A little brute force but with a calculator : 3^7 = -13 mod 100 3^(7^2) = (-13)^7 = -17 mod 100 3^(7^3) = (-17)^7 = 27 mod 100 3^(7^4) = 27^7 = 3 mod 100 so the remainders repeat over a cycle of 4 terms, with -13-17+27+3 = 0 which is not to diminish Michael's proof using phi. Consider this a confirmation via a different path.
From Michael's binomial expansion 13:16, (3^4)^5 = 1 mod 100 3^(49) = ((3^4)^5)^10 x 3^9 = 3^9 mod 100 = (-13 x 9) mod 100 = -17 mod 100 3^(343) = ((3^4)^5)^85 x 3^3 = 3^3 mod 100 = 27 mod 100 3^(2401) = ((3^4)^5)^600 x 3 = 3 mod 100 No need for a calculator :)
@@ScarredFrost Yes, this is doable without calculator, thank you! (Just two small fixes: 3^(49) = ((3^4)^5)^10 x 3^9 should be 3^(49) = ((3^4)^5)^2 x 3^9 and 3^(343) = ((3^4)^5)^85 x 3^3 should be 3^(343) = ((3^4)^5)^17 x 3^3)
You can skip the calculations at the end once you have that N is congruent to 5*(3^7+3^9+3^23+3) mod 100 because by inspection it is 0 mod 5 and in mod 4 it is 1*(-1+-1+-1+-1) since all the powers of 3 are odd, so -4 mod 4 = 0 mod 4.
@@NoNameAtAll2 I just rewatched the video and I'm not sure where I was going when I wrote that comment. Fortunately we just want to check 3+3^7+3^9+3^23 mod 5. We know from the discussed theorem that 3^4 = 1 mod 5, so this boils down to 3+3^3+3+3^3 = 60 == 0 mod 5. I hope that helps.
There is another shortcut that can be taken in here. If x=20q+r, then 5x=100q+5r. Therefore, you only need to calculate 3^7+3^9+3^23+3^1 mod 20. Here, you can use Euler's Theorem again and find that you only need the exponents mod 4. The result is then This is equivalent to 2(3^1+3^3) which is congruent to 2(3+7) which is congruent to 0 (mod 20). Multiply by 5, and the result is 0 (mod 100).
Since we have a 5, we can say N/5 = (3^7 + 3^9 + 3^23 + 3) (mod 20), giving us phi(20) = 8, allowing us to write this as 3^7 + 3 + 3^7 + 3 (mod 20) Since 3(3^7) = 1, 3^7 must be 7 mod 20, making the sum 0, and that's a good place to stop.
I find it amazing that these problem creators can come up with problems like this. I just don't see how they back into a problem like this where the huge sum ends up being divisible by 100 and at the same time involves a large exponent equal to the current year.
11:33 As 20*5=100 Only need to consider mod20 And 3^4=81=4*20+1 (3^7+3^9+3^23+3) =3(3^6+3^8+3^22+1) Mod20 obtain 3(3^2+1+3^2+1)=20*3=60 Mod 20 obtain 0
You need to use comprime mods so you will finish with CRT. If you get that n is 0 mod a and 0 mod b you are just proving that n es divisble by thr mcm of a and b
I started this a little differently just by looking at powers of 3 mod 100 and noticing that 3^20 is 1 mod 100 (didn't work that out directly, but I have 3^10 memorized and know it ends in 49, and 49 is a square root of 1 mod 100). Rest goes the same way.
11:21 Good place to start 16:11 Good place to stop Loading 94.6%... Hello Michael, hello world! I have you’re having a great day! Don’t forget to take care of your mental health. Homework... For which real α does the curve y = x^4 + 9x^3 + α x^2 + 9x + 4 contain four collinear points? (Putnam 1994 - Problem B2)
I like that you name the theorems that you’re using. So if I don’t know them I can google them and then attempt the question P.s. feel free to link any videos you have done on any of the theorems you name in the description.
I miss those exercises that ask you to find the last numbers of something. Even though you can usually solve it using mod 10^x ( for some natural x) , it stills funny
From an engineer (not a mathematician) step 1: 3^7 looking at only the last two is 87 (27*27*3) step 2: 3^7^2 = (3^7)^7 is like 87^7 looking only at the last 2 digits is 83 step 3: 3^7^3 = (3^7^2)^7 is like 83^7 looking only at the last 2 digits, which is 27 step 4: 3^7^4 = (3^7^3)^7 is like 27^7 looking only at the last 2 digits, which is 3, step 5 is the same as step 1 Step 1 to 4 (87+83+27+3) add up to x00 and repeat 505 (2020/4) times, so the last 3 digits are either 500 or 000.
