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Solve the Diophantine Equation 2^a + 2^b + 2^c = 2320
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[arghadeepdas]

Please pin my comment sir
Your teaching and the sum both are awesome

[zsoltszigeti]

In binary format it looks like this: 10^a+10^b+10^c=100100010000, so a=4, b=8, c=11

[sharonmarshall]

I just wrote the powers of2 up to 2*11 which is 2048, then subtracted from 2320 which gave a difference of 272 and from there split this into two powers of 2 namely 16 and 256 ie 2*4 and 2*8. Since a<b<c, a=4, b=8 and c=11.

[qstzhhg]

2320 (decimal system)=100100010000 (binary system), so a=4, b=8, c=11.

[Qbe_Root]

Comments are saying 2320 is 100100010000 in binary like it's trivial, really you can just find the greatest power of 2 less than 2320, which is 2048, and subtract it off to get 272. Then repeat to get 256 and 16, so now you know 2320 = 2048 + 256 + 16. Find the exponents (for big ones, it helps to know that 2^10 ≈ 1000) and you 4, 8 and 11.

How do you know powers of 2? Just play 2048 a lot 🙃

[pavlozemskyy]

1) If numbers a, b and c is a solution to the equation, then any of their permutations will be a solution as well (e.g. b, c, a). Therefore, let us assume that solution is always written in descent order, a>=b>=c;
2) a < 12, because 2^12 = 4096 > 2320;
3) a should be either 10 or 11, because if a<=9 then max sum of 2^a + 2^b + 2^c = 3 * 2^9 = 3 * 512 < 2320;
4) if a = 10, then b should be 10 as well, because otherwise max sum of 2^b + 2^c = 2 * 2^9 = 2 * 512 < (2320 - 2^10) = 1296;
5) but a = 10 and b = 10 gives us 2^10 + 2^10, which is 2048, 2320 - 2048 = 272 which is not a power of 2;
6) if a = 11, then 2^11 = 2048, 2320 - 2^11 = 272, 272 = 256 + 16 = 2^8 + 2^4;
Therefore a=11, b=8, c=4

And there are no other solutions since a can be 11 only and the last digit of 272 is 2.

The only way to get 2^x + 2^y mod 10 = 2 is to have 2^x mod 10 = 6 and 2^y mod 10 = 6.
The only two options in the range of 0 - 272 which fit the conditions are 16 and 256.

[pavelshank]

honestly it can be solve very easily without this operations. Simply take 2320 and split like sums of numbers. You’ll see that 2 power 11 = 2048 then 272= 256+16, therefore we can replace it with sums of 2 power 8 = 256 and + 2 power 4 = 16. At the end we got correct result. Anyway your method is great for scholars.

[d.yousefsobh]

Sir thanks for your approach.l can solve very easily.let b=a+n. C=b+m=a+n+m.

[j.r.]

So much "manipulation" is used to restate this equation that I'm not sure it's any less legitimate to simply "chunk" 2320 as 2048 + 256 + 16. Then the values for a, b, and c are obvious. It's certainly a faster approach.

[MrPaulc]

I don't know about Diophanine equations, but do know that 2048 is a power of 2 and that leaves 272. 256 and 16 are also powers of 2.
2^11 + 2^8 + 2^4. a, b, c are 11, 8, and 4 in any order. There may well be other solutions.
Now I will watch the video to see what Diophantine is about.

[keithwood]

Sir, When you see a problem like this for the first time, is your approach to solving it apparent to you immediately? Because I look at a problem like that and have no idea how to proceed, and when you start your initial work on it I have no idea why you are doing your first steps. I follow your work as though walking through a forest without a trail, following someone that knows the way already.

[quirinoswensson]

Math is magic. Congratulations for you nice solution. Thanks.

[alster]

Tricky at first but after seeing the technique in solving for a, it became easy

[ujwalsmanhas]

if a > b > c then c =4, b =8, a = 11 .

[dgvqyer]

If (a, b, c) є N is true,
But if (a<b<c) fail logic.

[GirGir]

3:40 Ho can you assume that 2^a on the left side is equal to 2^4 from the right side?

[allangriffiths]

Sorry to be pedantic but you need to define a, b and c to be integers, otherwise there are an infinite number of solutions. E.g. if a = 1, b =2, c would equal log 2314/log2.

[wesleysuen]

no way, the answer can be spotted in an instinct and it has to be unique!!

[ganeshrajk]

IF 2023 and all the exponents are different, how will you solve

[parvjain]

I try to solve without your solution after some efforts i finally solved it.🤗
I have an amazing problem which I am not able to solve how i can give it to you 😔
So that you will bring me the solution.😊