Why are there only 7 numbers like this? Proving it with Python code, with help from ChatGPT

This viral trick has tens of millions of views, and leads to a natural question: which numbers have cube roots exactly equal to the number's digit sum? We can tackle this question mathematically, with an assist from ChatGPT and Python code, and figure out all possible values!

Sum of cube root digits short video

Dudeney number

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OEIS Dudeney numbers

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[skattterdude]

Why am I not surprised that the invocation of ChatGPT results in an extra minute and a half of debugging/code optimization talk

[Voxel]

Here's few math puzzles I have made...

31²=961, reverse order of numbers in value to get 169. Square it to get 13. Reverse again to get 31 (original number)
If zero is banned there's only about 50 number that behaves like this. Largest is
...Easy to proof but fun to calculate on paper ;P

sqrt(x²-1) can always be cleaned with odd numbers. What I mean for that cleaning: Example sqrt(7²-1)=sqrt(48)=4*sqrt(3) <= Cleaned 4 out from sqrt()
... Proof of it is so simple, but hard to find.

Calculate (1+i)↑↑3 on paper
... Yes, I calculated that... Should I get mental health care because the answer is...
..
.

😁

[lnpmyvd]

Nice video, keep it up.
I want to show another proof:
when the sum of the digits is at most 54 (because there are 6 or less digits), so it is enough to check all the perfect cubes from 0, then 1 and up to 54^3.
When there is a natural number 2≤k so that 10^(3k)≤n<10^(3k+3)
Will set the cube root of n to be C(n).
Therefore, C(10^(3k))≤C(n),
So 10^k≤C(n)
Set the sum of the digits to be S(n). Because n has at most 3k+2 digits, so S(n)≤9(3k+2)=27k+18
Finally, for all 2≤k we get
S(n)≤27k+18<10^k≤C(n)
Which makes S(n)<C(n), Which means they cannot be equal, so all numbers n bigger or equal to a milliion do not satisfy what we want.

[Grassmpl]

Any number x^3 that works must have x^3=x(mod 9). This reduces the check to x=0, 1, 8 mod 9. To see this. 9|x(x+1)(x-1). Only one of these factors is a multiple of 3, which means that factor has to be a multiple of 9.

Now we only need x=0, 1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46, 53, 54.

19 possibilities.
The examples listed in the video indeed all are on my list.

[sparshsharma]

I hope that you took serious note of you not publishing general videos as before. Loved this video. I was missing your explanations. I don't like shorts, they are just for fun or pleasure but not for serious or education. Happy that you are back in the business.

[iydtxpo]

8:12 you can also write “num = num ** 3” as “num **= 3”

[MathOrient]

Nice number theory problem. There are many such problems. Great video. Concise and educative.

[pbenikovszky]

Nice video, but is this really a complicated proof? The homework proof is just showing that if f(x0)>g(x0), and both f and g are strictly increasing then f(x)>g(x) for all x > x0. The other part is just writing an equation for the sum of the digits of a number. And then finally ask AI to write code to test literally just 54 numbers for you :D

[Grassmpl]

To compute the digit sum, can also write

sum(map(int, digits))

[bjorntorlarsson]

I looked at the Wiki page on prime number groups. I like the Wall-Sun-Sun primes. They are the prime:th order of the Fibonacci numbers, plus or minus one (according to some rule I don't get) which are also primes. Anyway, they are easy to remember! Because none has been found yet. How is this helpful?

[MichaelPiz]

That incorrect space makes all the difference. 😁

[advaykumar]

There is a channel called Bhannat Maths, who steals all your videos
Edit: He copied this video as well

[JohnLeePettimoreIII]

i find it interesting that all of the numbers are in pairs, with the exception of 8.
0, 1
17, 18
26, 27

[mikeyallen]

Isnt there a more efficient proof? We know the sum of the numbers has to be less than 54 so cant we do
for n <= 54
cube = n**3
digitSum = sum(int(d) for d in str(n))
if (digitSum == n)
print(cube, "is equal to sum of digits:", digitSum, "^3")

Am i missing something or would that work too?

[aryanyadav]

729 cube root is not possible by your trick 😮😮

[gabrielpetre]

Any reason why the numbers come in pairs? 01, 17, 18 and 26 27?

[almightyhydra]

Does the final code work - if num is replaced by its cube, doesn't it short-circuit the loop? Not sure how Python works in that regard.

[cjslime]

I wonder if you can do this for higher roots,

[avinashsingh]

Can you prove that common chord of two circles bisects the intersecting portion?

[bruinjim]

A nice problem, but the solution is incomplete. The moment a number is not treated as a single concept, such as summing the digits, the base assumption of the numbering system becomes uncertain. What is the solution for base 9 numbers? Base 8? What if numbers start with numerous 0's, which invalidate your range assumption. Your solution is valid but it makes assumptions on how numbers are presented and their the number base. Part of this can be corrected by asserting that numbers are base 10.