a RARE mistake from Euler (AIME 1989)

[@blackpenredpen]

Ok now wtf? Why do I keep making mistakes in my videos! I am so sorry. Yes 27*6=162 not 154. I am in disbelief! Luckily, it didn’t affect my solution, phew. 😊

[@kartikeya9997]

Glad that you identified after there was a burst of comments on that

[@blackpenredpen]

Lol yea. Thanks to everyone who pointing out. Is it bc of this board or what that i just can’t multiply right. Lol 😂

[@hichamelyassami1718]

Yeah that is the only mistake you made in the video. I'm an electrical engineer but i used to love maths and especially math olympiads. Try to solve the birthday problem (a girl with his two male friends who try to find her birthday date given just some information and a very very short dialogue between the two male friends. Frankly the dialogue is very hard to understand). You can find the problem and its solution online but don't cheat lol. The problem doesn't need any math knowledge just some logic. Bye.

[@anuragguptamr.i.i.t.2329]

Hi @BlackPenRedPen, a better approach to solve this problem would be just to check out the last (UNIT) digit of the problem and the digital SUM. It would be much easier than REMAINDER Theorem's application. Let me explain my approach in detail. As per the first two observations, mentioned by you, we now know that 133< n 243 ==> 3. Last digit of 110^5 is: 0^5 ==> 0. Last digit of 84^5 is: 4^5 ==> 1024 ==> 4. Last digit of 27^5 is: 7^5 ==> 7. Hence, the last digit of n^5 is: 3+0+4+7 ==> 14 ==> 4. Thus, n^5 could be either 134^5 or 144^5. We have to check for just these two numbers, now. . Observation4) We could now simply take the 5th powers of 134 and 144 and check for the actual answer, directly. OR 134^5 is not divisible by 3, producing remainder 0. But, 144^5 is divisible by 3, producing remainder 0. Therefore, 144 is the answer, as per remainder theorem. OR . Check for the digital sum. Digital sum of 133^5 is ==> (1+3+3)^5 ==> 7^5 ==> (7^3)× (7^2) ==> 343x 49 ==> (3+4+3)× (4+9) ==> 10x 13 ==> (1+0)× (1+3) ==> 4. Digital sum of 110^5 is ==> (1+1+0)^5 ==> 2^5 ==> 32 ==> 3+2 ==> 5. Digital sum of 84^5 is ==> (8+4)^5 ==> 12^5 ==> (1+2)^5 ==> 3^5 ==> 243 ==> 2+4+3 ==> 9. Digital sum of 27^5 is ==> (2+7)^5 ==> 9^5 ==> 9. Hence, digital sum of n^5 is: 4+5+9+9 ==> 27 ==> 2+7 ==> 9. . Summary) n^5 could be either 134^5 or 144^5. n^5 should have the digital sum of 9. . Conclusion) Now, check for the digital sums of 134^5 and 144^5, separately. Whichever among these two numbers would result in digital sum of '9', would be the answer. [According to Observations 3 and 4.] 134^5 has digital sum of: (1+3+4)^5 ==> 8^5 ==> 2^15 ==> (2^10)× (2^5) ==> 1024x 32 ==> (1+0+2+4)× (3+2) ==> 7x 5 ==> 35 ==> 8. This is not same as 9. 144^5 has digital sum of: (1+4+4)^5 ==> 9^5 ==> 9. . Therefore, n^5 = 144^5. n= 144. . The best part about this approach is that, this can be done in brain without the need for pen and paper.

[@kartikeya9997]

@@anuragguptamr.i.i.t.2329 awesome

[@IamBATMAN13]

Imagine going to the AIME and one question literally asks you to prove Euler wrong

[@alexpotts6520]

Bit of trivia: this is the only conjecture Euler made which turned out to be false. It is also, to my knowledge, the conjecture which took longest to be disproven, 197 years between its statement in 1769 and disproof in 1966.

[@AlgyCuber]

what i would do : 1. multiply out the left hand side and get a huge number 2. try random numbers for n and manually multiply n out 5 times and hope i get it right on the first few tries 3. give up before even finding the solution and move on to other questions and never come back to that question again

[@xalluniverse9028]

Technically you could use binary search

[@lilyalan8522]

@@xalluniverse9028 when u can use Computer Science for math problems, nice

[@xavierlebeuf3061]

The song at the end is hilarious!

