One of the craziest number theory problems I have seen! Great job on the solution!

@pkh4840

3 жыл бұрын

yes sir

@aam2k6

3 жыл бұрын

Big fan here , shout out plz

@notpsy7298

2 жыл бұрын

69 likes

@Maths_3.1415

Жыл бұрын

​@@notpsy7298 now 128 likes

@greatrevan_

Жыл бұрын

​@@Maths_3.1415now 129

The last digit of x^13 is 9,so the last digit of x can only be 3,7 or 9. By making n^13 (mod10) we find that the last digit of x must be 9. By adding the digits of x^13 we prove that x is not multiple of 3,so can't be 99. Using the proof that x can't be lesser than 80, the only possible answer is 89.

@mathsandsciencechannel

3 жыл бұрын

GOOD.HE IS MY INSPIRATION

@mairisberzins8677

3 жыл бұрын

It's actually even easier. The last digit is 9 and x^13 always ends in the same digit that x ends with. Check it. They all loop so that the 13th order is always the starting one. So you just figure that it's less than 100 and not really all that much off from 100, so you would get it in 2 checks. 99 and 89. Took me like 3 mins to check the last digits. Ez

@banzaiboi9614

3 жыл бұрын

Also, 99=-1 mod 100, so 99^13=-1 mod 100 too but the last 2 digits of the given number are 69, so it's reminder mod 100 cannot be -1

@JoseFernandes-js7ep

3 жыл бұрын

@@banzaiboi9614 That is afaster and easier way to prove 99 is not a solution. I had to sum all the digits snd verify it a few times for ceritifying the number was not multiple of 3.

@banzaiboi9614

3 жыл бұрын

@@JoseFernandes-js7ep yeah I know I was only giving a new way of solvimg this

Imagine a non-mathematician's reaction when you look at this and say "oh, this shouldn't be too difficult"

@lightworker2956

2 жыл бұрын

Easy problem: if a solution exists, it must be x = 13th-root of the number on the right-hand side. Solved.

@lyrimetacurl0

7 ай бұрын

That's how I did it, had a sense of logarithms 100 --> 10^26 x --> 2.19*10^25 Looks to be about x=89 so I checked that and it turned out a lucky guess.

@epicm999

6 ай бұрын

​@@lyrimetacurl0There are no such things as lucky guesses. Just really cool series expansions that happen to be correct :v

Here's how I did it : 10^13 is 14 digits and 100^13 is 27 digits so x^13=26 digits implies 10

@farhatali2221

3 жыл бұрын

Pradeep bhai.. awesome

@sarmadsultan7981

3 жыл бұрын

Amazing man.....you approached it with an easier way! 💯

@dlevi67

3 жыл бұрын

I think the only 'trick' you missed is that the order of magnitude of the starting number also implies a number pretty close to 90 (so at most you needed to check 79 and 89). Other than that, much shorter than the method in the video!

@pdpgrgn

3 жыл бұрын

@@dlevi67 that's something that came to my mind too, but with a power as big as 13, I don't know the proper technique to actually prove that x has to be above something (like 80, which if I could prove would make it much easier).

@dlevi67

3 жыл бұрын

@@pdpgrgn Lg(6) = 0.78 , lg(7) = 0.85, lg(8) = 0.90 (lg is log base 10 and for '=' read approximately equal) Therefore, lg(60) = 1.78, lg(70) = 1.85, lg(80) = 1.90, which multiplied by 13 respectively gives ~23 ~24 and ~25, so 'something'^13 has to be higher than 80 to give a result of the order of 10^26. It does require one to know 'by heart' the approximate decimal logarithms of the numbers from 1 to 10 (which I do - old school engineer), or to have a calculator/log table.

You can easily see that x is 1 (mod 2), 2 (mod 3), 4 (mod 5), and 1 (mod 11) Thus, x = 1 (mod 22) by the Chinese Remainder Theorem Since x Then since x must be 2 (mod 3) the list of possibilities further simplifies to 23, 89 Then finally, since we want x = 4 (mod 5) the only option that works is 89.

@hfs-lk5ip

7 ай бұрын

I like this way the best, awesome

@israelruiz2906

6 ай бұрын

I love this solution. But I don't know how do yo says that this number is 1 (mod 11)

@ishaangupta2185

6 ай бұрын

@@israelruiz2906 One way is to add / subtract digits, then take the alternating sum mod 11 (this method is based off of a divisibility test for 11)

@Kettwiesel25

6 ай бұрын

You don't need 2 mod 3, you can directly cut the full list down using 4 mod 5

@chrisclub3185

6 ай бұрын

@@Kettwiesel25 true, but you don’t know that before doing the problem, so you might as well figure out the value of x mod p for as many small p as you can

After knowing x=11(mod13), we can immediately know x=89 as the ones of x^13 is 9, which means x can only be 9(mod10)

@mathsandsciencechannel

3 жыл бұрын

NICE. HE IS MY INSPIRATION.

@ishaangupta2185

6 ай бұрын

Same way I solved it!

This method has already been explained by many, but I think I can explain it more intuitively: The last digit of x^n will only depend on the last digit of x. Since we see x^13 is not even, we know x is not even. If x ends in 3 in 3, the final digit of its powers goes in cycles of 3,9,7,1. On the 13th step the last digit would be 3. If x ends in 5 it is always 5 If x ends in 7 it goes in cycles of 7,9,3,1, with the 13th step being 7. If x ends in 9 it goes in cycles of 9,1, with the 13th step being 9, so we see x must end in 9. We can also do some very simple checks to see that 80

@texchip977

7 ай бұрын

How do we know x is not divisible by 3 again?