Thanks Michael, another one I get because your steps are so clear. But is anyone else slightly troubled by the notation? Ordinarily it's not a problem for simple modular stuff, but using the same brackets (or braces or whatever you call them) for the powers of three and for the mod 100 to me is a bit ambiguous. I would have preferred to see different brackets around the mod 100, eg N=5(x+y+z){mod 100}.... or am I just Mr Picky?
Indeed if (a,100)=1 then a^20 is congruent to 1 mod 100 because 100=25*4, phi(25)=20 and phi(4)=2 then a^20 congruent to 1 mod 25, a^20=(a^2)10 congruent 1 mod 4. Therefore a^20 is congruent to 1 mod 100. Thus 3^23=3^20*3^3 congruent to 3^3=27 mod 100
13 th min : 3²³= 3²⁰ . 3³ (mod100) =(3⁵)⁴ . 3³ (mod100) =(43)⁴ . 3³ (mod100) =49². 3³ (mod100) =27 (mod100) This is easier than the way you treated it , since you've already calculated 3⁵
Challenge accepted! Use the Chinese Remainder Theorem to break the problem down into mod 8 and mod 125 to simplify the calculations. Since 3^2=1 (mod 8), it is quick to check that the entire sum is 4 (mod 8). Then use the facts that phi(125)=100 and the order of 7 mod 100 is 4, so every consecutive four terms have the same sum, which turns out to be 25, mod 125 (after a bunch of calculations by hand of powers of 3 mod 125). This implies that the entire sum is 0 (mod 125). Combining this with 4 (mod 8), we see that the whole sum is 500 (mod 1000).
@@riseciv7991 There is a relatively straightforward way to compute it using a formula from the Chinese Remainder Theorem, though it involves some modular arithmetic calculations. But in this particular example it is not too hard to figure it out directly: a number that is 0 mod 125 is a multiple of 125, so check 125, 250, 375, 500, etc. until you find the one that is 4 mod 8.
@@Notthatkindofdr The same number of digits as 3^7^2020, which is 1 + floor(log(3^7^2020)/log(10)), which is 597956502043767935135303320371262426014703635534155055142830258962780377290231766743500776054719394697087280766908482441274693523774407240513154475027306453642191680106730603124206675935327886069196832301974142608692202917564595561348305529405911352766324939445473098056728445509890990321479085236860102954866236825955515330882703149228480546160534839141354053103635812452944613596908024541713941342935811294399081501790307736513520570179117477440487705264618917983946166343228550787869625548275285778499834976864568423051259258114913483898293600963231089599470049905424724782678930178666111390775247163300079655601712026033441228932066650453147031817117872436167180890618402130716369818237715869561740984581132707906821908300326411194798955220926518902537533513075424542281067149826767664683370397865726424556106663715219483347560326575032884724955821624609207563025236406912675236657969941538578845681206893726601142636770579819368831957807114270908870896876232015092935508963563793248482471086216670752176695475885386471643640010490194150313723844630837902389323994910169030239809210243609658757001356971808482066692380846573616739198798773843743317269825149383368887937546937113424083407921366892569012420784559213783797774935886467262423808789639500249556622626759475356492829998029753370170387882678039038236889467817464337809575266286277698333584329197947713715705633777543434483781701762835412380122743334048622873443743174292261662079274561098376708024441011604561341407670431551434438228758915434921951731906821850410673072848244805717969876807206352094955199351420282560256297899162878851781586843696563023858538947869818638345063137419407629875072711962323802850151574052451413912441356777169017.
Man wowwww this type of question is my best beacause it show you how logic of algebra and arithmetic Reduce a huge problem to get the answer when most of calculator Can t do it I still weak in this type of problem but every video you upload make me better thank u
Let me explain. If N is the number we are studying, we can easily show that N == 0 mod 4 N == 0 mod 25 For the second one, prove that 5 divides the sum of any two consecutive terms.
Once you had 505 times the sum of four numbers, you could have divided by 5 and said 5X mod 100= X mod 20. That is much easier to calculate because 3^4=1 mod 20.
I just love how you keep saying "Fermat's little theorem" - I mean, there's probably a good reason (like Fermat has a "big theorem" or something) but it's still a very cute name for such a sharp and powerful theorem hahaha!