[@hymnodyhands]

Yes, that was bonus fun...

[@ffggddss]

From the thumbnail, you appear to have four specific 5th powers which add up to another 5th power, to be found. The four given are 133, 110, 84, and 27. Their 5th powers are supposed to add up to n⁵. Now any (positive) integer raised to the 5th power, retains its final digit. So n must end in 4. [This actually just amounts to determining n == 4 mod 10. Let's look at some other moduli.] Two of the four numbers (84 and 27) are multiples of 3. The others are +1 and -1 mod 3, each of which remains the same when raised to the 5th power. So n⁵ and therefore, n, must be divisible by 3. Two of the four numbers (84 and 133) are multiples of 7. The others are -2 and -1 mod 7. The sum of 5th powers must be (-32 - 1 == +2) mod 7. This requires n == 4 mod 7. Of course, n must also be > 133. The smallest integer meeting all these requirements is 144. The next smallest is 144 + 3·7·10 = 354, which is far too large to work. [354 > 2·133 = 266, so 354⁵ > 2⁵·133⁵ = 32·133⁵ ; while that sum of four 5th powers < 4·133⁵] So either n = 144, or there is no solution. [Whips out calculator . . . YES!!! 41,615,795,893 + 16,105,100,000 + 4,182,119,424 + 14,348,907 = 61,917,364,224 = 144⁵] ADDENDUM: Note that, had Euler's Conjecture been true, it would have instantly settled Fermat's Last Conjecture (in the affirmative). I'm thinkin that this is what Euler was thinkin. Fred

[@arkasarkar389]

n⁵=4(mod 5) then n=4(mod 5) by Fermat's little theorem.

[@hydrarl3869]

That is solid logic but you could have saved some time using the bounds. 133

[@Мих-ш6л]

Fred

[@davidemasi__]

How did you get to n=4 (mod 7) from n^5=2 (mod 7)?

[@Nothing_serious]

Kinda disappointed it's not true. I kinda like the idea behind it.

[@maskedman8368]

my father: what does your brother always do? my brother : i dont know hes always on youtube watching some asian guy write on two tiles (awkward silence)

[@ansper1905]

8:48 Fermat's little theorem showing up xD

[@wilderuhl3450]

Ansper interestingly there’s this nifty little proof I know of, but it will not fit in my allotted character count.

[@manamritsingh969]

@@wilderuhl3450 post it bruv

[@neilgerace355]

@@wilderuhl3450 hahaha

[@maxwellsequation4887]

@@wilderuhl3450 Only 1% get the joke!

[@dexter2392]

How to get a lot of views: "This disproves Euler"

[@erikkonstas]

In Euler's defense, he never said it's absolutely true...

[@alexpotts6520]

I believe this is the only conjecture Euler made that has ever been disproven.

[@samuelpaterson1045]

This guy is a noob he forgot to see if it’s a multiple of 1 lmoa

[@supercool1312]

SJFP And he forgot to see if it was a multiple of -1

[@kartikeya9997]

I think you all are 'fool' . How can he forget to see the multiple of "zero".....

[@blackpenredpen]

Yo my man, y'all forgot to check if it was a multiple of i

[@trueriver1950]

@@blackpenredpen The number i doesn't belong in the reals so it is hard to see what modulo i would mean. An example that, umm transcends these , would be mod pi

[@andrewzhang8512]

@@trueriver1950 mod e

[@davidgillies620]

Of course *in this particular instance* we can use the observation that n^5 mod 10 ≡ n mod 10(*) to see that n must be 4 mod 10 (this is itself a consequence of Fermat's Little Theorem). Thus we know straight away that n is 134 or 144 (given our upper and lower bounds, although as has been pointed out, 27 * 6 = 162). Then do the mod 3 divisibility check to rule out 134. * 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to the fifth power are 0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049 respectively, i.e. fifth powers end in the same decimal digit as the base.

[@youknowmeg7712]

6^5 = 7776

[@trueriver1950]

@@youknowmeg7712 Yeah, cunning how he snuck in the devil's number there...