@rbarreira2

7 ай бұрын

​@@texchip977You add the digits of the original number and check if the sum is divisible by 3.

@ahmadmuchlasabrar5856

7 ай бұрын

How to check x>80?

@derickdntss

7 ай бұрын

⁠@@ahmadmuchlasabrar5856consider 80 as 2 to the 3rd power * 10, then raise to 13th power and check that it is smaller than 2 to the 26th, therefore it cant be smaller than 80.

@Blackrobe

7 ай бұрын

Are you sure if x ends in 7 it goes in cycle of 7 1 7 1 7 1 etc? I checked 7...9...3...1

Finally a question from my country. Respect from iran :)

@soumyajitraha7682

3 жыл бұрын

🇮🇳💓💓🇮🇷

@abraham4124

3 жыл бұрын

As it should be earlier. Thanks for your great videos @letsthinkcritically

@aashsyed1277

3 жыл бұрын

I LIVE IN PAKISTAN. CAN YOU DO A VIDEO WHERE YOU SOLVE A PAKISTANI MATHEMATICAL OLYMPIAD?

@ryanjagpal9457

3 жыл бұрын

@@soumyajitraha7682 Wut

@Paapi_purush

3 жыл бұрын

@@aashsyed1277 😂😂😂😂 Pakistan m sirf terrorism ka course h ye Olympiad wagerah tumhare bas ka nahi

Very nice problem and very nice solution. I love the different parts of mathematics - well arithmetic and approximations - that make this work, and the nice reasoning behind the choices like checking 80 to the 13th. BRAVO!!!

The way you approached towards finding the soln is mindblowing. It is an eye-opener if you are innovative. The only thing I want to add is finding remainder when 26digit is divided by 13.Just beneath the 26digit number write down the digits from right to left ............4,3,-1,-4,-3,1 repeating again & again upto 26th digit.Multiply up and down digits &add them up. Finally divide the sum by 13 to get remainder 11.

Not only the problem but also your solution is very interesting! ❤️

Remember that if a is 1,3,7, or 9, then a^4 has last digit 1. Thus, a^12 has last digit 1, and a^13 has last digit a. Thus a = 9. Then, since 25 is close to 26, you can check quickly how close 90^13 is to the number. 9^4 = 6561 ~ 2/3 × 10^4, so 9^12 ~ 8/27 × 10^12 and 9^13 ~ 8/30 × 10^13 ~ 2.6 × 10^12. Thus the answer must be 89.

Here is another approach: The number of digits of N (assuming N is not a power of 10) is ceil(log(N)). So the number of digits of x^13 is ceil(13 log(x)). So 258^13 so 10^25>80^13 and finally 10^(25/13)>80. Thus x is 89 or 99. If x=99, then x^2=9801=1 (mod 100), and so x^13=99 (mod 100). Actually, if we write x=10a+b then x^13=30a+29 (mod 100) by binomial theorem and looking at powers of 9 (mod 100) (note that 9^10=1 (mod 100)). So we must have that x=89.

An elementary method: First, we know x < 100 by counting digits. Second, we know that if x^13 = 9 (mod 10) then x =. 9 (mod 10). Third, expand (10n + 9)^13 in binomials. Only the 9^13 term has non-zero ones digit. Only the 13*10n*9^12 and 9^13 terms have non-zero tens digit. Keeping only the tens and ones, we calculate by hand 9^13 = ...29 and subtract from ...69 in the question, we have 130n*9^12 = ...40. Since 9^12 = 1 (mod 10), we have 13n = 4 (mod 10). So n = 8(mod 10) and since x = 10n+9

Solved without using paper. 1. Obviously it is odd number. 2. Multiplication patterns for 1 is 1111, for 3 is 39713971, for 5 is 5555, for 7 is 79317931 and for 9 is 9191, so the last digit is 9 as it is the only one to give 9 in the 13th position in the pattern. 3. It is smaller than 99^13 and bigger than 80^13, so it must be 89. In the third step I made out of my hand estimates 99^13 approx 100^13 - 13*100^12 (Pascal triangle) and 8^13=2^39 =2^40/2 approx 1000^4/2

Just from using the binomial theorem and truncating at linear order you can estimate x to be around 90.

Awesome solution! Congratulations!

To do quickly in you head only: Number is 26 digits, so we know our number is approxumately one-fifth of 100^13. So surely x is something just a little under 100. Also, we know the last digit is 9. So our number is probably 90s or 80s, maybe 70s and ending in an odd number. But it cannot be ending in 1 or 5 for obvious reasons. 3 ends with pattern 3, 9, 7, 1, 3... so the 13th iteration is 3. So 3 fails. 7 iterates with 7, 9, 3, 1, 7... so the 13th iteration is 7, also fails. Our number is guaranteed to be oen of 99, 89, or 79 then. 69 is, by inspection, surely too low. We can also cross out 99 as being too high, as 0.99^13 is surely greater than 0.2. Our answer is probably 89 or 79. But 0.8^13 reduces rather quickly, as 0.8^2 is already 0.64, and 0.8^4 is already around 0.36, so we will reach 0.2 far too soon. Thus 89 is the answer. This whole time we use quick estimates and intuitive guessing, with only a tiny bit of mental math in iterating 3s and 7s.

@romeisbig6485

2 жыл бұрын

Sorry But, why did u take 0.2?