The last term in the sum is far larger than all the others, so the first digits are just determined by 3^(7^2020). Inspired by a calculation by @OscarCunningham in answer to another question in these comments, we can use software to calculate 7^2020*log(3) and remove the integer part, which tells us that 3^(7^2020) = 10^(0.120362185117...) times a huge power of 10, so calculating this we find the first several digits are 1319356573....
I'm not super confident with this kind of problems, but is there a reason behind the nice fact that the sum of 7^1, 7^2, 7^3, 7^4 mod 40 is 40? Or it's just coincidental?
It's because 7^4 = 1 (mod 40). From group theory that means 7 is a fourth root of 1 mod 40 and the sum of all roots gives 0. If we use algebra and the geometric sum formula instead, 7^1 + 7^2 + 7^3 + 7^4 = 7^0 + 7^1 + 7^2 + 7^3 (since 7^4 = 1 = 7^0 (mod 40)) = (7^4 - 1)/(7 - 1) (by the geometric sum formula) = (7^0 - 1)/(7 - 1) (7^4 = 1 = 7^0 (mod 40) again) = 0/6 = 0 = 40 (mod 40)
Tell me what is the largest known full reptend prime ? Why 7 is the only known mersenne prime that is also a full reptend prime ? (M_57885161 , M_74207281 , M_77232917 and M_82589933 are not full reptend)
This could have been simplified by noting that 505 and 100 are both divisible by 5. Which would mean we could solve this mod 20, vastly simplifying the calculations.
We take Z/100Z~Z/25Z+Z/4Z then (Z/100)*~Z/20Z+Z/2Z therefore if (a,100)=1 then a^20 congruent to 1 mod 100. (The generalization of Fermat theorem with the Chinese theorem of residue.) And if n>=1 then (Z/(2^(n+2))Z)*~Z/(2^n)Z+Z/2Z then if a is odd then a^(2^n) congruent to 1 mod 2^(n+2) indeed if n>1 then 5^(2^(n-1)) is not congruent to 1 mod 2^(n+2). if p is odd prime and n>0 then (Z/p^(n+1))*~Z/((p-1)p^n)Z
We are given 100 non-collinear points on a plane. Prove that the amount of acute triangles don't exceed 70%. A hint is to consider groups of five points first
3^7=2187. Sum n=1 to 2020 of (2187^n). 2187*(1-(2187^2019)/1-2187) = A 6,477 digit number. 1.4285372797592599083258606177361e+6743 what your calculator gives out but does not show the last 2 digits. We have to use the proof in the video to get the last 2 digits.
I don't know about that complicated Euler and phi stuff; I just calculated the first 4 terms mod100. 3^7=2187, and then the second term is 87^7 which is 83. The third term is 83^7 which is 27, and the fourth term is 27^7 which takes us back to 03. The first 4 terms add up to 200, and there is a multiple of 4 terms in the sum, so it is just 00. I did the whole thing in about 2 minutes without using any more complicated stuff.
We want to know N mod 100 Using the formula for the sum of the terms of a GP, N can be written as 3^7*((3^7)^2020-1)/(3^7-1) Inspecting 3^7 mod 100 = 2187 mod 100 = -13 mod 100 Inspecting powers of 13: 13 = 13 mod 100 13^2 = 169 mod100 = 69 mod 100 13^3 = 2197 mod 100 = -3 mod 100 Now we can manipulate 3^7*((3^7)^2020-1)/(3^7-1) mod 100 replacing 3^7 mod 100 by (-13) mod 100 and 13^3 mod 100 by (-3) mod 100. In the end, we obtain: N = 13 * (3^6*13^2-1)/(14) mod 100 Then we factor (3^6*13^2-1) as (3^3*13-1)*(3^3*13+1), obtianing N = 13 * (3^3*13-1)*(3^3*13+1)/(14) mod 100 Computing 3^3*13 = 27*13 = 351, which implies: N = 13 * (350)*(352)/(14) mod 100 N = 13 * (7*50)*(2*176)/(7*2) mod 100 N = 13 *50*(176) mod 100 N = 13 *50*(2*98) mod 100 N = 13 *100*98 mod 100 Since N mod 100 is a multiple of 100, the rest is 0. The answer is: the 2 last digits of N are 00.
Dr penn , since you love floor functions so much , here's another floor function problem from the Indian national maths olympiad 2014 problem 2 Let n be a natural (positive integer) number , prove that floor(n/1) + floor(n/2) + floor(n/3) + floor(n/4) + .... + Floor (n/n) + floor(√n) is even . I'm suggesting INMO problems for such a long time , please check them out , it's a humble request , thank you.