[@davidgillies620]

@@youknowmeg7712 Indeed. Fixed

[@shohamsen8986]

at 3:39, I was like Waaat? Then when you revealed the method I was like "Good one".

[@samsonblack]

At 8:40 "Okay, I did the math. It's not just erase the power." *Pierre de Fermat has entered the chat*

[@trueriver1950]

... assuming that there's space for him in our margins ...

[@timsmith8489]

Another way to get a bound on a sum of powers as a single number to the same power is to use the binomial theorem. To bound a^n + b^n where a > b, consider (a+t)^n = a^n + n a^(n-1)b + stuff. If you pick t so that n a^(n-1) b >= b^n, you will have (a+t)^n as a bound on a^n + b^n. Rearranging, we want t >= b/n (b/a)^(n-1). Note that since b/a < 1, if we don't mind the bound possibly being a little larger than it has to be t >= b/n works. You can use that 3 times on 133^5 + 110^5 + 84^5 + 27^5. Divide 27/5 and round up, giving 6. That gives us the bound 133^5 + 110^5 + 90^5. Divide 90/5 and we get 18. That gives us the bound 133^5 + 128^5. Then 128/5 rounded up gives us the bound 159^5.

[@mismis3153]

You look so enthusiastic in your videos and I absolutely love it

[@IoT_]

As a university teacher, I really appreciate your way of teaching. Your scrupulous explanations are off the charts.

[@MathIguess]

You are one of the youtuber mathematicians whom inspired me to start making content as well! I love your work, keep it up!

[@blackpenredpen]

Thanks!!!

[@aflobwelrooms]

Thanks!

[@blackpenredpen]

Thank you!

[@SD-mc9xm]

3blue1brown in your patrons list made my day

[@drpkmath12345]

One of the very common techniques comparing with near values to come up with a value that is super big with high powers or difficult in calculation. Nice presentation!

[@maxhaibara8828]

Actually, when you calculate 3^5 + 0^5 + 4^5 + 2^5 mod 5, you can talk a bit about the other Fermat's theorem. The little one :D

[@BlueEdgeTechno]

oh yeah, a^n = a (mod n) for n being co-prime with a

[@swenji9113]

@@BlueEdgeTechno actually, it's a^p congruent to a mod p for p prime and any a, you only need to ask that a is coprime to p for the statement a^(p-1) congruent to 1 mod p

[@bshubho]

Last digit remains the same if you raise any number to the 5th power. The last digit has of n has to be the last digit of 3+0+4+7 which is 4. That leaves 144 as the only possibility within the bounds.

[@shivpatel8288]

It hard to understand your language, but your contenet is gold. Thank you for putting up the efforts, It has almost been an year following your content

[@zanti4132]

How about just doing this in essentially one step? The mod 30 value of any integer n is the same as the mod 30 value of n^5. So take the remainder of your four integers when you divide them by 30 and add them up: 13 + 20 + 24 + 27 = 84. Your number has to be 84 + 30n for some integer n, and we know n > 133. 144 is the logical guess.

[@TheMiningProbe]

There is an easier way to do it once you have the constraints, just see that all which are equal to 4 mod 5 and even are 134 and 144, then check the divisability by 3

[@parasgovind6271]

6:27 "Because its cooler like this!"

[@barthennin6088]

Loved the math reasoning as well as the added in humor!

[@MultiDman2011]

This is such a treat. I’m reading a book about Fermat’s last theorem, I got to a part where it mentioned Eulers conjecture and how it was disproven. That ultimately led me here to this video. I had recently discovered what the Chinese reminder theorem was and how to use it to solve system of linear congruences, and then subsequently the method you used in this video. It’s just refreshing that everything I’ve been independently studying, you mentioned all in this video and delivered them comprehensively. Thank you 🙏🏽

[@lenochw]

Díky!

[@AniketKumar-lw6su]

3:40 I was so stunned and confused like how you did it. You really got me

[@egillandersson1780]

I found it the same way. Thank you very much for all the progress I made. Using négative numbers in the congruences makes them easier.

[@Kumar-oe9jm]

At 4:10, 27×6=162 not 154

[@neilgerace355]

3:35 hahaha rigorous

[@blackpenredpen]

Yup, I had to.