@georgebrantley776

2 жыл бұрын

@@romeisbig6485 The number we are finding is x^13 ~= 0.2 × 100^13

Beautiful problem. Thanks a lot.

1. Observe that x^13 ~ 2x10^25 2. Observe that x^13 ends with 9 only if x ends with 9 3. Check log (base 10) 90 = 1+2log3 = 1+2×0.477 = 1.954 Hence 13 log 90 ~ 13×1.95 ~ 25.35 Therefore 90^13 is between 2×10^25 and 3x10^25 So naturally, it's 89.

In power 13, the last digit remains the same. This number is less than (100*13) which is a 1 following by 26 zeros, this one is 26-digit-long. We can use bounds to notice it is between 80^13 and 90^13. So, it is 89.

@GauravPandeyIISc

3 жыл бұрын

How do you prove that the last digit remains the same in power 13?

@alainrogez8485

3 жыл бұрын

@@GauravPandeyIISc i use arithmetics in N/10N and check all the remainders. 0^13 = 0 (mod 10) 1^13= 1 (mod 10) 5^13 = 5 (mod 10) 6^13 = 6 (mod 10) For the obvious. For 2, we have the série 2-4-8-6 which it repeats itself. The 13th is a 2. So 2^13= 2 (mod 10) For 3, we have 13 threes, with 2 of them, we have a 9. With two 9, we have a 1. The 13th 3 which is alone is the remainder. So 3^13 = 3 (mod 10) For 4, we have 13 fours. Two of them give a 6. All 6 together give a 6. With the final 4, we have 6*4=24 so the remainder is 4. 4^13= 4 (mod 10) 7 = -3 (mod 10) so 7^13 = -3^13 = -3 = 7 (mod 10) 8^13 = -2^13 = -2 = 8 (mod 10) 9^13 = -1^13 = -1 = 9 (mod 10)

@m4riel

3 жыл бұрын

@@GauravPandeyIISc A better way is Euler's theorem. if a and b are coprime, then: a^(totient(b)) = 1 (mod b) where totient(b) is a function which gives the number of coprimes under b. •totient(10)=4 (only 1,3,7,9, are coprime to 10) so: a^4 = 1 (mod 10) (a^4)³ = a^12 = 1³ = 1 (mod 10) a^13 = a (mod 10) Since 2 and 5 are not coprime to 10, you can check them individually and notice how their powers will never end in a 9.

@zanti4132

3 жыл бұрын

@@m4riel It follows from your analysis that for any positive integer n, the last digit of n^k remains the same for any k that is 1 more than a multiple of 4. Hence, it would follow that n^(m+4) - n^m is divisible by 10, which isn't hard to prove: factor this as n^m (n^4 - 1) = n^m (n² + 1)(n² - 1) = n^m (n² + 1)(n + 1)(n - 1). Then show that at least one of these factors must be divisible by 2, and also that one of these factors must be divisible by 5.

@GauravPandeyIISc

3 жыл бұрын

@@m4riel This is awesome :)

watching and learning, well done

The last 2 digits of 69 are equivalent of -31 in mod 100. Power of 13 (last digit 3) means x must be equivalent of -11. As you say has to be less than 100 so answer must be 89. I've never studied any of this other than in my own head. Using the method above I was able to work it out in my head in about 15 seconds. But admittedly many other similar size calculations would be a lot harder but this was a nice straightforward set of numbers to make the calculations easier.

Solved in head even without a pen. 1. 26-digit result means that x has 2 digits. 2. The last number could be only 9, because 3 and 7 in 13 power give 3 and 7 respectively. Other digits never give 9 at the end position. 3. Sum of all 26 digits is equal to 107. 107 = 8 (mod 9). So, the sum of digits in x also has a reminder 8 when divided by 9. This is possible only when the first digit is 8. So the only possible value for x is 89.

@zachariastsampasidis8880

Ай бұрын

Most simple yet efficient solution. Bounding and checking for other reminders is moot. Well done

I did it in my head using logarithms. Since I grew up in the era before electronic calculators, I have memorized a couple of logarithms: lg(2) = 0.3010 (lg being log base 10) ln(2) = 0.693 (ln being the natural log) I started by noticing that the big number is approximately 0.2*10^26 This means that x can be written as a * 10^2, where a^13 = 0.2 (approx). So a is slightly smaller than 1. How can we calculate a? Since lg(2) = 0.3, we know that lg(0.2) = 0.3 - 1 = -0.7 Hence lg(a) = -0.7 / 13 = -0.05, roughly. Actually, I would like to have ln(a) instead of lg(a). I could get that by multiplying by ln(10), but unfortunately I never memorized ln(10). (Or I have forgotten it.) But wait! lg(2) * ln(10) = ln(2) => ln(10) = ln(2) / lg(2) = 0.69 / 0.3 = 2.3 Hence ln(a) = -0.05 * 2.3 = -0.115 Knowing that exp(y) is approximately 1 + y if y is small, we get that a = 1 - 0.115 = 0.885 So if x = 100 * a, then x should be either 88 or 89. It obviously can’t be an even number, since x^13 ends with a 9. But 89 fits perfectly. It is easy to check that an odd power of a number that ends with a 9 will also end with a 9.

@mayam3072

3 жыл бұрын

That’s really nice

3 жыл бұрын

You are boring. Nerd

@mayam3072

3 жыл бұрын

@ This was a video on math idk what you expected

@BenDRobinson

7 ай бұрын

Rather than decimal approximations to logarithms, I used what I knew from music, where the different intervals give you rational approximations to powers of 2^(1/12), or conversely rational approximations to various base-2 logarithms.

