Is 81 the ONLY Two-Digit Number with This Property?
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- Опубліковано 11 кві 2024
- In this video we try to answer the simple question using basic knowledge on quadratic equation and number theory.
Notice that √81 = 8 + 1. Are there any other two-digit numbers satisfying this property? (square root being equal to the sum of two digits)
#NumberTheory #DiophantineEquations #IntegerRoot
The follow-up video is released, dealing with √(4-digit) = (2-digit) + (2-digit) pattern: • √81 = 8+1. What About ...
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Check out the follow-up video of this video, where I attempt to find 4-digit numbers xyzw satisfying √(xyzw) = xy + zw: ua-cam.com/video/2q1ruls5vU8/v-deo.htmlsi=HR2j6soLsQp3-ke5
You should also mention that this result is in decimal system. More interesting would be to generalize this to any number system, in which case i suspect you'd end up with an equation as a result.
Well at least I can say that for an arbitrary base n, two-digit number (n - 2)1 base n (n - 2 as n's digit and 1 as 1's digit, base n) ALWAYS satisfies the given property - that is,
√(11 base 3) = 1 + 1, √(21 base 4) = 2 + 1, √(31 base 5) = 3 + 1, and so on.
This is because (2-digit number (n - 2)1 base n) = n(n - 2) + 1 = (n - 1)^2, so its square root is n - 1 = (n - 2) + 1.
I don't know if there are other possible numbers though, and I don't think I will explore that ever (the equation gets way complicated due to the additional variable for the base).
I enjoyed your method of proving the question. Thanks for sharing!
Thanks! This is one of the most well-known methods to deal with quadratic equation with integer roots.
Or you could have just tested all perfect 2 digit squares (there's only 6 of them)
Yeah even we don't get to explore the theory of quadratic equations with integer roots, that is perhaps the fastest way to do it.
The answer isn't the point. Learning how to solve the problem is the point.
@@tonyennis1787 I get what you mean but my solution is also a way to solve it. However, the solution shown in the video becomes increasingly better for larger numbers but this "obvious" method is more efficient for 2 digit squares
Very good
Thx❤
Can you do something similar for 3 or 4 digits?
Because the solution to this problem is too easy to find by just checking in your head.
Well, I wasn't going to dive too deep into this, but since you asked:
I haven't tried cases like √(xyz) = x + y + z or √(xyzw) = x + y + z + w, because they already seem too complicated due to the increased number of variables,
but I managed to handle √(xyzw) = xy + zw pattern, because this too only requires 2 variables and can be expressed as
√(100a + b) = a + b, where a and b are now two-digit numbers maximum.
Squaring both sides and doing similar process, we obtain
a = - (b - 50) ± √(2500 - 99b)
so we have
2500 - 99b = k^2 (k is integer) and 2500 - 99b ≥ 0 (equivalent to b ≤ 25.xxx)
This time there are slightly more cases, and it is actually beneficial to use b = (2500 - k^2)/99 = (50 + k)(50 - k)/99 instead, testing integer values k that makes b integers.
Then we can find that only possible values are k = 5 and 49, which gives b = 25 and 1, which in turn gives
(a, b) = (20, 25), (30, 25), (98, 1)
Therefore, there are three cases (if you include single-digit number such as 01):
√2025 = 20 + 25 = 45,
√3025 = 30 + 25 = 55,
√9801 = 98 + 01 = 99.
Now I'm wondering if I should make an additional video with these results.
@@CornerstonesOfMath Can't you show that max face sum of a 5 digit number is 54 and even (54)^2
@@ShivanshPachnanda It can be easily shown that it is impossible from 5 digits because for a 5-digit number abcde, √(abcde) ≤ 9+9+9+9+9 = 45 and thus abcde ≤ 45^2 = 2025.
And I think I can show you that even 4-digit numbers (abcd) are impossible since abcd ≤ (9+9+9+9)^2 = 1296, so the value of a is limited to a = 1 only, but then we have 1bcd ≤ (1+9+9+9)^2 = 784, which is impossible.
If we use cube root ³√ or 4th root ⁴√ then there can be different stories, but again, I just don't think that this problem is worth expanding.
@@CornerstonesOfMath
edit: I was 1h late with the N>=5 proof
there are no solutions for N-digit numbers where n > 2. I checked up to 15 with a python script.
1: [1]
2: [81]
3: []
4: []
5: []
6: []
...
Proof for N >= 5 is easy, because
(N * 9)^2 < 10^N if N >= 5
But the cases for 3,4 are less trivial.
I would love to see a video on this conversation thread
very noice
Thank you, you're noice too