[@icew0lf98]

raising a number to the fifth power has the same remainder when dividing by 10 in decimal base

[@obibellowme]

iCEW0LF wow, why does this work?

[@obibellowme]

Aneesh Saripalli wow that’s amazing, thanks for the clear explanation!

[@michaelwpannekoek]

for which n is k^n equivalent to k (mod n)?

[@michaelwpannekoek]

fermats little theorem guarantees it for prime n, but does it work in other cases?

[@michaelwpannekoek]

composite numbers where this holds are called Carmichael numbers

[@anuragguptamr.i.i.t.2329]

Hi @BlackPenRedPen, a better approach to solve this problem would be just to check out the last (UNIT) digit of the problem and the digital SUM. It would be much easier than REMAINDER Theorem's application. Let me explain my approach in detail. As per the first two observations, mentioned by you, we now know that 133< n 243 ==> 3. Last digit of 110^5 is: 0^5 ==> 0. Last digit of 84^5 is: 4^5 ==> 1024 ==> 4. Last digit of 27^5 is: 7^5 ==> 7. Hence, the last digit of n^5 is: 3+0+4+7 ==> 14 ==> 4. Thus, n^5 could be either 134^5 or 144^5. We have to check for just these two numbers, now. . Observation4) We could now simply take the 5th powers of 134 and 144 and check for the actual answer, directly. OR 134^5 is not divisible by 3, producing remainder 0. But, 144^5 is divisible by 3, producing remainder 0. Therefore, 144 is the answer, as per remainder theorem. OR . Check for the digital sum. Digital sum of 133^5 is ==> (1+3+3)^5 ==> 7^5 ==> (7^3)× (7^2) ==> 343x 49 ==> (3+4+3)× (4+9) ==> 10x 13 ==> (1+0)× (1+3) ==> 4. Digital sum of 110^5 is ==> (1+1+0)^5 ==> 2^5 ==> 32 ==> 3+2 ==> 5. Digital sum of 84^5 is ==> (8+4)^5 ==> 12^5 ==> (1+2)^5 ==> 3^5 ==> 243 ==> 2+4+3 ==> 9. Digital sum of 27^5 is ==> (2+7)^5 ==> 9^5 ==> 9. Hence, digital sum of n^5 is: 4+5+9+9 ==> 27 ==> 2+7 ==> 9. . Summary) n^5 could be either 134^5 or 144^5. n^5 should have the digital sum of 9. . Conclusion) Now, check for the digital sums of 134^5 and 144^5, separately. Whichever among these two numbers would result in digital sum of '9', would be the answer. [According to Observations 3 and 4.] 134^5 has digital sum of: (1+3+4)^5 ==> 8^5 ==> 2^15 ==> (2^10)× (2^5) ==> 1024x 32 ==> (1+0+2+4)× (3+2) ==> 7x 5 ==> 35 ==> 8. This is not same as 9. 144^5 has digital sum of: (1+4+4)^5 ==> 9^5 ==> 9. . Therefore, n^5 = 144^5. n= 144. . The best part about this approach is that, this can be done in brain without the need for pen and paper.

[@akumar7366]

Good on you, hope you get global recognition.

[@panyachunnanonda6274]

Thank you, I love this problem+ your solution.

[@erayyertass]

for x in range(133,154): if 133^5 + 110^5 + 84^5 + 27^5 == x^5: print(x)

[@AaronWGaming]

Simpler terms Combine the 0 Mod 2 (has to be even) and 4 mod 5 (4 or 9) rules... You are looking for a Number ending in 4... This with the first 2 rules makes the number either 134 or 144... Only 144 is a multiple of 3 and fits the 0 Mod 3 rule any time you do Mod rules for 2 and 5 Combine them and you can find the 1's place.

[@A6la3MnRasi]

At min 14:02 he said " next time you get asked this question, you can answer in no time" again, at min 14:02...

[@A6la3MnRasi]

jokes aside, thank you for this entertaining video. we love you!

[@Kory718]

Just finished a yr binge with these vids

[@tumak1]

Nice presentation...as usual. The way you showed the proof was quite delightful. Thank you. Cheers

[@rajatgupta4494]

Can be done using cyclicity of last number. It will be 4 leaving only 134,144 & 154 to check.