Also since 1001 is -1 mod 1000; you can alternate addition and subtraction every 3 digits 369-013+484-330...

Last digit is 9 and it’s something about 90 (can be easily seen by logarithm). Thus it’s 89. Couple minutes even without paper.

Thx for a nice video. I think, at the end you should also check that x=89 actually satisfies the equation, after you found that no other solution except 89 exist. Also, if I encountered such a problem on a olimpiad, I would just devide this long number by 13 to find the remainder, as 26 digits is not so much and did not required any "notice, that:" stuff except Ferma theorem. In video solution required like 10 small divisions, which is not my easier.

This is also a good Chinese Remainder Theorem problem. We know x < 100, so computing x mod 100 is enough. But knowing x mod 4 and 25 is sufficient to know it mod 100, so solving x^13 = 1 mod 4 and x^13 = 14 mod 25 is fairly straightforward.

@chrisclub3185

7 ай бұрын

x^13 = 19 has multiple solutions mod 25 2 and 11 being two at least, I didn’t go through everything though so there might even be more.

Note that for any x one has 1) x^13 = x (mod 5) 2) x^13 = x (mod 8) Now assume that x^13 = (this monstrous number) This gives us that x has to be congruent to 4 (mod 5) and congruent to 369 = 1 (mod 8). This is equivalent to x = 9 (mod 40) by chinese remainder theorem. So the only possible values are x = 9, 49, 89 (note that x

The faster way of doing this is just approximating the log, dividing by 13, then approximating 10 to the result. That gets you right around 90. Checking the last digit is all it takes to confirm the solution is 89.

@BenDRobinson

7 ай бұрын

This is what I did, helped by some musically informed logarithms. Specifically, x^13 is easily seen to be quite close to 26 semitones below 100^13, so x is really close to 2 semitones below 100, which gets you almost exactly to 89. And as you say, last digit check does the rest.

Wow u guys are so good at this😮

Since x is a positive integer, Taking square roots, and if necessary taking the cube root using long division. 13 = 2*6 + 1 6 = 2*3 3 = 2*1 + 1

x^13=9 mod 10 so x is coprime to 10. Thus x^4=1 mod 10 by Euler's theorem, so x^13 = x = 9 mod 10. By adding up the digits, we conclude x^13= 107 = 8 mod 9. As before, this means x=8 mod 9. By the Chinese Remainder Theorem this has a unique solution mod 90, which we see is x=89 mod 90. but 179 is clearly too big because the number has 26 (if I counted correctly lol) digits, and 100^13 already has 27 digits. So x=89

How can you divide by 10^6 when it is not a multiple of 13?

It has been shown by you that x^13=x (mod 13). Finding the remainder when RHS is divided by 13 comes 11.It means that x= 11 (mod 13). In otherwords we may write x= 13n+11 where n = 1,2,3,4,5,6,7. as x

Knowing that the last digit of x has to be 9. And that that nunmber has to be something close to 100 I checked if the big number was a multiple of 9, as it wasnt then x couldnt be 99, and 79 was just too far away to intuitively give a 26 digit number.

If you could use logarithm table, you quickly find its about 89. And if you analyse last digit, you also qiuckly find, that 9 is the only one possibility.

after you find last digit of x must be 9 and x^13 > 20^13. you can see that x^13 congruent to 5 mod 7, which implies that x is congruent to 5 mod 7. you can then check all numbers ending in 9 from 29 to 99 which satisfy this property. turns out only 89 does.

As you said x

watching last digit, you can come to conclusion that x must end with 9, then you look amount of digit to restrict what approx. value has x

I noticed that the units digit of x had to be 9, as others have mentioned. Then I let. x=10k-1. Then, I'll skip the details, I just had to look at the last two terms of the binomial expansion of (10k-1)^13 to get a 6 in the tens position.

_So x power 2 and 3 is just casual so we give u x power 13 XD_ - Olympiad literally

I started with the same bounding condition (

@zachariastsampasidis8880

Ай бұрын

You would have to prove by hand that the number is equal to 89^13, even if you found the suspected 13th root to be possibly only 89

No need to reduce mod difficult numbers like 13 or 1001. Using Euler's generalization of Fermat's little theorem, we know that x^13 = 9 (mod 10), so reducing the exponent mod phi(10) = 4 gives x = 9 (mod 10). To get x^13 mod 9, just add up the digits and reduce them mod 9, which gives 8 mod 9. Phi(9) = 6, so reducing the exponent gives x = 8 (mod 9). Here's the trick: For both, x is -1 modulo that number. Since 9 and 10 are coprime, then x is also -1 mod 9*10 = 90. So x is 89 mod 90.

Wow! What a great vid :3

The number 2.198 * 10^25 is quite close to 10^26 which would is equal to 100^13 (x=100). However, we need only approximately 22% of that. The number in question has to be odd, but to end with 9 it cannot be 1 or 5 either. We can exclude 3 and 7 by calculating the last digit of the 12 multiplications while ignoring the higher digits. We get "3" as the last digit for 3, and "7" for 7. So the last digit of x has to be 9. 99 is too close to 100^13 (every one of the 13 multiplicands lower by 1%, so at maximum 13% lower). 89 is worth a test and maybe 79 as well. However, 79 is way too low: 0.79^2 is less than 0.8^2 which is 0.64. 0.79^4 therefore is less than 0.7^2 which is 0.49. 0.79^8 is less than 0.25. Another 0.79^4 gets us down to 0.125 for 0.79^12 and therefore less than 0.12 for 0.79^13. The number is less than 12% of 100^13, but we need 22%. So in fact only 89 remains and that works.

