a RARE mistake from Euler (AIME 1989)

Here's a rare mistake from Euler and they made this a problem on the AIME exam in 1989. Read more on Euler's sum of powers conjecture: en.wikipedia.o...
Out song by motormusic: • Math Bros - We Love Ca...
#blackpenredpen #math

[@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

[@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

[@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!

[@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)

[@shohamsen8986]

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

[@IoT_]

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

[@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

[@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!!!

[@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!

[@mismis3153]

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

[@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!"

[@neilgerace355]

3:35 hahaha rigorous

[@blackpenredpen]

Yup, I had to.

[@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.

[@SD-mc9xm]

3blue1brown in your patrons list made my day

[@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.

[@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).

[@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

[@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

[@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.

[@AniketKumar-lw6su]

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

[@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.

[@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 :)

[@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.

[@rajatgupta4494]

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

[@panyachunnanonda6274]

Thank you, I love this problem+ your solution.

[@nicholasjohnson3542]

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

[@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.

[@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).

[@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.

[@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

[@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.

[@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.

[@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

[@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

[@gabrieljohnson6304]

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

[@barthennin6088]

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

[@AndDiracisHisProphet]

3:47 you sneaky bastard^^

[@modenaboy]

Thanks for another great video! I really enjoy your channel, and this was a really fun problem, with lots of food for thought too. (I loved how you offhandedly mentioned 6^5 is 6*6*6*6*6 and you can just replace with 7s and add 6 at the end) I might be missing something, but once you determined the congruences (is that the right word?), I assumed since the number must be evenly divisible by both 2 and 3, then it must be divisible by 6. Furthermore, the remainder when divided by 5 is either 4 or 9 since it is 4(mod 5), but 9 is out (since that would be an odd number and not divisible by 2), so that leaves 134, 144, 154 as candidates and 144 is the only number divisible by 6. Is that reasoning incorrect?

[@Kory718]

Just finished a yr binge with these vids

[@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

[@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

[@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

[@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.

[@akumar7366]

Good on you, hope you get global recognition.

[@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.

[@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.

[@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.

[@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.

[@roderickwhitehead]

Someone please ship this man a full sized whiteboard for home use!

[@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

[@dirklitter]

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

[@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

[@Quasarbooster]

Not that it affects your solution, but I thought I'd just point out that you put 27^5 * 6^5 = 154^5 at 4:08. I believe it would actually be 27^5 * 6^5 = 162^5. Still, this was a sweet video!

[@kartikeya9997]

I think 1 comment was enough

[@Quasarbooster]

Kitkat the first one I tried had a request time out. I didn't think it posted

[@kartikeya9997]

@@Quasarbooster That's ok bro/sis. Dont take that too seriuosly

[@blackpenredpen]

QuasarBooster I have no idea why or how I got 154....... but yes, luckily it didn’t change the answer phew. Thanks for pointing out : )

[@Quasarbooster]

blackpenredpen haha, I'm glad that it didn't mess anything up

[@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.

[@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.

[@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

[@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).

[@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.

[@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.

[@balthazarbeutelwolf9097]

All you did is reduce the number of possible values to 1 value - you still need to check that that single value actually works.

[@serenecereal5167]

preview picture is very scary

[@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

[@alonm6232]

Am i the only one who just found that rpbp have an ending song? Sorry for my English

[@waldemarmoskalecki7891]

let me translate it to English: "Am I the only person believing that blackpenredpen has his end soon (runs out of interesting topics)"?

[@sylowlover]

You've actually calculated the residue classes of n^5 modulo 2, 3, and 5. You need a bit more justification to say those are the same residue classes of n mod 2, 3, and 5. For example, if you were calculating modulo 7, you would get n^5=2 mod 7, but this admits n=4 mod 7. You can't just ignore the power on the right side. It luckily works out here, but it is not generalizable. You should reference Fermat's Little Theorem (Or Euler's Theorem for non prime cases).

[@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

[@trueriver1950]

This question would come out to a nice round number in base 12...