[@billprovince8759]

Interesting approach. Seems reasonably efficient. A binary search in the space [133 ~ 162] would also work, but this is less efficient. The approach I took was to let n = (133 + epsilon), and then use binomial theorem to expand n^5 as 133^5 + 5*epsilon*133^4 + ... epsilon^5 Now, Note: 133^5 + 5 * epsilon* 133^4 + ... epsilon^5 = 133^5 + 110^5 + 84^5 + 27^5 I can subtract 133^5 from both sides, so the result is a bit simpler: 5 * epsilon * 133^4 + 10 * epsilon^2 * 133^3 + 10 * epsilon^3 * 133^2 + 5*epsilon^4 + epsilon^5 = 110^5 + 84^5 + 27^5 Now, the idea is to simply estimate the value of epsilon. The largest term will be from 5 * epsilon * 133^4, so picking a value for epsilon so that this term is less than the rhs but where (epsilon+1) would make it greater should be sufficient. So, basically, divide the rhs by 5*133^4. It involves a lot of calculator work still, but is better than a brute force binary search. Using modulo arithmetic definitely seems the better approach.

[@MajaxPlop]

For the third step you can already work out congruence by 10 because for each integer, their fifth power has the same last digit as them (just a few to work out with 7 and 8 to prove that, others are easy) and then you can say that n must end in a 4, and 133 < n < 154, so we have two potential values: 134 and 144, and we just need to try both, or verify that 3 divides n, which it does, to argue that n is 144

[@creature_from_Nukualofa]

the last digit of a 5th power ends with the same digit - if the above is true (given) the last digit of n must be 4 (7+4+0+3) . given the bounds, 144 is the first guess (134^5 not big enough)

[@ssdd9911]

how can u tell not big enough

[@adiprime4147]

I figured out that n ends in a 4 by adding up the last digit of the four numbers then I was left with 134 or 144. The answer is 144 because it was divisible by 3.

[@Stat201atUTK]

Thanks for all the educational content you make! You were a big inspiration behind my channel and you've also taught me mathematics better than almost all my teachers (Gotta give credit to Dr. Witold Kosmala).

[@trueriver1950]

8:09 I would take 84 as -1 mod 5. It makes the calculation even faster. Then a bit later, we find that N is congruent to -1 mod 5, leading to a slightly different equation that N = 30 M - 6. (Obviously my M differs from BPRP's). Personally I find that it's easier to work with a small number (6 in this case) than to add a larger one, despite the disadvantage of having to contact instead of adding... I usually use -1 in clock arithmetic whenever it crops up. Some folk would go further than me, and swap to negative numbers as soon as you get half way round the clock. That's like saying ten to blah" instead of "fifty after blee" when thinking of the minutes on the clock. Advice if preparing for a contest where speed matters: Try a few examples: which do you find easiest when speed is important? Focus on what works for you, not what works for me. The options are: a) keep to positive integers and zero like BPRP does? b) use -1 when you see something one less than the size of the clock, like I do? c) change to negative as soon as your past half way? d) change to negative at some other point (like you might go negative at day -2, or some other small negative number)

[@jeanphillippes2196]

If "n" was Zero in mod2 and Zero in mod3 it had to be a multiple of 6. And to satisfy 4 mod5 it must end in either 4 or 9. Six times table from 132 (no) gives 136 (no), 140 (no) - 144 YES !!!!!! Yep and in real time (phew!).

[@saraqael.]

Jean Phillippes that‘s not the six times table

[@ОлегКогут-б6с]

@@saraqael. well, it is. 6×23=132, then 138 and 144

[@Joseph_Joshua]

kray2410 the times table can go beyond the first ten multiples though

[@saraqael.]

Joshua Yeah but its still multiples of 4 in his comment

[@jeanphillippes2196]

No, Your're right I meant 138 - he he - but still beat teacher on this one. Thanx

[@alecorsmatem4845]

For congruence 3,4 mod3, counts are easier if you know that 3=(-2)(mod3) and 4=(-1)(mod3), the power works as usual.