@BenDRobinson

7 ай бұрын

You are spot on... with a bit of practice at this type of estimation it really doesn't take long to narrow it down. For someone to prefer going entirely down the number theory path suggests a certain fear of "getting one's hands dirty" with a little bit of rough computation.

This can be done a lot simpler using Euler's Theorem. x^13 is congruent to x mod 10 and 9. Our number is clearly congruent to 9 mod 10, so our answer ends in 9. The sum of the digits is 8 mod 9, so the sum of digits in the answer is also 8 mod 9. Our answer is 89.

@lgooch

2 жыл бұрын

Why is the sum of digits in our answer also 8 (mod 9), does that hold with any number?

@dujas2

2 жыл бұрын

@@lgooch A decimal number is equivalent to the sum of its digits (mod 9). For example, 453=4(10^2)+5(10)+3. Since 10 is equivalent to 1 (mod 9), those 10s can be replaced by 1s to get 453 is equivalent to 4+5+3 (mod 9). 9 and 10 were chosen for 2 reasons. 1. It's easy to figure out what large numbers are equivalent to with these moduli and 2. their totients divide (13-1). x^13 = x(x^6)^2 which is equivalent to x (mod 9) by Euler's Theorem, assuming x and 9 are coprime which they are since x^13 and 9 are coprime.

@lgooch

2 жыл бұрын

@@dujas2 ohhh thanks so much for clearing this up, I deleted my reply asking another question but I just read your reply again

Solve for the trajectory of the missile, you can ignore friction.

Two cool facts about 13th powers is that a) the answer is approximately 13 times the length of the number; and b) that the last digit is always the last digit of the number we’re after… Given this, we can confidently say this is a two digit number ending with 9 (26 digits long, very convenient!), given 100^13 has 27 digits! The digit sum is not divisible by 3 or 9, so that leaves 39, 69 and 99 out… Another trick to know is that for 13th powers the second to last digit (which we can call x) can be used to determine the second to last digit of the root… in the case of a digit ending with 9, the value of the root’s second to last (technically first) digit would be 7(x-2) mod 10… equal to 7 x (6-2) mod 10 = 8… that would mean in our case that the 13th root of our number is 89…

It has 26 digits, meaning x is a two-digit number. The power is a 4k+1 odd number that can also be written as 5k+4, so x is also 4k+1. According to Fermat's little theorem, x^4-1 is always divisible by 5 if x isn't, making x^12 a 5k+1, so x == x^13 (mod 5). We can figure out x can be written as 20k+9, and only 89 is large enough for the equation.

@leif1075

2 жыл бұрын

Woah assumption outbof nowhere..how do you know it's 2 digit just from seeing its 26 digits? You're just guessing right? Or youbfirst tried 10 to the 9th or something

@handanyldzhan9232

2 жыл бұрын

@@leif1075 26-digit numbers are between 10^25 and 10^26 (not included), so x must be less than 10^2, but obviously more than 10 if we are to divide powers into 13.

@leif1075

2 жыл бұрын

@@handanyldzhan9232 bitnyoubsaidnthat or assumed it like it like was obvious when it's not..you know what I mean?

@handanyldzhan9232

2 жыл бұрын

@@leif1075 I just assumed people would understand how I got there. I just couldn't bother to write the explanation above. Sorry about that.

at 6:00 you can also check it by cyclic power of unit digit 😃

Rather than working mod 13, i believe Fermat's little theorem should be used for small primes 2, 3 and 5. It immediately gives x = x^13 = -1 mod 30 and then you can discard 59 and 29 for being too small

@swenji9113

6 ай бұрын

No computation needed

The last digit of x^13 is a 9, it is easy to figure that therefore x must also end by 9. Also x has to be less than 100, therefore only 2 digits. Now, x^13 is approx 22.10^24, and 100 x^13 is 22 .10^26. With some simple power law maths you get that x is approx 100.(22/100)^(1/13). A crude first order taylor expansion gets x is approx 94, and actually less than that because the next order taylor terms are all negative. That makes 89 a good contender. Finally you can check that 80 is smaller than x. 80^13=8^13 . 10^13. 8^13 is 2^39. 2 ^10 is approx 1000 therfore 2^39 is approx 0.5 x 10^12, and finally 80^13 is about 5 10^24. 80 is smaller than x, therefore it has to be 89.

When you divided by 10^6, the results were equivalent to portions of the original value…?

My solution (done mentally): 1) x^13 > 100^13, because 100^13 has 27 digits. 2) Nine is the only mod 10 solution that ends in 9 when raised to the 13th power. 3) x != 99, because 99^13 = 99 (mod 100). 4) x > 79, because 79^13 Therefore, x=89.

I found it easiest to compute x^13 = 8 mod 9 (just add the digits mod 9) and x^13 = 4 mod 5. By Fermat's little theorem we have that x^13 = x mod 9 and x^13 = x mod 5, so we therefore have x = 8 mod 9 and x = 4 mod 5. From these we get that x = 44 mod 45. Then since 10^25 < x^13 < 10^26, it's trivial that 50 < x < 100. Therefore x can only be 89.

@pedrojose392

3 жыл бұрын

Very beautiful solution! I have to walk a lot. But to show tha x>50, we have to show that 25/13 > 1,7 (log50). It woul be easier if we note that x is odd.

