=21(mod 32) we see that 5^5*k=21k(mod 32) and we are looking for k such that 21k=21(mod 32) which in this case is trivial but generally would require finding an inverse of 5^5 mod 32. In this case, instead of extended Euclid one can opt to use Euler's theorem - since 5^16=1(mod 13) we see that 5^5*5^11=1(mod 13) i.e the inverse is 5^11=(5^5)^2*5=21^2*5=441*5=25*5=125=-3(mod 32) - the fact that this inverse is just 5^3 stems from the fact that not only 5^16=1(mod 32) but 5^8 as well.
5^5^5 is already 10^2184. That means that the number of digits of the number 5^5^5^5 is around 10^1526. And that means, that the number of digits, for the number describing the number of digits of 5^5^5^5^5 is around 10^1067. Numbers that big is not comprehensible, at least not for me.
@@Evilhippie64 It's insane. Even the n-1:th iterate of the logarithm of (n↑↑n) grows superlinearly with n. That is log(2↑↑2), log(log(3↑↑3)), log(log(log(4↑↑4)))), log(log(log(log(5↑↑5)))), ... (unless I made a mistake). I wonder what an asymptotic of this series would be.
Try explaining isi entrance exam questions mainly subjective It's the exam given by indian students who r 17 to 18 to enter indian statistical institute which we have only 50 seats in entire india
Am I the only one confounded when he said "fifth digit from the right" thinking that he knows that there would be 10004 digits in the exact answer of the power of tower? But okay, he meant ten-thousand, the place value, not ten-thousandth, the position
Here is a problem I couldnt solve, someone interested?? 729 , 71289 , 7112889 ,711128889 , 71111288889 ... ... ... Prove that each term of the sequence will always be a perfect square. PS I am a high school student.
Again, the solution in the video is much too complex. Simply: 5^8 ≡ 1 mod 32 and 5^2 ≡ 1 mod 8, so that 5^(2k+1) ≡ 5 mod 8 and 5^5^5^5^5 ≡ 5^5 ≡ 3125 mod 100000.
I think a mistake some mathematicians make is to assume shorter solutions are better (or less complex) than longer solutions. If the longer solution contains individual steps that are easier to follow, or ideas and process that can be applied to solve other problems then it is arguably better. In this example I have no idea why one would start by saying “simply 5^8=1 (mod 32)”, but then I’m no doubt an inferior mathematician to you!
@@AlephThree I didn't include the classical steps. But here are details and some general explanations after the answer. One seeks for N modulo 100000. Since 100000 = 2^5 × 5^5, we just need to consider N modulo 2^5 and modulo 5^5 (due to the Chinese remainder theorem). The modulo 5^5 will be easy, so let's focus on N modulo 2^5, i.e. N modulo 32. [Not much difference with the video until here.] Then... One has 5^8 ≡ 1 mod 32, so we just need to find 5^5^5^5 mod 8. And since 5^2 ≡ 1 mod 8 and 5^5^5 is an odd integer, 5^5^5^5 ≡ 5 mod 8. Hence N ≡ 5^5 mod 2^5. And N ≡ 0 ≡ 5^5 mod 5^5. Therefore, N ≡ 5^5 ≡ 03125 mod 100000. No need for Euler's theorem (in particular in simple cases like here, one can often do better, and this simplifies a lot); well, it can be useful in order to know that the order of 5 is a power of 2, but one can see that during the computation (I think that's just like Hensel lifting here). Moreover, in the video, 5^5 mod 16 could have been computed in a faster way with "fast exponentiation" (a.k.a. exponentiation by squaring). And no need for the costly system of linear congruences; the bad idea in the video was to "simplify" 5^5 by 21 instead of just keeping 5^5 for the conclusion like above.
The computation after 8:00 can be replaced by a much simpler one. Since N = 0 (mod 5^5), we can write N = (5^5)n for some integer n; inserting this in N = 5^5 (mod 2^5) we obtain (5^5)n = 5^5 (mod 2^5). Since gcd (5^5,2^5) = 1, we can divide both sides by 5^5, obtaining n = 1 (mod 2^5), or n = 1 + (2^5)m for some integer m. Therefore, N = (5^5)n = (5^5) [1 + (2^5)m] = 5^5 + (10^5)m = 5^5 (mod 10^5) = 3125 (mod 10^5).