Με αυτό που διευκρινιζεις στην αρχή τι εννοείς ως προς τους εκθέτες δεν το συμβολίζεις σωστά, είσαι λάθος , κλπ κλπ το καταλαβαίνεις φυσικά γιναυτο κάνεις την διευκρίνιση
0:28 Aaaaand this is where you lost most people, hahaha. Still looks like a lot of fun, so I'll keep watching :-) EDIT: Looks like a sick powerful theorem though, I might want to check your proof just for fun too! EDIT2: Turns out it wasn't that hard to follow! Thanks for the great video!
I just went with the series of 3^(7^n)) mod 100 using the fact that 3^(7^(n+1)) = (3^(7^n))^7 3^7 ≡ 87 (mod 100) 3^(7^2) ≡ 87^7 ≡ 83 (mod 100) 3^(7^3) ≡ 83^7 ≡ 27 (mod 100) 3^(7^4) ≡ 27^7 ≡... 3 (mod 100) Then it loops around since 3^(7^5) ≡ 3^7 (mod 100), so we can rewrite the sum a bit: sum(n = 1; 2020; 3^(7^n)) = sum(n = 0; 504; 3^(7^(4n+1)) + 3^(7^(4n+2)) + 3^(7^(4n+3)) + 3^(7^(4n+4))) ≡ sum(n = 0; 504; 87 + 83 + 27 + 3) ≡ sum(n = 0; 504; 0) ≡ 0 (mod 100)
11:20
Michael: That's a good place to ...
Me: No it's not!
Michael: ...start
Me: Oh, I see
I felt like you :D
@@mohammadsalehi765 Me too)
@@lyudvigfeyerbax8659 :D
11:20
16:11
This is wonderful
thank you good sir
A little brute force but with a calculator :
3^7 = -13 mod 100
3^(7^2) = (-13)^7 = -17 mod 100
3^(7^3) = (-17)^7 = 27 mod 100
3^(7^4) = 27^7 = 3 mod 100
so the remainders repeat over a cycle of 4 terms, with -13-17+27+3 = 0
which is not to diminish Michael's proof using phi. Consider this a confirmation via a different path.
From Michael's binomial expansion 13:16, (3^4)^5 = 1 mod 100
3^(49) = ((3^4)^5)^10 x 3^9 = 3^9 mod 100 = (-13 x 9) mod 100 = -17 mod 100
3^(343) = ((3^4)^5)^85 x 3^3 = 3^3 mod 100 = 27 mod 100
3^(2401) = ((3^4)^5)^600 x 3 = 3 mod 100
No need for a calculator :)
@@ScarredFrost Yes, this is doable without calculator, thank you!
(Just two small fixes: 3^(49) = ((3^4)^5)^10 x 3^9 should be 3^(49) = ((3^4)^5)^2 x 3^9
and
3^(343) = ((3^4)^5)^85 x 3^3 should be 3^(343) = ((3^4)^5)^17 x 3^3)
@@AndrejPanjkov Ah yes! Thanks for fixing my typo
You can skip the calculations at the end once you have that N is congruent to 5*(3^7+3^9+3^23+3) mod 100 because by inspection it is 0 mod 5 and in mod 4 it is 1*(-1+-1+-1+-1) since all the powers of 3 are odd, so -4 mod 4 = 0 mod 4.
why mod 4 and not mod 20?
@@NoNameAtAll2 if something is 0 mod 5 and 0 mod 4, then it's 0 mod 20. I hope that helps.
@@fmakofmako yeah, but if x = 0 mod 4, then 5x = 0 mod 20, but not always =0 mod 100
@@NoNameAtAll2 I just rewatched the video and I'm not sure where I was going when I wrote that comment. Fortunately we just want to check 3+3^7+3^9+3^23 mod 5. We know from the discussed theorem that 3^4 = 1 mod 5, so this boils down to 3+3^3+3+3^3 = 60 == 0 mod 5. I hope that helps.
I appreciate that you say φ like phi and not phaj
There is another shortcut that can be taken in here. If x=20q+r, then 5x=100q+5r. Therefore, you only need to calculate 3^7+3^9+3^23+3^1 mod 20. Here, you can use Euler's Theorem again and find that you only need the exponents mod 4. The result is then This is equivalent to 2(3^1+3^3) which is congruent to 2(3+7) which is congruent to 0 (mod 20). Multiply by 5, and the result is 0 (mod 100).
That's a nice observation! The math would be relatively quite simple.