@xiaoyang1519

3 жыл бұрын

Did you mean the generalised Fermat little theorem? The one using Euler orient function?

@charlottedarroch

3 жыл бұрын

@@xiaoyang1519 Sure. Whatever you'd like to call it. I just think about it as Lagrange's theorem applied to the group of units modulo n.

@charlottedarroch

3 жыл бұрын

@@pedrojose392 Showing that x > 50 is easy. 50^13 = 100^13/2^13 = 10^26/2^13 < 10^26/2^4 = 10^26/16 < 10^26/10 = 10^25 < x^13

@pedrojose392

3 жыл бұрын

@@charlottedarroch , good evening! I did not say that it is not easy, but it is easier to choose a odd or a even number. That is the point.

After figuring out the numbers in the last one can argue that the last digit of the x^13 is 9 I.e. an odd power of x is ending in 9... That is possible only if x itself ends in 9... So the only answer is 89... As it was supposed to be done by hand 2^13*3 need not be calculated... And we don't need to prove that it is greater than 80...

@mathsandsciencechannel

3 жыл бұрын

GOOD.HE IS MY INSPIRATION

@clickrick

3 жыл бұрын

That final digit was certainly where I started too.

@nitish18tayal

3 жыл бұрын

@@clickrick #metoo... ;)

Here's how i did it. 1st part: I approximated x^13 to 2.2*10^25, logx~25+log2+log11-1 because log2~0.3010, log3~0.4771, log7~0.8451, log11~1.0414 (we were taught to memorize this in high school in Taiwan, don't ask me why lol), so log(x^13)~25.3, logx~25.3/13~1.946, 1

All of the stuff with 1001 seems really involved... is that really faster than just doing 26 steps of long division to determine the remainder?

@zachariastsampasidis8880

Ай бұрын

It's a lot faster to write down, you can split a 26 digit number into 9 parts of at most 3 digits, to get at most a 4 digit number where long division can be applied comfortably

I wonder when the DBS students would come and raid the channel lol :) but anyways nice video

The number is between 10^25 and 10^26, with a log of about 25.3, so the 13th root will have a log of about 1.95 or so. The last digit must be 9 because φ(10)=4, so x^13=x (mod 10), if x is co-prime (which it has to be because x=/=5), digits don't sum to a multiple of 9 so can't be 99. log(79)

Number is 0.219*100^13, let y^13 be 0.219 (x will be 100*y). upon checkin it is clear that x ends with digit 9, also hand calculting 0.8^7 gives 0.209 which is less than 0.219 => y is greater than 0.8 and as the number is not divisible by three x cannot be 99 so y =0.89 and hence x=89

Why can we ignore the remainders from the powers of 10?

Just wondering. Just because 89 is the only number less than 100 and greater 80, that satisfies the modulo arithmetic, doesn't mean 89^13 is actually the number on the rhs, or? Shouldn't we really check whether 89^13 is the number in question? Adding 130 to the right hand side, would give the same conclusion, but it is not a 13th power...

I got the last digit of the solution easily, and the boundary of the first digit (10-90, with the right answer leaning to the high end.) I found 8, but not with the proper rigor. I am not seeing how people are rigorously determining the 8. For example, if the 8th digit from the left was an 8, who here would have noticed the problem had no solution?

My first clue is to find the last digit. For this case, only 9 fits that requirement.

THANK YOU VERY MUCH SIR FOR YOUR INSPIRATION AND YOUR TEACHINGS.

I said that 3^2 is 9 so 3^4 last digit is 1. Then, 3^12 also ending with 1, 3^13 last digit is 3. If 7^2=49, so same with 7.. 7^12 ends with 1, 7^13 last digit must be 7. after showing number must be smaller than 100, i showed for each case why it cant be the number above, untill i left with 89.

Im not very good but i solve it using last digit analysis and divisibility test.. that sure reduce a lot of possible answer..

what if the answer is not integer

We can find unit digit and we don't have to do the last part.

by mod 10&using euler phi Fnc, x=9(mod 10),,, 09,19,29.........99 probable x.(x

No need to break the number down into 6-digit blocks if you observe that x^13 = 9 (mod 10) implies x = 9 (mod 10). Then using your proofs that 80 < x < 100, the only choices are 89 and 99, and 99^13 = -1 (mod 100) so it has to be 89. But of course, I wouldn't have thought of this on my own. :-)

Well, this is my way: I remember that log2=0.3 and log3 is slightly less than 0.5,so I take log base 10 and you get logx is about (25.3)/13=1.946, and I can esitimate that log80=1+3*0.3=1.9, and log90=1+2*log3 which is slightly less than 2, so we know that x is greater than 80 but smaller than 90. and we can know from the unit digit and the exponential rule that this number should end with 7 or 9, plus this number is not divisible by 3 which is easy to check, so we know that this number must be 89.

