Would have gone with for a in range(1,10): for b in range(0,10): for c in range(0,10): if a*100+b*10+c == a + b**2 + c**3: print(f'{a}{b}{c} = {a} + {b}^2 + {c}^3') But your solution is fine too :)
The three digit problem has four 'proper' solutions (135, 175, 518, 598) along with three improper solutions (1, 43, 63). Extending the problem to four digits with the sum being f(n=abcd) = a+b^2+c^3+c^4 gives three proper solutions (1306, 1676, 2427) along with the trivial improper solution of 1. How many solutions might there be for the equivalent five digit problem?
Just for fun, I employed Mathematica to find higher numbers that satisfy this pattern. Surprisingly, no 5-digit 6-digit numbers with this property, but I found a single 7-digit number: 2,646,798 = 2+6^2+4^3+6^4+7^5+9^6+8^7. I wonder if there is infinitely many numbers that satisfy this. I find it more logical to make the ones carry power of 1 and increase the power with the decimal place to the left, so like f(123) = 1^3 + 2^2 + 3. I found no numbers that would satisfy f(n) = n with this definition, up to 10,000,000 (except 1 through 9 of course). Is it possible that no such number exist for n > 9?
I like the idea of developing a formalized version of the problem, where for any m digit number, we take the sum: leftmost non-zero digit raised to the first power, plus second from the left digit raised to the 2nd power, and so on... until the final term in the sum is the r-th power of the r-th digit, i.e., the rightmost digit. If r is the number of digits of the positive integer whose leftmost digit is non-zero, then for any number m, r is the number such that Floor[log_10(m)] = Floor[ln(m) / ln(10)] = r. The leftmost digit is then a_1 = Floor[m / 10^r]. Now let b_0 = m and and define also b_1 = b_0 - (a_1 * 10^r) The second digit in m would then be a_2 = Floor[log_10(b_1)]. To find the third digit (assuming r≥2 here), we would first take b_2 = b_1 - (a_2 * 10^(r-2)) and then have a_3 = Floor[log_10(b_2)] and so on until the final term is found from a_r = Floor[log_10(b_(r-1))] where b_(r-1) = b_(r-2) - a_(r-1) * 10^1. Then the sum for f(m) would be given by Sum from k=1 to k=r of (a_k)^k. This a lot to unpack and it's super late right now, but I imagine (or hope) there might be some kind of functional equation that could be found to determine a_k without the need to compute each b_k, but since the Floor function is involved in the process, it might be either to messy to be practical, or perhaps not even possible. But I love the idea of the generalized version of the function with a domain of N (all of the natural numbers) and what properties and limits we could determine about it.
@@Kapomafioso Excellent! I also though of running the powers in the same direction as the powers of the base for the number - the attraction is it is slightly more intuitive to program ... but think of the mathematics for a second. It simply can't have anything other than the unit digit solutions. Take abcd = a(10^3)+b(10^2)+c(10)+d f(abcd) = a^4+b^3+c^2+d For abcd=f(abcd): a(10^3)+b(10^2)+c(10)+d = a^4+b^3+c^2+d =a(a^3)+b(b^2)+c(c)+d But 10^3>a^3, 10^2>b^2, 10>c, so for all abcd>9, f(abcd)>abcd.
@@Hiltok ohhh that's right! Okay, I looked into what if we shift the powers up by one (start with the second power for 1s, third power for 10s etc..) And I found a spooky coincidence! 1676 = 1^1 + 6^2 + 7^3 + 6^4 (just like in the video), but also! 1676 = 1^5 + 6^4 + 7^3 + 6^2 Whoah. Another one is 4,975,929 = 4^8+9^7+7^6+5^5+9^4+2^3+9^2 Again, wonder if there's infinitely many of these.
@@Kapomafioso There won't be infinite solutions. The right hand side (the digit power sum) grows slower than required. For instance you could show a bound of 23 digits: Even if it were all 9s, the f-value is upper bounded by 9^23 + 9^22 + ... which is smaller than the smallest 23 digit number (10^22). Intuitively, I'd expect the solutions in higher digits to be rarer, wouldn't be surprised if the 7-digit one is the last one.