Since we have a 5, we can say N/5 = (3^7 + 3^9 + 3^23 + 3) (mod 20), giving us phi(20) = 8, allowing us to write this as 3^7 + 3 + 3^7 + 3 (mod 20)
Since 3(3^7) = 1, 3^7 must be 7 mod 20, making the sum 0, and that's a good place to stop.
Amazingly clear explanation! Thank you for this.
Nice haircut
He releases videos so often you can pretty much pin down the day he sees the barber.
@@Tiqerboy exactly
i don't think the barber did a very good job, it's patchy lol.
I find it amazing that these problem creators can come up with problems like this. I just don't see how they back into a problem like this where the huge sum ends up being divisible by 100 and at the same time involves a large exponent equal to the current year.
11:33
As 20*5=100
Only need to consider mod20
And 3^4=81=4*20+1
(3^7+3^9+3^23+3)
=3(3^6+3^8+3^22+1)
Mod20 obtain
3(3^2+1+3^2+1)=20*3=60
Mod 20 obtain 0
That ia wrong statement. For example 170(mod100)=70 but 170(mod20)=10
170=5×34
You need to use comprime mods so you will finish with CRT. If you get that n is 0 mod a and 0 mod b you are just proving that n es divisble by thr mcm of a and b
The another tip would be that if you are using non coprime residues then just use the generalization of CRT
@@Accusan
am( mod mn) = mb
mean am=Kmn+mb=m(Kn+b)
so that
a(mod n)=b
am (mod mn)=mb
these two are equivalent
I just forget to multiple both side by 5
At 16:14, get 7^4 (mod 40) by squaring (7^2 (mod 40) = 9 (mod 40)). 9^2=81=1(mod40).
Please upload your research papers and explanation as your 100 k special.whoever agrees hit a like
Lol he can’t do that
F
why do some arses always want more?
I started this a little differently just by looking at powers of 3 mod 100 and noticing that 3^20 is 1 mod 100 (didn't work that out directly, but I have 3^10 memorized and know it ends in 49, and 49 is a square root of 1 mod 100).
Rest goes the same way.
11:21 Good place to start
16:11 Good place to stop
Loading 94.6%... Hello Michael, hello world! I have you’re having a great day! Don’t forget to take care of your mental health. Homework...
For which real α does the curve y = x^4 + 9x^3 + α x^2 + 9x + 4 contain four collinear points? (Putnam 1994 - Problem B2)
Umm where do you get these problems
It's now 94.7% :D
Would it work finding those alpha which make the first derivative have three zeros and the sencond derivative two?
I like that you name the theorems that you’re using. So if I don’t know them I can google them and then attempt the question
P.s. feel free to link any videos you have done on any of the theorems you name in the description.
find rational number a,b,c such that (2^(1/3)-1)^(1/3)=a^(1/3)+b^(1/3)+c^(1/3)
@@angelmendez-rivera351 thanks
I miss those exercises that ask you to find the last numbers of something. Even though you can usually solve it using mod 10^x ( for some natural x) , it stills funny
I'm going back to 9:49
If it's a 7-hour flight or a 45-minute drive
fantastic explanation
Never learned about this tool in my number theory class. Very cool
thanks for the tips and short example, was enough to let me solve the rest by myself which is a lot more fun
I am pleased that I got all of that on my own except the 3^23 bit and on.
It's very nice work you went through super details of explication
From an engineer (not a mathematician)
step 1: 3^7 looking at only the last two is 87 (27*27*3)
step 2: 3^7^2 = (3^7)^7 is like 87^7 looking only at the last 2 digits is 83
step 3: 3^7^3 = (3^7^2)^7 is like 83^7 looking only at the last 2 digits, which is 27
step 4: 3^7^4 = (3^7^3)^7 is like 27^7 looking only at the last 2 digits, which is 3,
step 5 is the same as step 1
Step 1 to 4 (87+83+27+3) add up to x00 and repeat 505 (2020/4) times, so the last 3 digits are either 500 or 000.
Really nice review of this thm. Good stuff
That was satisfying
Thanks Michael, another one I get because your steps are so clear. But is anyone else slightly troubled by the notation? Ordinarily it's not a problem for simple modular stuff, but using the same brackets (or braces or whatever you call them) for the powers of three and for the mod 100 to me is a bit ambiguous. I would have preferred to see different brackets around the mod 100, eg N=5(x+y+z){mod 100}.... or am I just Mr Picky?
It's just standard convention, though I won't try to argue that it's a good convention.