[@ushasingh6204]

Well you could have also solved it by another method(Which I did). Any number n raised to a power of 5 would have the unit digit same as the unit digit in n. So u could have calculated the unit digit in the sum 133^5+110^5+84^5+27^5 to be 3+0+4+7 =4(mod10). If the sum is a number n raised to 5 then it must have a 4 in the units place. Since we have calculated the lower limit and the upper limit the only possible ans are 134,144.(I would choose 144 among the two because 134 is too close to the lower limit 133).

[@James-le8gd]

There was a video i remember from numberphile that was about powers of 5 i knew that n has to end with a 4 because any number to the power of 5 will have the same last digits so i added the last digits of all the terms on the left hand side

[@deidara_8598]

Solving the system of congruences in this case is super easy. Notice that since n = 4 mod 5, n has to be one less than a multiple of 5, thus n has to end in either 4 or 9. Since n is even, it must end in 4. We now have two options, 134 and 144. We know that n = 0 mod 3, so all we have to do is add up the digits to check if it's divisible by 3. 1+3+4=8, which is not divisible by 3, but 1+4+4=9, which is. Therefore, n=144. QED

[@aidancheung7264]

When you had written up and proven so much theroems and conjectures, people simply overlooked your mistakes lol.

[@IvanWongHongKong]

I mean it just meant no one was able to disprove it so far (its a conjecture so there's no actual proof)

[@okaro6595]

From 5 we get that the last digit is either 4 or 9 but as it is even it has to be 4. This leaves options 134 and 144. Using the rule to check a number is divisible by 3 we can discard 134 so it is 144.

[@hamanahamana3799]

At 4:12 , shouldn't that be 162 ^ 5? Doesn't change the answer

[@Alians0108]

0:14 REEEEEEEEEEEEEEEEEEEEEEEEE

[@amitotc]

The last part of the solution reminded me of Chinese Remainder Theorem. Great question and good solution :)

[@muckchorris9745]

In 8:23 its not "randomly" the same numbers! Its because the Group (Z/5Z, *) , where * is multiplication,, has group power (is this right in english? German word is: Gruppenordnung) of five, and you take everything to the power of 5, so for every element E in a Group with grouppower of n you have E^n = E

[@ScorelessPine]

Not that it was necessary, since the number of values to guess and check for m in the end were so small, however when you have that set of inequalities, you can extend it further to calculate it, by doing the math on the inequality itself. If 133 < 30m +24 < 154, then you can subtract 24 from all sides to get 109 < 30m < 130, and then continue by dividing all sides by 30 with remainders to show 3r19 < m < 4r10, so the only integer above 3 and some remainder, and below 4 and some remainder, is the integer 4. Then plugging m back in to solve for n. Technically speaking, you're also 'guessing and checking' with the last inequality as well, but the bounds of the inequality more intuitively define the value of m than guessing and checking on the original inequality.

[@hurdler]

For the upper bound, you can consider it mod 7 and use fermat's little theorem to get it mod 210. Then you don't need the upper bound :)

[@ricardoguzman5014]

Any number to the 5th power ends with the same digit. Add the last digits of the numbers, 3+0+4+7=14, So the number will end in 4. So choices are 134, 144, 154, 164, etc. As 5th powers grow rapidly, it's gonna be somewhere in that range. Just test a few and you got the answer.

[@TrimutiusToo]

Interestingly enough if you then go to 7 it will be 210p+144 so the plus part becomes the answer, though if you go further you might need to plug in negative values so not like there is always a solution where you can just plug in 0

[@bot24032]

10:15 it's allowed because (2, 3) = 1

[@damyankorena]

8:50 it actually is just erase the power, since n^(4k+1) mod m = n mod m, and bc 3,0,4 and 2 are already smaller than 5, you just remove the powers

[@deldarel]

Average person: "1. An even number divisible by 3? Then it's divisible by 2x3=6 2. it ends in a 4 or a 9, but 9 isn't even so it has to be 4, leaving 134 and 144 3. 144 is 12², which is (2x6)² so this one fits all criteria. 10 isn't divisible by 3 so 134, being 144-10, does not fit the criteria" Mathematician: 9:33 to 13:37

[@gabrielaloisi5556]

When you said: “you could just plug in now” I wrote a python program in 8 seconds and found the result instantly ahahah.