If you start by listing 5^x (mod 100000), you'll find that it cycles every 8 times. Then you just need to deal with 5^5^5^5 (mod 8)
Yep. You can easily prove (e.g. by induction) that the cycle will continue, and 5^5^5^5 mod 8 is easily determined as well (those values just alternate between 1 and 5).
Whst if you dont think of that mod crap because most people dont use it?? And i mean most including very smart people
That would be great, but realistically how would you come across that observation? What would motivate one to check it?
@@angelmendez-rivera351 that's different though and much less explicit we sont use or think mod x or something like that..so my point still stands..so why dont you not talk out of your ass for once ..
@@tomatrix7525 5 is a special one to exponentiate. It already loops every 2 times for mod 1000
I have a somewhat easier solution:
I put the numbers 5^0, 5^1,5^2,5^3 ,... mod 100,000 in a row and what I realised is, that there is a cycle of length 8 starting at n^5:
3125 (n^5)
15625
78125
90625
53125
65625
28125
40625
3125 (n^13)
Thus I only need to know what 5^5^5^5 mod 8 is. But for 5^k there is a cycle of length 2 mod 8 , that is 5 mod 8 for odd k and 1 for even k. As 5^5^5 is clearly odd, we have 5^5^5^5 = 5 mod 8. And that tells us, that 5^5^5^5^5 = 3125 mod 100,000.
Exactly what I did sir. I was about to write it down 😅😅
me too
I did it this way too. I always try to use a kiss (keep it simple stupid) approach first and it worked here so why not?
you need to prove that. noticing patterns isn't a proof
@@tmpqtyutmpqty4733 patterns is also an absolutely acceptable proof, you just need to know the modulo concept.
@Michael Penn
@11:30, be careful. 0 . 293 . 2^5 is not a n y as you showed on the blackboard, but b m x
where b = 0, m = 2^5 and x = 293
6:22 Let’s maybe go ahead and do that
12:52 That’s a good place to stop
Homework. Number theory. Putnam 1963.
Find all integers n for which x^2 - x + n divides x^13 + x + 90.
I just realised that you have over 500 comments on this channel. Lol, if that's not dedication then I don't know what is.
@@at7388 yes
@@at7388 Nope
@@at7388 Do you really think Michael have time to waste on a meme account? He has better things to do 😂
My solution:
If n
12:52 And that's a good place to stop........Its really awesome.
N = any + bmx. Nice.
When you are getting ready to find the values of n at 6:33, I gasped. That's the same combination on my luggage! What a coincidence, that's amazing! I'd better change it.
Spaceballs taught us that the only combination you should use for luggage (Or the shield to stop your planet's atmosphere being stolen) is 12345.
Now you do, yes!
2:20 What?!?! You meant (5^5) ^ (5^5^5) because that is 5 raised to 5*(5^5^5) and that is different from N. You just made a common mistake when it comes to towers of powers, which is thinking that a^b^c^d is EQUAL to (a^b)^(c^d), which is not.
The tower is divisible by 5^5 because it can be written as 5^(5+n) with n = non-negative integer, and any number in this format has no remainder when divided by 5^5, and in fact the quotient is 5^n.
Typically, exponentiation is right-associative, e.g.
n^n^n = n^(n^n),
n^n^n^n = n^(n^(n^n)),
n^n^n^n^n = n^(n^(n^(n^n))), etc.
This is the best convention, since the left-associative interpretation can be reduced:
(n^n)^n = n^(n^2),
((n^n)^n)^n = (n^(n^2))^n = n^(n^3)), etc.
Look up "tetration" for more on these power towers!
@@samsonblack Yep. Think about it: a left assoc power towers would be pretty useless. Like you said, a left assoc exponentiation would give 5^5^5^5^5 = 5^(5^4) which has "only" 437 digits. The right assoc power tower 5^5^5 already has 2185 digits. Clearly, right assoc is the way to go if you want to reach those absurdly large numbers. :D
@@emanuellandeholm5657 I'm not sure there are only 10,000 digits in 5
5
@@jamirimaj6880 For sure not. 5^^4 already has many, many more digits than 10k. On the order of 10^1000 something digits, so clearly 5^^5 has even more than that. And by by m^^n I meant double Knuth's up-arrow notation, ie. tetration. 5^^5 = 5^(5^(5^(5^5)))
Quick solution. Observations: 5^5=03725 and the last five digits of 5^13 are also 03725. Therefore, the last five digits of 5^n is cyclical by the period of 8 when n>=5. Then question boils down to find the value of (5^5^5^5-5)(mod 8). Easy to see it is 0 using Euler's theorem. Hence, the last five digits of 5^5^5^5^5 is 03725.