When you were working on the 3^23 mod 100 section, I realized the number was a multiple of 100 and went "Yooooou bastard... " at the problem. XD
Indeed if (a,100)=1 then a^20 is congruent to 1 mod 100 because 100=25*4, phi(25)=20 and phi(4)=2 then a^20 congruent to 1 mod 25, a^20=(a^2)10 congruent 1 mod 4. Therefore a^20 is congruent to 1 mod 100. Thus 3^23=3^20*3^3 congruent to 3^3=27 mod 100
Incredible resolution!
13 th min : 3²³= 3²⁰ . 3³ (mod100)
=(3⁵)⁴ . 3³ (mod100)
=(43)⁴ . 3³ (mod100)
=49². 3³ (mod100)
=27 (mod100)
This is easier than the way you treated it , since you've already calculated 3⁵
Homework: Find the last three digits!
Challenge accepted! Use the Chinese Remainder Theorem to break the problem down into mod 8 and mod 125 to simplify the calculations. Since 3^2=1 (mod 8), it is quick to check that the entire sum is 4 (mod 8). Then use the facts that phi(125)=100 and the order of 7 mod 100 is 4, so every consecutive four terms have the same sum, which turns out to be 25, mod 125 (after a bunch of calculations by hand of powers of 3 mod 125). This implies that the entire sum is 0 (mod 125). Combining this with 4 (mod 8), we see that the whole sum is 500 (mod 1000).
Now for another homework question: how many digits does the number of digits of this sum have?
@@Notthatkindofdr the process of combining 4 mod 8 and 0 mod 125 into 500 mode 1000 is relatively easy? Conceptually or computationally?
@@riseciv7991 There is a relatively straightforward way to compute it using a formula from the Chinese Remainder Theorem, though it involves some modular arithmetic calculations. But in this particular example it is not too hard to figure it out directly: a number that is 0 mod 125 is a multiple of 125, so check 125, 250, 375, 500, etc. until you find the one that is 4 mod 8.
@@Notthatkindofdr The same number of digits as 3^7^2020, which is 1 + floor(log(3^7^2020)/log(10)), which is 597956502043767935135303320371262426014703635534155055142830258962780377290231766743500776054719394697087280766908482441274693523774407240513154475027306453642191680106730603124206675935327886069196832301974142608692202917564595561348305529405911352766324939445473098056728445509890990321479085236860102954866236825955515330882703149228480546160534839141354053103635812452944613596908024541713941342935811294399081501790307736513520570179117477440487705264618917983946166343228550787869625548275285778499834976864568423051259258114913483898293600963231089599470049905424724782678930178666111390775247163300079655601712026033441228932066650453147031817117872436167180890618402130716369818237715869561740984581132707906821908300326411194798955220926518902537533513075424542281067149826767664683370397865726424556106663715219483347560326575032884724955821624609207563025236406912675236657969941538578845681206893726601142636770579819368831957807114270908870896876232015092935508963563793248482471086216670752176695475885386471643640010490194150313723844630837902389323994910169030239809210243609658757001356971808482066692380846573616739198798773843743317269825149383368887937546937113424083407921366892569012420784559213783797774935886467262423808789639500249556622626759475356492829998029753370170387882678039038236889467817464337809575266286277698333584329197947713715705633777543434483781701762835412380122743334048622873443743174292261662079274561098376708024441011604561341407670431551434438228758915434921951731906821850410673072848244805717969876807206352094955199351420282560256297899162878851781586843696563023858538947869818638345063137419407629875072711962323802850151574052451413912441356777169017.
Man wowwww this type of question is my best beacause it show you how logic of algebra and arithmetic
Reduce a huge problem to get the answer when most of calculator
Can t do it
I still weak in this type of problem but every video you upload make me better thank u
Why did tou choose 5 raised 575 mod 8..isnit random or some connection with the problem?
well the 5^575 (mod 8) shouldn’t be that complicated. But it does it job well for the demonstration part.
I don’t known why you don’t like to use the Chinese Remainder Theorem.
Gougu theorem when
Let me explain. If N is the number we are studying, we can easily show that
N == 0 mod 4
N == 0 mod 25
For the second one, prove that 5 divides the sum of any two consecutive terms.
I am really glad to see such people on youtube :)
5^2 is 1 mod 8,5^574 is 1 mod 8 so 5^575 is 5 mod 8
Once you had 505 times the sum of four numbers, you could have divided by 5 and said 5X mod 100= X mod 20. That is much easier to calculate because 3^4=1 mod 20.
