Math Olympiad Great Challenge Without a Calculator | Geendle

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[@play_with_Vpp]

X = 2, solved in 10 sec in my mind.

[@danielyousif8186]

Next to useless, if you don't have a systematic approach.

[@play_with_Vpp]

@@danielyousif8186 bro your point of view is absolutely right, but for competative exam you have to solve question as quick as possible. I am from India and I have cleared CUET and got rank 3 college for MCA. In this exam you have to solve 1 question in a minute so logically I have to solve the question as fast as I can.

[@scottpelak1856]

Same!

[@Ikkarson]

@@play_with_Vppany proper comp math exam should look at whether you proved your result. Result without proof are worthless. Usually, these problems would be designed so that : - one can’t just pull the answer without a proper proof ("the proof is in the pudding" case) - obvious answers mask other hidden cases that are required to get the points In this case, merely saying that 2 is one answer gives no clue whether it is the only one, until you can prove it.

[@Geendle]

The idea was to find a way to show that x=2. Thanks bro!

[@frenchmarty7446]

You can guess this one pretty quickly as others have pointed out. We know X is an integer. If you understand how negative exponents work you'll know a negative integer won't work and our answer will be a unique positive integer. X can't be very big because X^X^X explodes very rapidly. X = 4 is 4^256 for example. X^X doesn't grow nearly as fast, so the difference is going to be around the same order of magnitude as the first term: X^X^X. This puts a very low upper bound on values worth trying. X has to be greater than 1. It's trivial mental math to see X = 1 results in 0 for the left side. Pretty quickly you'll guess X = 2 and if you check you'll find that it is. You might try X = 3 but you'll see at a glance that 3^27 - 27 >> 12 without having to go further.

[@Geendle]

Thanks, bro. I like you analysis.

[@farnorthbear3046]

My first thought was to write 12 as 'A - B' to directly compare components, where A could be written as c^c^c, and B could be written as c^c. After some initial guessing, you can conclude that 16 - 4 would be a good match as 16=2^2^2 and 4=2^2, by comparing the left and right side of the equation, where x^x^x - x^x = 2^2^2 - 2^2, x must be equal to 2.

[@Geendle]

Thank you, bro! That's nice!

[@tejpalsingh366]

On solving 4^2-4&(-3)^2-(-3) gives solns .. but power value cant be -ve hence first x=2 only works

[@Geendle]

Yes! That's the answer. Thanks!

[@NicholasBertollo]

I argued that x^x is an increasing function for x >= 1, and then x^(increasing function of x) must also be a increasing function of x, and for x >= 1, x^x 12, in which case you know that x = 2 is the only solution. So yeah not too hard of a question tbh.

[@abhirupkundu2778]

Or take x^x = t. Now we have a variable polynomial in terms of t. The equation is: t^x-t-12=0. Through observation, x=2 satisfies this polynomial. Because as soon as you put x=2, you get 2 factors of the quadratic in t, which are: t=-3 and t=4(through middle term factoring). Now, since x is an integer, we will check for which solution satisfies the term for an integer. Any integer to the power of itself never is equal to -3. And x^x = 4 is satisfied only by x=2 when x is an integer. So our observation was correct, and the ans is x=2

[@user-yt4fn9pj8l]

@@abhirupkundu2778 it's not t^x-t-12=0, it's x^t-t-12=0, it's another equation.

[@abhirupkundu2778]

@@user-yt4fn9pj8l tf u mean, did u even read what I said. I said x^x as t. Then we get this equation. Ofc we can get another one which is x^t -t=12 which is another equation but it is extremely complex when solving.

[@NicholasBertollo]

⁠@@abhirupkundu2778 that user is correct x^x^x != (x^x)^x and so you cannot substitute t=x^x into x^x^x to get t^x because this would imply x^x^x = (x^x)^x. Note that exponentiation is right associative, so x^x^x = x^(x^x). So you can only get x^t after the substitution, not t^x. But other than that your solution doesn’t prove uniqueness of x=2.

[@Geendle]

Thanks for the analysis bro!I appreciate!

[@handlebar4520]

x=2

[@Geendle]

www.youtube.com/@handlebar4520 thank you!

[@chadlj]

can you find the value of x^2+1/x^2 if x-1/x=2?

[@whtveru]

(x - 1/x)² = 2² x² + 1/x² -2 = 4 x² + 1/x² = 6

[@Geendle]

www.youtube.com/@chadlj, www.youtube.com/@whtveru has answered your question in he comment above. Thank you. Whatever doubt, please send me.

[@Geendle]

Dear www.youtube.com/@chadlj, thank you!

[@chadlj]

@@Geendle Alright!

[@abhirupkundu2778]

​@@whtveruUh where did you get -2 from? Shouldn't you be getting (x-1)^2/x^2 on squaring both sides. Then expanding (x-1)^2 we get x^2+1-2x and dividing by x^2 we get x^2+1/x^2 -2/x.

[@jacopomaccione7791]

Hi, you could have removed a few steps at the last by saying that since x^x^x and x^x must both be even or odd then x^((x^x)/2) and x^(x/2) must also both be even or odd and for this the you could use the step with 2 and 6

[@Geendle]

Thanks for the analysis bro!I appreciate!

[@jacopomaccione7791]

@@Geendle ❤

[@Sasha123-d1q]

X=2

[@Geendle]

www.youtube.com/@Sasha123-d1q Thank you!

[@gazamidori2866]

A=x^x A^2-A-12=0 (A+3)(A-4)=0 => A= 3,4 => x^x = 3,4 x^x=4 x=2 | x^x = 3 √2 < x < 2 Idk how to solve it better than that

[@jacopomaccione7791]

bro x^x^x is not (x^x)^2

[@abhirupkundu2778]

​@@jacopomaccione7791bro, it was a variable polynomial in terms of A. When you put x=2 through observation, you get 2 roots of the quadratic in A which are -3 and 4. Now an integer solution would be x^x=2 ; x=2 and hence our assumption was correct. The reason you cannot put any number higher than 2 in x, is because you cannot get integer solutions for the polynomial in A.

[@user-yt4fn9pj8l]

​@@abhirupkundu2778 hey bro, the rewriting of the statement x^x^x - x^x into the form (x^x)^2 - x^x is incorrect. If you don't see it immediately, just check for x=3: 3^3^3 - 3^3 = 7625597484960 (3^3)^2 - 3^3 = 729 - 27 = 702 The guy in the main comment solves another math problem, not the given in the video.

[@gazamidori2866]

@@jacopomaccione7791 yea you right mb I totally missed that. I tried and failed but I still learned thank you for correcting my error

[@jacopomaccione7791]

@@gazamidori2866 ❤

[@user-yt4fn9pj8l]

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.

[@Geendle]

I like your analysis. Thank you for sharing your ideas!