I guess not but is there a general solution for this problem? The 10^n digit of N = n↑↑n?
Honestly, when I saw the title I was excited how Prof. Penn wanted to find the 10,000th digit of that number. And then I got a n interesting video how the find the fifth digit from the right. :-) Thank you for this interesting insight into number theory.
Maybe I missed that step, but why is the 10000th digit equal to the 5th digit from the right? (Edit: Ok, language barrier ^^ there is a difference between ten-thousands and ten-thousandth ...)
I had the same misunderstanding first and was pondering, how on earth one is supposed to conclude that 10,000th digit :-D
Same
Yeah, as a non native that sure could be confusing!
@@tomatrix7525 huh? What language difference? He said ten thousands and then ten thousands again..
It was indicated in the book that the 10000th digit is also equal from the 5th digit to the right. Maybe he should've started by saying how many digits the number has.
I was thinking a lot about this problem but I got nowhere. Turns out I misunderstood the problem. I thought we were looking for the Nth ordinal digit, N=10000, not the 10000 radix valued digit. There is a huge difference between the two... So glad I finally unpaused so I could discover my mistake.
what's 10000nds, coz I'm confused too?
@@jamirimaj6880 10000 valued digit is the fifth digit from the right. Sorry for being unclear.
Well. I shall patiently wait for the English translation someday
? The Wohascum County Problem Book is in English originally, is it not?
Please solve questions from the book
Challenge and Thrills of Pre College Mathematics
Ioqm ka test kaisa gaya
I solved this problem based on observation.
First and foremost, observe the first few powers of 5 mod 100000.
(5^1) mod 100000 = 5
(5^2) mod 100000 = 25
(5^3) mod 100000 = 125
(5^4) mod 100000 = 625
(5^5) mod 100000 = 03125
(5^6) mod 100000 = 15625
(5^7) mod 100000 = 78125
(5^8) mod 100000 = 90625
(5^9) mod 100000 = 53125
(5^10) mod 100000 = 65625
(5^11) mod 100000 = 28125
(5^12) mod 100000 = 40625
(5^13) mod 100000 = 03125
We can observe that the last 5 digits of the 5th power and 13th power are the same.
Hence, there will be a cycle of 8 where the last 5 digits of the powers remains the same.
Therefore, it is enough to determine what the power 5^5^5^5 evaluates to mod 8.
Assume that k=5^5^5
(5^5^5^5) mod 8 = (5^k) mod 8
= ((4+1)^k) mod 8
= (k*4 + 1) mod 8 {Using the binomial theorem}
We know that k is odd. Replacing k=2*m+1, we get
(5^5^5^5) mod 8 = ((2*m+1)*4 + 1) mod 8 = 5
Therefore the last 5 digits of 5^(5^5^5^5) are the same as that of 5^5, i.e., 03125.
Thank you, that is very great!!
Can you do the question 6, IMO 2020?
well, it's not that hard to prove the for any integer n where n>0, 5^n(mod 10) =5
Yeah just take Mod 2 and mod 5
I have a problem, mau you can help me to solve it?
If we have a triangle ABC, and point D inside the triangle, then can we find the maximum length of AD+BD+CD mathematically?
Thank you..
4:40 He wants to reduce a^b modulo c. Why can he do this by reducing b modulo phi(c) ? I don't see the connection to a^phi(n) congruent to1 modulo n.
Would be nice if someone can explain this to me.
OK, I think I got it: a^(phi(n)+d) = a^phi(n) * a^d = 1 * a^d = a^d (modulo n), therefore the result is not changed when reducing the exponent modulo phi(n). I think Michael should have explained this in more detail. Also, his reasoning at 2:30 is not really correct, but for sure the conclusion is.
But for sure still great videos where there is always a lot to learn from!
We could use the fact that the last 5 digits of 5^(k+8) are the same as of the number 5^k, because 5^(k+8)-5^k is divisible by 10^5 (for k>=5). And now we have to figure out what is 5^5^5^5 congruent to mod 8. It is true that 5 is congruent to 5^5 mod 8 thus 5^(5) is congruent to 5^(5^5) mod 8 and similarly 5^(5^(5)) is songruent to 5^(5^(5^5)) mod 8. This means that 5^5^5^5 is congruent to 5 mod 8. And thus the last five digits of number 5^5^5^5^5 are the same as of the number 5^(13)=1220703125. And from here we can see that the ten-thousands digit is 0.