[@connorhorman]

Gabriel Aloisi True. I would have used a range binary search, which is lg(153-134)=lg(19)~4

[@martinepstein9826]

Small optimization at 2:25 : Instead of calculating 5^5 then doubling you can calculate 5^4 then multiply by 10 :)

[@onebeing4164]

Bruh

[@connorhorman]

144 was actually obvious based on the congruences. The last digit had to be 4, because congruent to 4 (mod 5) and even (which disqualifies the last digit being nine). You get 134 and 144 being the only possibilities, and 144 is divisible by 3 so 134 can't be

[@MikehMike01]

I wish you had done the inequality. You made it that far without jumping to guess and check! 133 < 30m + 24 < 154 109 < 30m < 130 3.63 < m < 4.33 m = 4

[@olivergrimes6145]

You are really good with numbers by the way

[@cmilkau]

You can use Fermat's little theorem to reduce the powers as well, if they are larger than the modulus. That is also why you actually are allowed to just erase the powers in one of the cases.

[@anubhabghosal8496]

You can simply delete the powers at 8:41 due to Fermat's Little Theorem

[@neighbourtwin3381]

Every time I see your mic Im laughing so loud....😂😂😂

[@Nick-kg7sk]

Why

[@kienthanhle6230]

My way of doing it: Step 1: Calculate the left hand side Step 2: Use binary search to find n If this was a multichoice question, this could be usable.

[@PankajTiwary]

Great solution and explanation. I do feel like the last part can be simplified a bit. Once you established the 5 facts on the right hand side (which is the hard part), from (3) and (4) you know that the number has to be divisible by 6. There are only 3 numbers in the range that are divisible by 6 (138, 144 and 150), only one of them is equal to 5 mod 4, so the answer is 144.

[@lucaslucas191202]

Sure, but is a great general example of what you can always do. Sure you can figure this problem out faster through different simplifications, but you can't always do that in other problems. This method always works.

[@nguyenduynam7398]

3:48 how ? it so....

[@jayantverma2136]

BPRP solve this interesting integral- √(-x^2+2ex+99e^2)dx from e to 11e. With and without calculus

[@1llum1nate]

A general approach for the systems of congruences is cool, but this exact one can be solved pretty easily in the head. Since x=0(mod 2) and x=4(mod 5), we clealy see that x=4(mod 10). And then all we have to do is to check which of the numbers 4, 14 or 24 (all the numbers less than 2*3*5=30 that are 4 mod 10) is divisible by 3, which is clearly 24.

[@zanmaru4302]

Just calculate the sum of the numbers' last digits to the fifth power; that'll be 3^5 + 0 ^ 5 + 4 ^ 5 + 7 ^ 5 = 18074, but you only need to look at the last digit - that'll be the last digit of the number n, so it's either 134 or 144; easy to proof that 134 won't fit

[@MikeBTek]

Actually consider the last digit of each addend to the 5th power. For any digit to the 5th power the resulting last number will be equal to the digit. That is 1^5 yields last digit 1, 2^5 yields last digit 2, ... 8^5 last yields last digit 8, 9^5 yields last digit 9. So looking at the addends to the fifth power we know the last digits will be 3, 0, 4, and 7 which add to 14, last digit 4. Hence we could first guess 144 as the answer.

[@sunilbamal2948]

After having inequality on n we can approach by unit digit of given value

[@SJrad]

I know you gotta write it down formally, but i find it funny thst you ho through the work to find that n is divisible by 6 when you know automatically that the divisibility rule for 6 is that it must be divisible by 2 and 3 which we found out in observation 3 and 4

[@goseigentwitch3105]

I think you made the last step more complicated. If n == 4(mod 5) then n ends in a 4 or a 9 If n == 0(mod 2) then n is even and ends in a 4 Our only optins are 134, 144, 154 if n == 0(mod3) then the only option is 144. The others aren't divisible by 3. bang! 144

[@timurpryadilin8830]

The last part, where you got that all such n's are separated by 30 (24, 54, 84,..., 144,...) is a good link to a Chinese remainder theorem. Consider making video about it? :)

[@blackpenredpen]

Oh yea I have one of those videos already.

