Відео

I like the use of that lemma.

I actually guessed the correct answer using a few memorized logarithms: The base ten logs for 1-10 are approximately 0, 0.3, 0.48, 0.6, 0.7, 0.78, 0.85, 0.9, 0.95, 1 Looking at the big number, it's basically 22x10^24, so its base 10 log is 24 + log(11) + 0.3. The 11 log isn't in the list I memorized, so I used linear interpolation between 12 and 10: log(12) ~ 1.08, so log(11) ~ 1.04. Then log(big number) ~ 25.34. Dividing by 13 to get the 13th root logarithm, I just do 3 digits mentally to get 1.94. This is extremely close to what I would compute for log(90) ~ 1.95, so my first guess would be that the 13th root of big number is 89.

Hello from iran

(x ➖ 3x+2)

Another approach without using modulo and log (this is "no-pen-no-paper" method): 1. the last digit of x must be 9, because only 9 goes in cycles (9,1) with the 13th power being 9. 2. 100^13 is 27-digit number, so x must be 2-digit number ----> 26(digits) : 13 = 2. 3. x can't be 99, because the given 26-digit number is too small (it's obvious). 4. 8 = (2^3)^13 = 2^39; let's calculate 2^40 5. 2^10 = 1024, let's use "1000" 6. 2^40 = approx. 8^13 = approx 1000 ^4 7. 80^13 = approx 1000 ^4 * 10^13 = 10^12 * 10^13 = 10^25 8. It is less than the given 26-digit number 9. therefore x = 89

1. Since 100^23 has 27 digits, x must be a two digit number. 2. Last digit of x^(4*3+1) must be same as last digit of x^1, therefore, last digit of x must be 9. 3. Digital sum of the given number is 8, therefore, the digital sum of x^13 must be 8. Solving which we get x=89.

i think you can stop and show contradiction at M(1/2)^4 < M(1/2)^k, since from the observation its implied that k > 4 but obviously in that case the inequality does not hold.

Value of X must be odd, since the result is odd, and X is not 1 or 5 (where the result has unit 1 and 5) For K=1,2,3,4,5,..., then ?3^K, the unit result is 3,9,7,1,3,9,7,1,... ?7^K, the unit result is 7,9,3,,1,7,9,3,1,... ?9^K the unit result is 9,1,9,1,9,1,9,.... For K=13, then ?3^K has unit 3, ?7^K has unit 7 and ?9^K has unit 9 So, possible answer is 09,19,29,39,49,59,69,79,89,99 Number of digit 99^13 is about 36 (less than 36) so 99 is rejected, then X=88 (** if there is no typo like this 21,982,145,917,038,330,487,013,369 😊😊😊)

I just thought... Ends with 9 -> The original number must end with 3, 7, or 9 3 -> Final digits of 3 9 7 1 that repeats itself, will be 3 in 13 (Fail) 7 -> Final digits of 7 9 3 1 that repeats itself, will be 7 in 13 (Fail) 9 -> Final digits of 9 1 that repeats itself, will be 9 in 13 (Success) There are 26 digits, close to 10^26 with 27 digits (AKA 100^13), but not too close I'm now left with 89 and 99 as the most likely answers, and thought, "It shouldn't be too close" and picked 89...i was right...

Last digit is 9, it's easy. 80^13 < 10^25, so it can be 99 or 89. If it 99 then the given number mast be divided by 11. Easy to se that it doesn't. So it is only 89

Rather easy to check that by sheer size x it has to be between 80 and 90 and has to end with 3 7 or 9. Of which only 9 can work. So x is 89.

For people who doesn't know, this equation was the first counterexample, to Euler's conjecture: If the sum of n many kth powers of positive integers is itself a kth power, then n≥k This conjecture was proved wrong in 1966, by L. J. Lander and T. L. Parkin. Their example was 27⁵+84⁵+110⁵+133⁵ = 144⁵

0:08 X is a real number 0:18 X is an integer number WTF!? How did you figured out (in just 10 seconds) that solution is an integer number?

Despite your content not being serious, it is very enjoyable to watch you.

Hahahahahahaha.

I counted digits and figured it had to be 2 digit number. Last digit had to be 9. Then I compared to 100 and saw we ended up at 22% of what 100 would have. This means we haved a little over two times so a half life of roughly 6 years. Law of 72 says we are decreasing by 12% per year to have a half life of six years so answer should be close to 88. Since last digit is 9 the answer must be 89.

