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Excellent approach. I like the way you factor the left hand side of the equation. You could have considered all factorizations of 1241 to show that 3 is the only solution. You used the factorization 17x73 to show that 3 is a solution of the equation.
The solution as presented needs to exclude the points at x=4, x=12, and x=4/3. But it also needs to separately include the cases for x=-2 and x=2, which are missing as is.
Hi, I also solved it with the same method but how do you prove that 823543=7^7? In my solution I use this idea: the last term is 3 so the repeated number could be 3 or 7 because they are the only numbers that if repeated generate a 3: for 3 the pattern is 3^(1+4k) and for 7 it is 7^(3+4k). So 3^3 is not a solution because 1+4k cannot be 3 (if k is natural), but if you do it for 7 you find the solution because 3+4k is equal to 7 when k is 1, so the result is 7 ^7. The idea could also be extended to numbers that contain 3 or 7 as a final term such as 13, 27 or 10023...I thought that with 10^10 we exceed the digits of 823543 and this allows us to stay under 10
because f(a,b) = f(b,a), i started by using a = 2+ x, b=2 -x ... solutions for x will be in equal and opposite pairs, and this substitution into the second given will result in a quartic. that quartic must have factors of the form (c - x^2) only because of the first observation, meaning it is really quadratic in x^2. distributing and combining (2+x)^5 + (2-x)^5 is demeaning and time-consuming, but is most of the work in the problem.
www.youtube.com/@walterwen2975, Wow! That's a nice solution!!! I'm saving it and I will post a video solving a similar problem in two different ways soon. Thank you!
@@walterwen2975 ok you found a solution after quite a bit of mental juggling and how do you know you got all the solutions? That part is still missing tho.
If a person learns sequences of quartics, just like most remember sequences of squares, {1,4,9,...}. By mental recall of memorized arithmetic: Whatever the sum be, the sum of 6 and 7's as squares will be too small, as a sum of quartic's, HA, a single 6 OR 7 is too much. Edit start: If that has been said already: I paused at @0:03 - took longer to type then solve.
yeah who was, at secondary school, advised to memorize • all squares from 11 up to 20 . • all powers under 1024 of 2 up to 10 . so incl. 2¹⁰ , 3⁶ , 4⁵ , 5⁴ , 6³ ,7³ , 8³ , 9³ . that is , in the question is immediately visible 343+216. then indeed considering that 6^m and 7^m are monotonously increasing functions of m, meaning that the found m = 3 should be the only possible answer .
To verify you can just multiply with the standard algorythm: 34^2 = 30*34+4*34 = 1020+136 = 1156 . There's no need of looking for algebraic identities since it's an arithmetic problem that doesn't require further generalisation.
@@lucho2868 of course you can, but if you know rules for nearby perfect squares than my way is quicker. You might as well say that you didn't need to use your knowledge that 32^2 = 1024 in the first place!
@@kicorse You also used 1024^.5=32. In fact you wrote a proof around the fact that that 32^2=1024. And your answer is formally incorrect since you didn't even proove that 32^2=1024. Your solution is useless because it requires differences, elementary algebra and (over all that) a long mutiplication whilst mine only require the long multiplication to check at the end. Making it shorter, more rigurous, and more generally applycable than yours.
@@kicorse for example, if you wanted approximate the value of any square root by hand the long multiplication's result of a nearby approximation could be used as an input in a Bakhshali-kind formula to get a good approximation result whilst the method you suggest would work less generally.
1156 is between 900 and 1600 and squared numbers that end in 6 are either X4 or X6 so the answer is either 34 or 36 and after that you just need to check these
20=(8343-7^a)^(1/a), now i think this, if 20^k (k is a natural number) finishes with a 0 for all k, it means that 8343-7^m has to finished with a 0. So 7^m has to finished with a 3 because the last 3 of 8343 minus the last term of 7^m has to be equal 0. Now I study the pattern of 7^k and it is: 7^1=7, 7^2=...9, 7^3=...3, 7^4=...1, 7^5=...7. Now i know that every 4 times the pattern restart and i know that for 7^3 i have a number that finished with a 3. So 7^a=7^(3+4n) where n is natural. But for n>1 8343-7^a is negative so my only option is n=0 that means a=3 and if you try to put a=3 in 20^a+7^a it becomes 8343
Niiiiiiceeeee solution, i find another process: 6=(559-7^m)^(1/m), now i think this, if 6^k (k is natural number) 6^k has to finished with a 6, for example 6*6=36, 36*6=216,..., it means that 559-7^m has to finished with a 6. So 7^m has to finished with a 3 because the 9 of 559 minus the last term of 7^m has to be equal 6. Now I study the pattern of 7^k and it is: 7^1=7, 7^2=...9, 7^3=...3, 7^4=...1, 7^5=...7. Now i know that every 4 times the pattern restart and i know that for 7^3 i have a number that finished with a 3. So 7^m=7^(3+4n) where n is natural. But for n>0 559-7^m is negative so my only option is n=0 that means m=3 and if you try to put m=3 in 6^m+7^m it becomes 559
1156 is a very long way from being a "huge number". As others have pointed out, this answer takes only a few seconds to work out in your head (divide by 2, divide by 2, left with 289 which is 17^2, answer is 2*17). Also, you didn't need to multipy by 4/4, what you had was (10^2 + 2) ^ 2, giving the result of 102/3 after the square root which is 34. Also, also, what's with the unrelated #-tags? The only one relevant to this video is #squareroot...