I first noted that, sinceit thad 26 digits, x would have to be between 10 and 100. Then I considered the cases (n*10)^13>10^25 for n between 1 and 9, reducing it to n^13>10^12. Since 12 and 13 are very close, I figured n would have to be almost 10, like 8 or 9, and decided to check n=7: 7^3=343

@mathsandsciencechannel

3 жыл бұрын

GOOD.HE IS MY INSPIRATION

By examining the last digit of x^13, we know that the last digit of x is either 3, 7 or 9. Powers of a number that ends with 3 go: 3 ; 9 ; 7 ; 1 and so on, which means to the power 13 it'd be 3. For 7, it goes 7 ; 9 ; 3 ; 1, it would be 7. 9 goes 9 ; 1 ; 9 ; 1 etc... this one works. The last digit of x is 9. Starting here, I've went for estimations. 10^25 50^13 = 10^26 / 2^13. We know that 2^10 > 10^3, then we know that 50^13 I tried something similar with 80: 80^13 = 2^39 x 10^13, but 2^39 I checked that by writing (2^20)^2 That means 80^13 x is either 89 or 99, but 99^2 = 9801, which means the last two digits of 99^n alternate between 99 and 01, which is not the case of x^13. Then, x = 89.

I first proved that that number is -1 mod 3 and -1 mod 10. Hence it is -1 mod 30, and so is its thirteenth root. So the number is 29^13, 59^13 or 89^13, since 100^13 has 27 digits and this number has 26 digits. But 50^13 = 100^13 / 2^13 is a 27 digit number divided by 8192 (a 4-digit number), so 50^13 has at most 24 digits. This rules out 29^13. Moreover, 60^13 = 10^13 x 2^13 x 3^13. I worked out 3^13 by hand which gave me a 7 digit number, and of course 10^13 x 2^13 = 8192 x 10^13 having 17 digits. Thus 60^13 has at most 17+7=24 digits, so it is ruled out as well. The number must thus be 89^13.

I got to the answer quite quickly using some music theory... We have x^13 ~= 0.22 x 100^13, so x/100 is the 13th root of ~0.22, which is about 7/8 of 1/4. That is to say, musically the ratio 0.22 is about negative 2 octaves + 2 semitones, or -26 semitones. Therefore x is very close to 2 semitones (26/13) below 100, i.e. x~=89. Looking at the tail of x^13, obviously x has to be odd and some simple number theory shows that it must in fact end in 9, so indeed x=89.

We desire two primes p and q with Euler totients a factor of 12 so that by Euler's Theorem x≡ x (mod p) and x≡ x (mod q) both hold; and that have convenient divisibility rules. The pair 5 (with φ(5) = 4) and 9 (with φ(9) = 6) work. Then note, using their well known divisibility rules, that: x≡ 4 (mod 5) and x≡ 2 (mod 9); and hence by the Chinese Remainder Theorem x≡ 44 (mod 5*8 = 45). Since x Hence x = 89.

@TheodoreBrown314

2 жыл бұрын

I’m curious now: Chinese Remainder Theorem?

@pietergeerkens6324

2 жыл бұрын

@@TheodoreBrown314 A basic result in introductory Number Theory: "... if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime." - en.wikipedia.org/wiki/Chinese_remainder_theorem

@TheodoreBrown314

2 жыл бұрын

@@pietergeerkens6324 Ah, that's neat! I (somewhat) intuitively knew that, but I haven't seen the mathematical definition before. I will keep it in the back of my head! For anyone else who's having troubles reading the wiki page, I suggest this YT video as a better explanation: kzread.info/dash/bejne/pKlrz7mTes-4gco.html

@pietergeerkens6324

2 жыл бұрын

@@TheodoreBrown314 I've rewritten my comment to be correct. I'm not sure what I was smoking when I thought 11 worked, as it's totient 10 is not a factor of 13--1 = 12. However use of 9 and 5 is even simpler than use of 9 and 11. Thank you for drawing my attention back to this comment - so I could correct it.

@leif1075

2 жыл бұрын

@@pietergeerkens6324 but whybthinknof primes at all? Youbdont know if x is prime so seeks out of nowhere..because you are trying to decompose it?

Yay a question from my country :D

@mathsandsciencechannel

3 жыл бұрын

YEP.HE IS MY INSPIRATION

0.2…*10^26, therefore the answer is a little less than 100, because 26/13=2, and 10^2=100

i think i have the weirdest solution here it is ...1)unit digit cyclicity 2)found unit digit of x must be 9(out of 3 choices 3,7,9).3)checked 20^13 does not have required no of digits and 100^13 has 27 digits so 20

It took me 20 seconds to guess 89. The last digit must be 9 and it's close to 100^13 but not too close as 99.

I calculated it in my head - calculated roughly a logarithm of 2E27 to know where to look for answers, less than a 100 and later found out that last digit must be 9 to match the 9 in the result, 99 is too much - around 85 E26 (99% ^13) so it must be 89 or no solution if the composer was mean :). 90% ^13 is roughly 0.9 * 0.8^2 * 0.64^2 ~= 0.9 * 0.2

the equation could have no solution. a canidate that might be a solution is easily found, but the canidate has not been checked by hand as the title asks. is there a typo in the rephrasing?

@dudacasagrande

3 жыл бұрын

Rightly put. It's still a long way to prove that 89^13 is indeed that big number. And I don't think there's too many shortcuts to this calculation...

@ihti20

3 жыл бұрын

@@dudacasagrande Not a big deal. Multiply 89 by numbers from 2 to 9 to use as a hint: *2=178 *3=267 *4=356 *5=445 *6=534 *7=623 *8=712 *9=801. Then elementary school division(I don't know how it's called in English, we call it column division) gives: 21 982145 917308 330487 013369 /89 246990 403565 262140 303521 /89 2775 173073 766990 340489 /89 31 181719 929966 183601 /89 350356 403707 485209 /89 3936 588805 702081/89 44 231334 895529/89 496981 290961/89 5584 059449/89 62 742241/89 704969/89 7921/89 89. Proven.