10<x<100, since 10^13 would be too small and 100^13 is too big. x cannot end with an even, 1, or 5, because then x^13 would end in an even, 1, or 5, respectively. 7^2 and 3^2 end in 9, so 7^4 and 3^4 end in 1, so 7^12 and 3^12 end in 1. So 7^13 ends in 7 and 3^13 ends in 3. So the last digit cannot be 7 or 3. 9^2 ends in 1, so 9^12 ends in 1. So 9^13 ends in 9. The last digit can be 9. So we know that x is in the form 10y+9, with 0<y<10. Looking at the first 13 digits, y^13 <= 2198214591730, or about 2.198 * 10^12 because if it was bigger, the most significant 13 digits would have to be bigger. 7^2 < 50, so 7^12 < 50^6. So 7^13 < 50^6*7, or 1.09375 * 10^11. This is far too small. 8^3 = 2^9 > 500. So 8^12 > 500^4, and 8^13 > 5^4 * 8 * 10^8. So 8^13 > 5*10^11, which is in the right ballpark. 9^2 = 81 > 80, and 9^13 > 80^6 * 9 = 8^6 * 9 * 10^6 > 500^2 * 9 * 10^6, or 25*9 * 10^10, or 2.25*10^12, which is too big. So y=7 is proven too small, y=9 is proven too big, and y=8 looks plausible. So the answer is 89.

Прикольно

Strangely, x^13 = my phone number multipled by my wife's number plus our door number to the power of the numbers in my post code.

Thank you very much for this proof. There aren't many out there actually proving the existence of the limit, by using lim inf and lim sup

Gonna guess like 91?

Sir, doesn't this depend on the choice of e? We could choose e>1/2, and then a_n <= M(2e)^k -> inf, implying a_n is unbounded. Am I missing something here?

It's obviously 89. Because 9^9 has the last digit of 89 and same as 9^(10n+9). Which means that 9^9^9^9=9^9^(....89)=9^(....89)=.....89.

89

some classic hard anal. great video

Can we split fraction to _a_(n+1) / a_n_ and _a_(n+2) / a_n_ so at least one of them goes to infinity? Then it may be easier to prove that _a_n_ is unbounded if _a_(n+1) / a_n_ is unbounded.

It doesn't go neither from the paper...

I used 10^13 < x^13 < 10^26 -> 10 < x < 100. Then n^k (mod 10) == n^(k (mod φ(10)) -> n^13 == 9 (mod 10) == n^(13 (mod 4)) -> n == 9 (mod 10) From there, a binomial expansion gives 90^13 = 10^13 * (10-1)^13 ≈ 10^26 - 13*10^25 + 78*10^24 = 52*10^25, whereas 10^26 - 13*10^25 + 78*10^24 - 286*10^23 = 23.4*10^25. The remaining terms are alternating and are decreasing in magnitude, so 90^13 > 2.34×10^26. Testing 80^13 tells us that 80^13 < 10^26 - 26×10^25 + 312×10^24 - 2288×10^23 + 11440×10^22 ≈ 0.3×10^25, which is too small (calculations might be a little off since I did them in my head). Therefore, since 80<x<90, and x==9 (mod 10), then if x € Z, then x = 89. I like that your way gives a small set of values that we can test, though.

x = 13th root of the original number *drops chalk as explosion destroys board*

Great video sir! What is the source of this exercise? Thank you.

I am starting to suspect Mr. Jiasheng Jin is playing with us. Because this is too much for this question ... The first natural thing to do is to try some numbers. And we "immediately" can see that for n=8, we have 8^(1/1) = 8 for n=9, we have 9^(1/2) = 3 For n=10, 10^(1/3) is NOT an integer. For n≥11, we have n-7≥4, then n < 2^(n-7) which implies n^(1/(n-7)) < (2^(n-7))^(1/(n-7)) < 2 Clearly, n^(1/(n-7))≠1, so the obtained inequality implies 8 and 9 are the only solutions. No need to use GM ≤ AM, or to manipulate fractions, etc ...

wow this is crazy

Thanks for the interesting problem!

This is way too complicated. It's a super easy problem: 1. check the last digit, it can only be 9 2. check the remainder mod 9, it's 8 3. check the total digit count, it can only have two digits therefore it's 89

A) Last digit of x^(4n+1) is the same as the last digit of x. So we know that the last digit of x has to be 9. B) 100^13 = 10^26, which is 27 digits long, so we know that x is less than 100. C) Fermat's little theorem tells us that a^p = a (mod p) for any prime p. Dividing the given number by 13 gives a remainder of 11. From A, B and C we know that we're looking for a number of the form 13k + 11 < 100 that ends on a 9. This means that 13k has to end on an 8. Looking at the times table for 13 find 78 = 13*6 as the only one that ends on an 8. 78 + 11 = 89. x = 89