I'm not sure if the last bit was a joke or not, but of course use 5^5 instead of 5^13 haha
Why 10^5 and not just 5^5?? And who the fuck would ever think of that anyway??
But presumably you need a computer to work out the ten-thousands digit of 5^13, in which case why not just solve the original problem directly on a computer? No offence intended, but seems a bit of a cheat in the context of a math competition!
@@AlephThree Thatas why 5^5 should be used instead, as it is pretty easy to calculate in a few ways mentally
@@minime1235able You are right, I have no idea why i used 5^13 instead of 5^5 maybe it was too late already for me.
So many of your videos, I can follow ALMOST to the end. So I guess past where I needed to stop. ;)
Sir can you make a concept aboit thue lemma in chapter 13 of elementary number theory...for msc maths
I have no idea where to start with problems like this. Your solution was well explained. But did you work it out yourself? How long did it take?
He is a professor so more like less than 10minutes i guess
@@imacup5047 Haha probably. I guess if you know the toolbox like he does, it's a great advantage.
Now find the last digit of G64 (Graham's Number)
I wondered of this too, but as you have a certain number of power, I think it would be difficult even for a computer using any existing function
@@nexio5454 We actually know the last 100 digits of that number, look it up on wikipedia.
Weirdly enough, mathematicians have already worked this out; we know it ends in a 7
It's not that hard to do. I have a program in python that calculates the last 500 digits of G64 in a couple of seconds.
I got the idea when I solved problem 188 at project euler.
@@chemicalbrother5743 Ok well, I didn’t know that! You know what tool was used for finding out those 100 digits?
10:50 It's not super hard, but.... *writes down solution which is impossible to guess*
It's "easy" with the extended euclidean algorithm
Hello professor michael , can you plz do some more AIME number theory / combinatorics problems ? they are really great problems and if u make a video about those , it will be the best thing ever :)
I started with a completely wrong interpretation of the problem. I thought it was 10,000th digit, versus the 10000's digit. Is there a reasonable way to approach this harder problem?
Amazing vid Prof
Find a and b both natural no such that
a^-1+b^-1=3(a+b)^-1+(ab)^-1
Very nice problem.
Recentemente conheci seu canal muito incrível
How did he know to look for the fifth digit from the right? How can you compute the number of digits for this Power tower?
Pardon; the first question is how to know that "the fifth digit from the right" is the same as "the ten-thousands digit"? I think you may need a refresher on place-value arithmetic....
@@la.zanmal. I think I need to work on my english, it's my second language. I translated it wrong and that's why I was confused ...
@@razvbir I made the same mistake :D 5^5^5 alone has 2185 digits so unless you find some tricks I doubt arbitrary digit extraction here is tractable.
@Paul O'Reilly Oh I see, I get it now. But beginners would be very confused that the 10,000 digit of the number is the 5th number to the right. Maybe he should've cleared this up at the beginning.
At the end we are essentially searching for a multiple of 5^5 that is congruent to 21 modulo 32. Since 5^5=(25)^2*5=(-7)^2*5=49*5=17*5=85=21(mod 32) we see that 5^5*k=21k(mod 32) and we are looking for k such that 21k=21(mod 32) which in this case is trivial but generally would require finding an inverse of 5^5 mod 32. In this case, instead of extended Euclid one can opt to use Euler's theorem - since 5^16=1(mod 13) we see that 5^5*5^11=1(mod 13) i.e the inverse is 5^11=(5^5)^2*5=21^2*5=441*5=25*5=125=-3(mod 32) - the fact that this inverse is just 5^3 stems from the fact that not only 5^16=1(mod 32) but 5^8 as well.
Is this Calculus or something?
@Paul O'Reilly That's a really challenging question.
7:13 HE DID IT
The book has pdf in an easy google search, but with the solutions removed 😞
Edit: Found a complete copy, djvu then converted it into pdf hahaha 👍
Is there any chance to find how many digits a number like this N might have? It must be huuuuge...
5^5^5 is already 10^2184. That means that the number of digits of the number 5^5^5^5 is around 10^1526. And that means, that the number of digits, for the number describing the number of digits of 5^5^5^5^5 is around 10^1067.