My approach is simply to find the prime factors. It's obviously divisible twice by 2, leaving us only to find the square root of 289. It's also obviously not divisible by 3. 5 can be rules out as 5^4 is 625 which means we must hop that there's an exact square root of 289, or it gets very messy. It's clearly greater than 13 squared (169) which most people know, and the next prime is then 17, the square of which is 289. So, we have sqrt(2*2*17*17) = 34. nb. I don't think merits being an olympiad question.
I used the "by hand" method, which only took two steps = less than a minute. Grade 4 students used to be taught this method, which is similar to long division.
It is waay simpler to solve by converting it to binary. Find the largest power of 2 that fits in 400 -> 2^8 = 256 Same for the remaining 144 -> 2^7 = 128 Same for the remaining 16 -> 2^4 The approach shown in the video serves no better purpose, it’s neither faster nor more general
I tried beginning to solve it before opening the video, my method was to try to find the prime factors of 1156 to if possible remove most of the work from the square root, found that you could divide it by 2 two times wich gives 2√289, then I had to think for a bit longer until I reached 17 wich fits perfectly, giving √1156 = 34, ngl I was not expecting the answer to be an integer
What?!?! I figured out that it is 34 in about 2 seconds, in my head. 30 x 30 = 900.....40 x 40 = 1600. So the number must fall between 30 and 40. Only 4 squared and 6 squared end in 6, so it has to be one of those two numbers in the singles position. Since 1156 is much closer to 900 than 1600, the singles number must be 4. Therefore 34 is the answer. Multiplied 34 x 34 to verify--also in my head. I am not a mathematician or an engineer, and I am 74 years old. Your method is far too complicated. I stopped watching after about a minute.
You msde a mistake, the real solutions are (5, 2) amd (-2,-5) You are also making a mistake with the complex solutions, x^2 -3x+19=0, x= 3/2 \pm i sqrt(67)/2.
how about this: let 3^(2x-1/x+1) = t 3^(3x-x-1/x+1)=t 3^(3x/x+1)=3t thus, 3^(5x-1/x+1) = t・3t =3t^2 which means that 3t^2+t-30=0 t must be real so the only solution is t=3 hence 2x-1/x+1=1 i.e. x=2
on the one hand I don't like problems where imagination has go soooo far. I am a teacher and would not have found every step. Would have solved the quadratic in a different way. However. This is an example I really like to teach my students. It's too complex to solve it alone. but its a perfect example to train all different kind of methods and keep an eye open to all of them at once. Learning this problem by heart is in itself a good lesson. So thank you. I copied it already in latex to make a lesson out of it.
27 and 3 are the only possible values. so equate powers to (3,1) and (1,3) and see which value yields a coherent solution. thjnk about it powers of 3 are 1,3,9,27. any power higher than that cannot yield a sum of 30. and we cannot have irrational numbers as the sum of two irrational numbers canot be rational(integer). so fractional powrrs are out of the picture. and, negative powers are also out of picture. you cannot possiblt have 1/3 and 89/3, both are not power of 3.
A Physicist(or any common sense person) will realize the positive part just crescent and so x should be small, just throwing integers will find x=2 and it is the only one. Even if x was not integer is just throw integers on it until the left size would be bigger than the left. Why spend 12 minutes at it? It takes about 7 to 10 seconds. You don't need to be a Ramanujan to solve this.
Sorry man, but when you are rewriting to (a-b)*(a+b)=12 and saying that (a-b) = 1 or 2 or 3 and (a+b) = 12 or 6 or 4 your are making mistake. (a-b) or (a+b) are integer values only for even values of 'x'. So you are solving the equation for even 'x'. If you take odd 'x' then (x^x)/2 will be non-integer, e.g. x=3 -> (a+b) = 3^13.5+3^1.5; or different value: x=5 -> (a+b) = 5^1562.5+5^2.5. This isn't integer. Yeah, you got lucky with this math problem that both your multipliers (a+b) and (a-b) are integers and that for growing x>1 you get ever growing function of f(x)=x^x^x-x^x. But for another similar problem like x^x^x-x^x=7625597484960 this your trick wouldn't work. (a+b)*(a-b)=7625597484960 doesn't imply, that (a+b) is integer and (a-b) is integer. For this case, where x=3, this will be 2761453.635828058 * 2761443.2435232126 = 7625597484960.
