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
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. 😊
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.
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.
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.
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
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
@@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
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)
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!
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.
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 🙏🏽
@@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
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.
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.
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
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
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).
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.
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? :)
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)
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
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 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.
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
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).
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.
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.
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
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
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.
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.
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?
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.
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
I neither watched the thumbnail,nor the topic, simply I just opened it,as I knew, it'ld be definately something very amazing and I was correct ♥️😌,sir u bring really awesome videos♥️♥️
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
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.
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).
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
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.
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
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
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.
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?
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.
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.
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.
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.
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.
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.
A number to the 5th ends in the same number, so n^5 ends in 3+0+4+7=14, or 4. Thus n ends in 4. Once you have the bounds, it’s only two numbers to check, and I can assume it’s not 134 because 110^5 is too large
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.
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
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.
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!
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
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).
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)