Knowing unit digit is 9 and answer is below hundred and not a multiple of 3 Final possibilities are 19,29,49,59,79,89 Now you can calculate how many digits 80^13 is by taking log with base 10 [log80¹³] +1 = 13(log8+log10) +1 = [13(1.9)] +1 which gives 25 digits but answer has 26 digits so 89 is answer

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

They way I did it was to check the ones digit. x^13 ends in a 9, and only 9^13 ends in a 9. This required a simple calculation for all cases. Then, taking the logarithm base 10 of both sides gave me 13log(x) = 25.something. x^13 has 26 digits, so its log base 10 is above 25 and below 26. Meaning 1

@leif1075

2 жыл бұрын

Why log base 10nim curiousbindid the seducing the last digit is 9 but why base 10 it seems random to me..

Let's see. We have 26 digits. This is slightly less than 10^26, so I know that the original number is less than 100, but not by much. The last digit is 9, this tells me that with the 13th power, last digit of x is also 9. So it realistically can only be 99 or 89. I would just check those, starting with 99.

So the method was basically trial and error, but less trial and error

I set x^13 = 69 mod 100 This gives us x^13 = 1 mod 4 and x^13 = 19 mod 25, which isn’t too difficult to do. After this, you get x = 89 mod 100, and since 0 < x < 100, x must be 89.

Good proof to illus cryptography...and based on 1001 you can take 3 digits %3 (alternate + - )

@wise_math

2 жыл бұрын

cryptography yes, nice.

Using arguments other people have already made, we can see that x must be a two-digit number where the second digit is 9. Now, what if we consider the final TWO digits of the numbers on each side? Express x as 10y-1 for some y. Expanding (10y-1)**13 modulo 100, we see that we can neglect any power of 10y above 2. Therefore: 69 = (10y-1)**13 modulo 100 Observing that 10y**n modulo 100 is 0 for n>1, we can discard most terms of the binomial expansion: 69 = 13 * (10y)**1 * (-1)**12 + (-1)**13 modulo 100 69 = 130y - 1 modulo 100 70 = 30y modulo 100 7 = 3y modulo 10 By inspection, y can only be 9, so x = 10*9-1 - 89

X^c = N Then, X is greater than Or equal to 10^[(D-1)÷c] And x is less than Or equal to 10^C[(D-1)÷c] Where D is number of digits in 'N' and C signifies ceiling function of [(D-1)÷c]. In this case, x is greater than Or equal to 84 and less than Or equal to 100.

I think 83. The sum of digits of our number M is 0 mod3, so the x must be some power of 3. Also we know M has 26 digids in base 10. Despite to the (3^(n*13) = 10^(log(3)*13*n), and log(3) is something abowe 0.5, we can write: 26=0.5*13*n, so n = 4. 3^4 = 81.

It took me longer but here is how I did it. I first assumed that x had to be more than 0 and less than 100. I knew that x could not end in 0 or 1 or 5 since the result ended in 9. Also it could not end in an even digit. So x had to end in 3, 7, or 9. Next I squared 3 twice to get 81 and dropped the 8 off to get a 1. I knew that if x ended in 3 and was put to the 12th power then that result would end in a 1 since 81 cubed would end in a 1. Then I multiplied 1 by 3 and concluded that if x ended in 3 then x^13 had to end in 3, too. That eliminated x from ending in a 3. When trying 7, I got 7^4 ending in a 1 and knew that 7^12 also ended in 1 and 7^13 ended in 7. That eliminated 7. When squaring 9, I got 81. If you put a number that ends in 1 to the 6th power, you get a 1 at the end of that result, so 9^12 has to end in a 1 and 9^13 has to end in a 9 and x has to end in a 9. I then added all the digits of the number up to get a sum of 107. Since 107 is not a multiple of 3; I figured that I could eliminate 9, 39, 69, and 99. I then took 19, 29, 49, 59, 79, and 89 and put all of those through the following process. I squared the number and dropped off anything past the tens place. I then did this two more times. I multiplied the results after the 2nd and 3rd times to figure out what 2 digits x^12 would end in. I then multipled that result to the number I started with and compared that to the last two digits of that large number. 19, 29, 49, 59, and 79 did not prove to end in the same two digits; but 89 did and I concluded that x = 89.

This problem can be solve in less than a minute using binomial expansion and mod 100

Quick proof: One has that x^13 is 1 mod 2, 4 mod 5 and 8 mod 9 Hence x is 1 mod 2 and since φ(5) and φ(9) both divide 12, the fermat euler theorem immediately gives that x is also 4 mod 5 and 8 mod 9 The number has 26 digits while 100^13 has 27, so is less than 100 All of these facts together immediately goves x = 89

Иззи. 1)Извлекаем корень столбиком.Разделим число на группы по значению степени.В данном примере две группы по 13. Значит число двухзначное. 2)Последняя цифра пятой степени числа равна последней цифре числа. Выводим закономерностьстепени+4 будут обладать таким свойством,то есть 9 и 13.значит последняя цифра числа 9 3)80 в 13 степени=10в 13 степени х 8 в 13 степени = 2 в 39 степени.Округлим до 40 .2 в 10 степени=1024 или примерно 10 в третьей степени.Итого 2 в 40 степени примерно равно 10 в12 степени 80 в 13 степени примерно равно 10 в 25 степени.Что меньше искомого числа. 4)делаем вывод искомое число больше 80,но меньше 99.Так число 99 будет близко к числу 100. Ответ искомое число 89.