Numbers that big is not comprehensible, at least not for me.
@@Evilhippie64 Thanks.
Be sure, numbers that big are not just for you not comprehensible.
@@Evilhippie64 It's insane. Even the n-1:th iterate of the logarithm of (n↑↑n) grows superlinearly with n. That is log(2↑↑2), log(log(3↑↑3)), log(log(log(4↑↑4)))), log(log(log(log(5↑↑5)))), ... (unless I made a mistake). I wonder what an asymptotic of this series would be.
I love the line 'generalisation of Fermat's little theorem' 😇
0 is such a satisfying answer 🤣🤣🤣
My word....
Any bmx 👍🏽
Hey, Michael! What I found out is that N is the fifth tetration of 5.
okay
It is like listening to an ancient Persian of 500 B.C. It sounds nice but is incomprehensible.
I misunderstood the problem and was trying to find the digit at the 10th thousand position, not the 5th position. Needless to say, I couldn't 🙃
Try explaining isi entrance exam questions mainly subjective
It's the exam given by indian students who r 17 to 18 to enter indian statistical institute which we have only 50 seats in entire india
In other words, the problem has "zero" solutions. :)
Anyone plzz give me concept about thue lemma
But i need some examples about it........
me: isn't 5^5^5^... divergent? how is this even possible?
*2 hours later*
Ha this guy's doing all this modulo stuff when I just used wolfram alpha, and it took like 3 seconds this guy isn't very fast at maths. /s
Lol
please turkish subtitle
N is any BMX
8:02 some kids enjoying the maths in the background?
Who got up with a calculator and ended up with a 🤨
Am I the only one confounded when he said "fifth digit from the right" thinking that he knows that there would be 10004 digits in the exact answer of the power of tower? But okay, he meant ten-thousand, the place value, not ten-thousandth, the position
5^5^5^5^5 ends in 03125
Nice good
Here is a problem I couldnt solve, someone interested??
729 , 71289 , 7112889 ,711128889 , 71111288889 ... ... ...
Prove that each term of the sequence will always be a perfect square.
PS I am a high school student.
🔥🔥🔥
hahaha, this is interesting!
Too many ads, skipping...
Sir, Wolfram just said that N has 10^10^2184 DIGITS, not just 10,000 lol
Again, the solution in the video is much too complex. Simply: 5^8 ≡ 1 mod 32 and 5^2 ≡ 1 mod 8, so that 5^(2k+1) ≡ 5 mod 8 and 5^5^5^5^5 ≡ 5^5 ≡ 3125 mod 100000.
I think a mistake some mathematicians make is to assume shorter solutions are better (or less complex) than longer solutions. If the longer solution contains individual steps that are easier to follow, or ideas and process that can be applied to solve other problems then it is arguably better. In this example I have no idea why one would start by saying “simply 5^8=1 (mod 32)”, but then I’m no doubt an inferior mathematician to you!
@@AlephThree I didn't include the classical steps. But here are details and some general explanations after the answer. One seeks for N modulo 100000. Since 100000 = 2^5 × 5^5, we just need to consider N modulo 2^5 and modulo 5^5 (due to the Chinese remainder theorem). The modulo 5^5 will be easy, so let's focus on N modulo 2^5, i.e. N modulo 32. [Not much difference with the video until here.] Then... One has 5^8 ≡ 1 mod 32, so we just need to find 5^5^5^5 mod 8. And since 5^2 ≡ 1 mod 8 and 5^5^5 is an odd integer, 5^5^5^5 ≡ 5 mod 8. Hence N ≡ 5^5 mod 2^5. And N ≡ 0 ≡ 5^5 mod 5^5. Therefore, N ≡ 5^5 ≡ 03125 mod 100000.
No need for Euler's theorem (in particular in simple cases like here, one can often do better, and this simplifies a lot); well, it can be useful in order to know that the order of 5 is a power of 2, but one can see that during the computation (I think that's just like Hensel lifting here). Moreover, in the video, 5^5 mod 16 could have been computed in a faster way with "fast exponentiation" (a.k.a. exponentiation by squaring). And no need for the costly system of linear congruences; the bad idea in the video was to "simplify" 5^5 by 21 instead of just keeping 5^5 for the conclusion like above.
@@vinc17fr thanks for sharing - much appreciated
What about the 9999th decimal?