Once you figured out that 1001 is a multiple of 13, you can calculate x^13 mod 1001 by subtracting multiples of 1001 until the remaining number is smaller than 1001. For example, subtracting 1001*k*10^n can be achieved by subtracting k simultaneously from the 10^n and the 10^(n+3) digits, i.e. you can subtract (or add) the same amount simultaneously from any two digits that are 3 positions apart from each other. This does not change the remainder when dividing by 1001 (or 13). For example, if x^13 starts with the digits 21982145..., subtract 2 from the first and the 4th digits, you obtain 1962145.... Then subtract 1 from the 1st and the 4th digit to obtain 961145... Now next is to subtract 9 from the 1st and 4th digit. If the 4th digit is not large enough, include the 3rd digit: 60245... If you keep doing this 21 times you end up with the remainder 89. In principle this is just a long division of x^13 by 1001, but the calculation is not difficult, and you don't need to keep track of the result of the division.

If we know that x is an integer: x^13 has 26 digits, so x must have 2 digits. x^13 is an odd number, thus x also has to be odd. The last digit of x cannot be 1 or 5, because then all powers also end in 1 or 5, respectively. Thus the last digit of x must be 3,7 or 9. 3^2 and 7^2 both end in 9, thus 3^4 and 7^4 end in 1, as well as 3^12 and 7^12. Thus 3^13 and 7^13 end in 3 and 7, respectively. Thus the last digit of x must be 9. It is easy to calculate x^13 mod 3, just add all digits and we find that x^13 mod 3 = 2. This means x mod 3 = 2 as well. We can exclude 0 and 1, because if x mod 3 was 0 or 1, then all powers of x mod 3 are also 0 or 1, respectively. Now we know the last digit of x is 9, and x mod 3 = 2. A possible value is x=89. The next smaller value is 59. x must be a 2-digit number, but it cannot be too small. 60^13 has already fewer than 26 digits: 60^2=3600, 60^4=36^2*10^4=1296*10^4. This is smaller than 13*10^6, thus 60^8 is smaller than 169*10^12, which has only 15 digits, so we already lost one digit, 60^13 can have at most 25 digits, most likely it has not more than 24. This leaves us with the only possible solution of x=89, since x=59 is already too small.

in the beginning of the video you say x from real numbers. then you can't assume it's an integer

If you give the 'a' series 2 indexes, and define a_k0 = sqrt(k) and a_{k,n+1}=sqrt(k+a_{k+1},n. Then you can prove by induction on n that a_kn<n+1, and the whole shabang follows.

Use the full cycle of 10 (mod 13), making no additional divisions by 13. We have 10¹ = -3 (mod 13) 10² = -4 (mod 13) 10³ = -1 (mod 13) 10⁴ = 3 (mod 13) 10⁵ = 4 (mod 13) 10⁶ = 1 (mod 13) The number is 21 982145 917308 330487 013369 Just sum the digits in each column and multiply by the respective value of 10ⁿ, n=0,...,5, (mod 13), (9+9+3+0)×4 + (8+1+3+1)×3 + (2+7+0+3)×(-1) + (1+3+4+3)×(-4) + (2+4+0+8+6)×(-3) + (1+5+8+7+9)×1 = 21×4 + 13×3 + 12×(-1) + 11×(-4) + 20×(-3) + 30×1 Mod 13: = (-5)×4 + 0×3 + (-1)×(-1) + (-2)×(-4) + (-6)×(-3) + 4×1 = (-5+2)×4 + 0 + 1 + + 18 + 4 = -12 + 1 + 5 + 4 = 1 + 10 = 11 No division of three digit numbers or 6 digit numbers required. It has to end in 9. So 13k+11 = ...9 This means k ends in 6, so k = 10n+6, 13k+11 = 130n+89 Then it is over. Of course, this reasoning only works because we are assuming the number has an integer 13th root.

You made some confusion, didn't you? Because you could divide the number in 3 digits, alternating plus and minus ...

9 in odd power ends with 9, so our x ends with 9. Now we need a number that in a power 13 reaches 26 digits. So we drop 13 right digits leaving a number with 13 digits. Now find number that in a power 13 is closest to it. (8^13 = 2^13^3): 2^13=8192, 8192^3 ~ 500000000000 12 digits. So our answer would be 89?

amazing, i love your vids specially about number theory, keep it up with the good stuff

Amazing!

Great. At the end you finished a bit quickly.

Thanks!

I wonder what is equals

I love how you always give detailed logical proofs of the sub-results used...helps a lot!! Thank you so much for these clear illustrations!

I want to be an ass and say, a,b,c,d = 1/4 and it's solved

Very nice equation Thank you Dr Jin