[@markmajkowski9545]

How about just determining the last digit - which is 4! So you have only 144 and 134. Seems you can rule out 134 by expanding (133+1)^5 and also it is too close - so 144.

[@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.

[@sunilbamal2948]

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

[@haflam.]

From knowing n between 133 and 154, why not just look at least significant bit. 3^5 gives a 3, 4^5 a 4, 0^5 a 0 and 7^5 a 7. Adding them means the least significant has to be a 4, which can only come from 134, 144 or 154. Just try these 3. Althought your solution is more elegant. Question: apperantly, if you do x^5, the least significant of the answer equals the least significant of x. Why?

[@ork0_095]

the doraemon background music is mood

[@nightshade7745]

Searching within a range is not going to take long, I won’t iterate down from 154 or up from 133, I will instead binary search in that area

[@suniltshegaonkar7809]

Hello bprp, why is there n< 154 ?; it should be 162 (=27x6) ? Extremely beautiful. Thanks bprp.

[@blackpenredpen]

Yes. That was my mistake. Thank you.

[@YoavZilka]

Take any number to the power of five and the last digit will stay the same. If the last digit stays the same, then the number%5 will stay the same. That’s why it looks like you just erased the powers.

[@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.

[@aflobwelrooms]

Thanks!

[@blackpenredpen]

Thank you!

[@peterromero284]

After you found the range, couldn’t you just find the last digit? Any integer to the fifth power ends in the same digit as the original number, so if you add the last digits of 133, 110, 84, and 27, you get 3+0+4+7=14, meaning the last digit of n^5 is 4. The only number in your range that ends in 4 is 144.

[@peterromero284]

D’oh, I misread the range as >134. My bad. This still narrows the search space to 2 numbers.

[@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

[@hamanahamana3799]

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

[@mikezilberbrand1663]

Alan, x^5 has the same last digit as x, thus the last digit of n is 4. So two candidates: 134 and 144.

[@fabscud]

3:42 6x6x6x6x6 “the easiest way to do this is by replacing 6 with 7...” 😂

[@iainfulton3781]

6:×27 = 154 does it. If you used Fermat's little theorem then 134 144 and 154 would be your 3 choices and 154 you would have seen isn't 0 mod 3 so you would have seen 27×6 isn't 154

[@L13832]

Hey! Calculators are not allowed in AIME!

[@Superman37891]

But Euler didn’t just simply assume that n was an integer

[@KillianDefaoite]

I wish this conjecture was true. It has a beautiful symmetry to it.

[@blackpenredpen]

Yea. And that will also imply FLT is true!

[@willmunoz1638]

4 is the cosmic number babyyy

[@mooshiros7053]

a lot of the modular arithmetic would have been easier if you used that 4 = -1 (mod 5)

[@saultube44]

Congruence, something I haven't seen it in a long time, since school, those problems are long but interesting

[@frozenmoon998]

Andrew Wiles left the comment section.

[@danielzitnik4247]

I'm surprised you didn't use the knowledge that x^5 has the last ones digit as x. Thus, 133^5 ends in 3, 110^5 ends in 0, 84^5 ends in 4, and 27^5 ends in 7. Add up our ones digits, 3 + 0 + 4 + 7 = 14. Mod 10 is 4. So, we know n^5 and n both end in 4. So, our only possible solutions (based on your earlier work) is 134, 144, 154. And we can confidently assume 134 is too low based on your methods, as well as 154 being too high because of all the rounding up. So, 144 seems like an obvious first guess, right?

[@rogerwang21]

Why is Observation 5 not n^5 = 4 (mod 5)?

[@violetasuklevska9074]

9:00 You are just gonna slide that deduction without mentioning Fermat's Little Theorem?

[@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 .

[@leonardovalente9772]

Honestly i think it would be faster to just computate the left hand side and binary search the values for n. If you set the lower bound to 134 and the upper bound to 200 you would need at most log2(200-133) = 6 attempts! Idk about you but these odds look pretty good to me

[@matthewware8973]

I remember when Euler said that

[@greatbritton4455]

Great video but isn't 27^5* 6^5 = 162^5?

[@blackpenredpen]

Yes, you are right