And also, congratulations , a few more subscribers and you will reach 100k!
last 3 digit: 500
c++ help calculate last 6 digit: 768500
Very good!
I just love how you keep saying "Fermat's little theorem" - I mean, there's probably a good reason (like Fermat has a "big theorem" or something) but it's still a very cute name for such a sharp and powerful theorem hahaha!
it's to distinguish it (FlT) from fermat's last theorem (FLT)
@@demenion3521 I see, thanks for the info!
@@demenion3521 👍
Incidentally his little theorem is much more useful.
Little Fermat's Theorem is quite useful!
So, what I have found out is:
I really need to work on my Number Theory proficiency.
Same here.
What about the FIRST two digits?
That is left as a homework assignment to the viewers.
The last term in the sum is far larger than all the others, so the first digits are just determined by 3^(7^2020). Inspired by a calculation by @OscarCunningham in answer to another question in these comments, we can use software to calculate 7^2020*log(3) and remove the integer part, which tells us that 3^(7^2020) = 10^(0.120362185117...) times a huge power of 10, so calculating this we find the first several digits are 1319356573....
@@haleshs66 Ah, I see you found a brilliant way to find them, but no room to write the solution in the margin ?
finaly a good place to start
Wow! Fantastic!
hmm. The given tool doesn't use mod of Phi, but the solution does.
I'm not super confident with this kind of problems, but is there a reason behind the nice fact that the sum of 7^1, 7^2, 7^3, 7^4 mod 40 is 40? Or it's just coincidental?
It's because 7^4 = 1 (mod 40). From group theory that means 7 is a fourth root of 1 mod 40 and the sum of all roots gives 0.
If we use algebra and the geometric sum formula instead,
7^1 + 7^2 + 7^3 + 7^4
= 7^0 + 7^1 + 7^2 + 7^3 (since 7^4 = 1 = 7^0 (mod 40))
= (7^4 - 1)/(7 - 1) (by the geometric sum formula)
= (7^0 - 1)/(7 - 1) (7^4 = 1 = 7^0 (mod 40) again)
= 0/6 = 0 = 40 (mod 40)
@@toriknorth3324 Thank you!
Tell me what is the largest known full reptend prime ? Why 7 is the only known mersenne prime that is also a full reptend prime ? (M_57885161 , M_74207281 , M_77232917 and M_82589933 are not full reptend)
This could have been simplified by noting that 505 and 100 are both divisible by 5. Which would mean we could solve this mod 20, vastly simplifying the calculations.
When will you bring the backflip back? It’s been a while.
phenomenal
All those 3s and 7s and the answer is a multiple of 100!
We take Z/100Z~Z/25Z+Z/4Z then (Z/100)*~Z/20Z+Z/2Z therefore if (a,100)=1 then a^20 congruent to 1 mod 100. (The generalization of Fermat theorem with the Chinese theorem of residue.) And if n>=1 then (Z/(2^(n+2))Z)*~Z/(2^n)Z+Z/2Z then if a is odd then a^(2^n) congruent to 1 mod 2^(n+2) indeed if n>1 then 5^(2^(n-1)) is not congruent to 1 mod 2^(n+2). if p is odd prime and n>0 then (Z/p^(n+1))*~Z/((p-1)p^n)Z
We are given 100 non-collinear points on a plane. Prove that the amount of acute triangles don't exceed 70%.
A hint is to consider groups of five points first
What if all 100 points are on a common semicircle, then every possible triangle of three points would be obtuse.
@@bsmith6276 sorry, I meant acute...
3^7=2187. Sum n=1 to 2020 of (2187^n). 2187*(1-(2187^2019)/1-2187) = A 6,477 digit number. 1.4285372797592599083258606177361e+6743 what your calculator gives out but does not show the last 2 digits. We have to use the proof in the video to get the last 2 digits.
(Spoiler alert)
Woww that entire sum is divisible by 100 🤯
I think it would be simpler to just notice it's divisible by both 25 and 4
I don't know about that complicated Euler and phi stuff; I just calculated the first 4 terms mod100. 3^7=2187, and then the second term is 87^7 which is 83. The third term is 83^7 which is 27, and the fourth term is 27^7 which takes us back to 03. The first 4 terms add up to 200, and there is a multiple of 4 terms in the sum, so it is just 00. I did the whole thing in about 2 minutes without using any more complicated stuff.
I saw the qn and I was like " aaahh it must use the Euler totient " I still wonder should I use the uppercase Phi or the lowercase one?
