How To Solve 2^m-2^n=8064 | Math Olympiad | Algebra 1 | Math Challenge. 

Thanks for this observation. Well received and it shall be implemented in subsequent videos. Respect sir. ...👍👍👍

Iam very weak in math. I must be doing something wrong. Is n= log base 2(8064/(2^k-1)) ? Where if K=6 in both n and m, 2 will be raised the power of 5? I could not get n=7 and m=13. Help! Thanks

@@billl3936 Yes, n=log base 2 (8064/(2^k-1)). If you plug in k=6 you get n=log base2 (8064/(2^6-1)) 2^6-1=64-1=63 and 8064/63=128=2^7 finally log base 2 (2^7)=7 So n=7 and you get m with the same method.

Absolutely! In each of 3 videos, which I watched so far - such antimath "maybes". Shitty stuff. Blocking.

yes, the condition about integers is mandatory

Thanks a million ma.

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In the factored form (@3:10) you know that 2^n must be a power of 2, and therefore from the set {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096}. You also know that 2^k - 1 is always an odd number. So, our factor pairs (2^n, 2^k - 1) are from the set {(2, 4032), (4, 2016), (8, 1008), (16, 504), (32, 252), (64, 126), (128, 63)} - the remaining factors are non-integer and are disqualified. The only factor pair that matchs the requirement for an integer pair (Even, Odd) is (128, 63) so it is the only factor pair that needs to be checked. J'espere que cela t'aider, et je te souhaite bonne chance mon ami.

By expressing 8064 as a product of its prime factors and putting base two into consideration because both bases are in base two.

Thanks for this reply on behalf of onlinemathstv @Andrew Layton. Thanks and much love sir...💕💕💖💖💖

@@andrewlayton9760 thanks a lot mon ami.

@@onlineMathsTV Thanks for reply

Sorry am lost at your procedure sir. I will appreciate it more if you can put more light to this approach because I can see a wonderful and a unique method from it sir. Thanking you in advance sir.

@@onlineMathsTV I'll try to explain in simpler terms :) (When the number is in binary, I'll prefix it with b, that is, the binary 1101 will be written as b1101) To write a number in binary, we can only use 0 and 1, for example, 2 in binary would be b10; 3 would be b11 and 5 would be b101. As in base ten we have 1234 = 1*10³ + 2*10² + 3*10¹ + 4*10⁰ In binary we have b1010 = 1*2³ + 0*2² + 1*2¹ + 0*2⁰ What hapens if we have a power of two in binary? Well, the same way in base 10 we have 10³=1000, in binary, 2³ = b1000. What if, in base 10, we subtract two powers of 10? For example, 10⁶ - 10⁴ = 1,000,000 - 10,000 = 990,000. Note that the answer has 6 digits in total, and 4 trailing zeroes. This will hapen for any two powers of ten (you can see the why trying to subtract by the subtraction algorithm). And the inverse? How to find m and n such 10^m - 10^n = 999,900,000? Well, m=9 and n=7, we can do by just looking at the number! (Note: if the equation was 10^m - 10^n = 9876, we don't have integer solutions) The same thing happens in binary: 2⁴ - 2² = b10000 - b100 = b1100, 4 digits and 2 zeroes (the subtraction algorithm will work in a similar way) and, in a more general way, 2^m - 2^n will have m digits and n trailing zeroes. So, if we convert 8064 to binary, finding m and n would be trivial: just count the number of digits and the number of trailing zeroes. (Note that if instead of 8064 we had 5 = b101, there would be no integer solutions) To convert 8064 to binary, we could simply use a binary calculator (not so elegant uh?), or use an algorithm to do so. The algorithm consists of continuously divide the number by two, then read the remainders backwoods. Its kinda hard to explain in text, but the process is pretty simple, there are many videos explaining how to convert decimals to binary. After converting, we find 8064 = b1,1111,1000,0000 13 digits, 7 trailing zeroes

What a nice solution!

great also

Respect sir. You are great at what you do sir. Respect!!!

Sure, it shows you are the master here🙏🙏🙏

Thanks a million and we are glad you gained some values from this video tutorial because, this is our utmost priority sir. We promise to give the best in all our teachings in order to serve you and other better by His grace. We love you sir...❤️❤️💖😍💕🙋🙋

Noted sir, thanks

Simply consider the base numbers on the LHS

Thanks

You alright bro?

Noted sir. Thanks for this educative comment. Respect 🙏🙏

Thanks a million sir/ma.

Thanks for the encouragement sir. We love you ❤️❤️💕💕

infinite

@@onlineMathsTV ok.

Bravo 👍👍👍 Respect 🙏🙏🙏

Thanks for asking this hidden question sir. We know that by putting into consideration the existing base on the LHS.

You are most welcome sir. We are glad to have you here and at this time happy that you gained some values from our video tutorial. Much love from everyone here sir.💖💖💖

Thanks for the complement sir. Much love and more grace to carry on the good works in our times. Love you sir....❤️❤️

@@onlineMathsTV Thank u for ur beautiful words.l hope u a lot of success and progress ❤️👍

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Thanks for these words of encouragement sir. Much love❤️❤️💕💕💖💖

@@onlineMathsTV a

A

Thanks sir. We are glad you gained some values. Much love 💖💖💖

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Many thanks to you sir

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