[@JonathonV]

I had an inkling it was 144 before you started your system of equations. The only numbers between 133 and 154 that are congruent to 4 (mod 5) are 134, 139, 144, and 149. Because they're even you know they can't be 139 or 149, and 134 isn't a multiple of three (easily shown by the 3 divisibility test).

[@Sench999]

The n=0(mod3) and that the sum of fifth powers end with digit 4 would be enough to say that the number is exactly 144.

[@gabrieljohnson6304]

love how they asked students a problem that stumbled/took top mathematicians years to solve

[@lunstee]

Rather than work out that 3^5=243 and 2^5=32 are, you could just point out that 3 ≡ -2 mod 5. So 3^5+2^5 ≡ (-2)^5+2^5≡ -(2^5)+2^5≡ 0, leaving n^5 ≡ 4^5 in mod 5. You jump from n^5 ≡ 4 to n ≡ 4. While the conclusion is true, I think it warrants a little caveat, or someone might think, say, that n^5 ≡ 5 implies n ≡ 5 base 7. A similar caveat would apply going from n^5 ≡ 4^5 to n ≡ 4.

[@albertmoulin]

Hey ! for the fifth power, it is just "erasing the powers" because 5 is prime so Fermat's little theorem tells us that for any prime p: a^p = a (mod p)

[@Wi11daThri11]

@8:40 "Okay, I did the math ... It's not just erase the top [the exponent]" He even said it in the video!!! =)

[@albertmoulin]

@@Wi11daThri11 Yeah I know, he's saying that he actually did the maths in the sense that it did do 3 to the power of 5 and then divided it by five. But du to ferma's little theorem it just wasn't necessary. Furthermore he say's it jokingly because one may be tempted just to erase the powers which could be false except in this particular case. You always have that a^p is a mod p when p is prime, you can test this it works.

[@Catman_321]

i've seen this before, but i wonder if there are any more examples? if it doesn't work for any more of 4 5th powers added together, than what about 5 6th powers, 6 11th powers, and so on?

[@andrasfogarasi5014]

A generalisation would be the Lander, Parkin, and Selfridge conjecture. Stated, it is as follows: If a sum of m like powers equals a sum of n like powers, where each power is distinct, the exponent must not be greater than m+n. The conjecture is unproven. So it's good homework.

[@aliakkari5742]

I think There is an easier way . I know that 3^5 it is sth that ends with 3 also 4^5 is sth which ends in 4 and so on . we can use this to figure that their are only 2 possible values that are 134 or 144 but 134 is not divisible by 2 but not 2 so 144 is the solution .

[@yassinezaoui4555]

I liked the video , indeed, it is an interesting topic , in general, at first when you see an equation or a system of equations that we want to find the integer solutions you perhaps think it will be too easy because we are dealing mostly with real and complex numbers so integers are pretty basic and low level but this may be misleading 😊

[@Ynook]

I think it would have been easier to check for a number between 133 and 154 that is a multiple of 2, a multiple of 3 and a multiple of 5 with a remainder of 4.

[@JohnSmith-vq8ho]

Great video! You could have reduced (mod 5) much quicker by noting 133 = 3 (mod 5) and 27 = -3 (mod 5), so they cancel out. So, you are left with the 84 term, and 84 is -1 (mod 5). So, everything is equivalent to (-1)^5 = -1 = 4 (mod 5).

[@trueriver1950]

There is a shortcut. From (3) and (4) we know the number is a multiple of both 2 and 3. Therefore it must be a multiple of 6, without having to do the underlying modular arithmetic. So I would write (6) n == 0 (mod 6) This is neat as (5) tells us the mod 5 rule and (6) tells us the mod 6 Then I would go into the detailed modular algebra like you did to combine 5 and 6 to get a rule in mod 30. In formal terms my method is the same as yours, but in exam conditions it is faster. However that also depends on how much working the marking scheme needs to see

[@nicholasjohnson3542]

I have seen 27, 84, 110 and 133 together before but I cannot remember where.

[@malcom91]

Fermat wouldn't be proud if he saw you doing the math at 8:30.

[@alexismandelias]

8:43 isn't that a property of numbers raised to the 5th power? Like a^5 (mod 5) = a (mod 5). I swear I've seen that before

[@kaynight64]

It's a property of any prime power, actually, not just 5! It's called Fermat's Little Theorem.