Lawl
Abeee inmo problems comment kar
Suggest MP sir
@@angelmendez-rivera351 thanks for the clarification
We want to know N mod 100
Using the formula for the sum of the terms of a GP, N can be written as 3^7*((3^7)^2020-1)/(3^7-1)
Inspecting 3^7 mod 100 = 2187 mod 100 = -13 mod 100
Inspecting powers of 13:
13 = 13 mod 100
13^2 = 169 mod100 = 69 mod 100
13^3 = 2197 mod 100 = -3 mod 100
Now we can manipulate 3^7*((3^7)^2020-1)/(3^7-1) mod 100 replacing 3^7 mod 100 by (-13) mod 100 and 13^3 mod 100 by (-3) mod 100.
In the end, we obtain:
N = 13 * (3^6*13^2-1)/(14) mod 100
Then we factor (3^6*13^2-1) as (3^3*13-1)*(3^3*13+1), obtianing N = 13 * (3^3*13-1)*(3^3*13+1)/(14) mod 100
Computing 3^3*13 = 27*13 = 351, which implies:
N = 13 * (350)*(352)/(14) mod 100
N = 13 * (7*50)*(2*176)/(7*2) mod 100
N = 13 *50*(176) mod 100
N = 13 *50*(2*98) mod 100
N = 13 *100*98 mod 100
Since N mod 100 is a multiple of 100, the rest is 0.
The answer is: the 2 last digits of N are 00.
Nice problem
Why do we consider 7;9;23;1 instead of just 7;9;3;1?
Oh wait, mod 40. NVM
If you want the last 3 digits, it's 900. It goes in a similar 4-cycle of 187, 083, 627, 003
16:11
Good
Dr penn , since you love floor functions so much , here's another floor function problem from the Indian national maths olympiad 2014 problem 2
Let n be a natural (positive integer) number , prove that floor(n/1) + floor(n/2) + floor(n/3) + floor(n/4) + .... + Floor (n/n) + floor(√n) is even .
I'm suggesting INMO problems for such a long time , please check them out , it's a humble request , thank you.
Try till you get this prob done lol....
@@shubhayubasak8909 yeah lol
@@shubhayubasak8909 ami eta prottek video te suggest korbo xD
Also profomar maths has done this
@@chhabisarkar9057 koris na
Eta dekhe onnyoder mone hobe je tui spam korchhis
👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍 wonderful
There's obnoxious static in the microphone.
Με αυτό που διευκρινιζεις στην αρχή τι εννοείς ως προς τους εκθέτες δεν το συμβολίζεις σωστά, είσαι λάθος , κλπ κλπ το καταλαβαίνεις φυσικά γιναυτο κάνεις την διευκρίνιση
0:28 Aaaaand this is where you lost most people, hahaha. Still looks like a lot of fun, so I'll keep watching :-)
EDIT: Looks like a sick powerful theorem though, I might want to check your proof just for fun too!
EDIT2: Turns out it wasn't that hard to follow! Thanks for the great video!
It is indeed a nice little theorem.
I'm a big rere, I somehow thought that 3^7^2 is 3^14 and went about solving that version. Welp I don't know totient func so I have my own excuses
Easy this time, solved in under than 3 minutes (I used a different approach, faster)
7^3 is 43 mod 100, not 23
mod 40, not mod 100
Anticlimatic omg
2020 wasn't hard enough?
❤️❤️❤️❤️❤️
I just went with the series of 3^(7^n)) mod 100 using the fact that 3^(7^(n+1)) = (3^(7^n))^7
3^7 ≡ 87 (mod 100)
3^(7^2) ≡ 87^7 ≡ 83 (mod 100)
3^(7^3) ≡ 83^7 ≡ 27 (mod 100)
3^(7^4) ≡ 27^7 ≡... 3 (mod 100)
Then it loops around since 3^(7^5) ≡ 3^7 (mod 100), so we can rewrite the sum a bit:
sum(n = 1; 2020; 3^(7^n)) = sum(n = 0; 504; 3^(7^(4n+1)) + 3^(7^(4n+2)) + 3^(7^(4n+3)) + 3^(7^(4n+4))) ≡ sum(n = 0; 504; 87 + 83 + 27 + 3) ≡ sum(n = 0; 504; 0) ≡ 0 (mod 100)
Nice
Another exercise
Find the last three digits of N
The video gives 2/3 of the work already done ;)
1 min in and i'm already lost....
Nice shave....
Can you just.. look into the camera for a second? And say nice things
A nice question but it was relatively easier ...... (solved it)....
Abeeee INMO problems suggest korrrr
nice hairdo
Nice haircut