When you wrote 99a+b(10-b)=(c-1)c(c+1) you can make the argument that 99a and (c-1)c(c+1) both are divisible by 3, the latter a product of three consecutive numbers, so b(10-b) has to be divisible by three also. Checking all cases can we get b=1,3,4,6,7,9. From there we have 3 cases, because we b(10-b) is the same for b=3,7 , b=1,9 and b=4,6. From there you can also do cases, use divisibility and inequalities
Since squares and digits are nowhere near a 3 digit answer, i suggest we check just the cubes only, given that the cubic can extend high enough to get in reach of a 3 digit sum. 19x (81+1) and then a cube, requires 19x-82 = 110~ cube root = 4.~ So we know we need at least a 5 at the end of the number to reach into the relevant 3 digit range. 5^3 = 125. We can use the other two digits to fill in the neccesary number, 1,3 = 10, therefore 135 is valid. 6^3 = 216. We need another +10. 2+what square = ends in a 0? None, so it doesn't exist, because 216+81 7 we need a +4. 3+what square ends in a 4?. 1^2,9^2 317 is too small,. 397 results in a number too big. Repeat question with 8 and 9 but with ending in a 3. 4+what square (8 or 9 only) ends in a 3? none. 8^3 = 512, 518, easy as pie. 9^3 = 729, no such number works as 7+square ends in 0. maybe 8 + square = 10? no. No number exists with 9^3 Finally, go back and check we didn't miss any duplicates, by checking the general formula now that we understand it better. 1+7^2 = 10, 175, 50+125=175, also works. 5+square = 6, 1^2, 9^2, 598. 512+81+5=598 So we have 135,175, 518,598.
Or generating n from its digits rather than extracting digits from n (I like this less): print([n for a in range(1, 10) for b in range(0, 10) for c in range(0, 10) if (n:=100*a + 10*b + c) == a + b*b + c*c*c])
But also 336 & 210 can be proposed to have solutions at a break off 39=333/πe then 336= 99.a+39 a=3 210=99+39πe/3 Or saying the equation becomes 99a+39(πe/3)^n For 210 a=1 & n=1 For 336 a=3 & n=0
It's as if the equation mutates as it goes on for 720= [(39+99/2) +39(π×e/3)^1]×4 That's why this is conjectural or you take the 2-6 solution & multiply it by the 5-120 solution . Thanks for your valuable input.
thats an interesting question, we need solutions of this equation a0*10^n + a1*10^(n-1) +... + a(n-1)*10 + an = a0 + a1^2 + a2^3 +...+ an^(n+1) to ur question, I think no, because 10^n>9^n, and the difference 10^n - 9^n increases, so for large enough n, we have LHS > RHS so we may have solutions for higher number of digits but it cant be infinite
First on-the-trot reaction (busy - don't ask) @ 1:15 just maybe ... maybe an instruction ... an order ... an appeal ... do a few more of these yourself. Do at least ten and if you notice a pattern emerging note it down and see if it matches pattern we see in video. Before trying to solve the problem do try to research it. Build up subconscious knowledge of the problem... gotta dash now ... hugz
Hi Michael, I've been watching your channel for a while and I've big fan of your videos! I have a problem suggestion that I thik could make for a good video idea. My Linear Algebra teacher had a couple of exercises about matrices and linear systems reduced mod p. 3 years later and I've never been able to solve one of these whch asked how many m by n matrices whose entries where reduced mod 2 had rank 1. I'm not sure if this is how you usually take problem suggestions, or if the problem is actually simple and I'm just being dumb, but I think the whole idea would make for a nice video.
This is fun for a 5-grader. Better tell us how to find a polynomial function whose values at all 3-digit numbers are equal to your function f(N)=a+b^2+c^3.
I couldn't find any additional periodic points for 2-digit and 3-digit functions. However, I did find the longest chains: 896 - 305 -> 128 995 - 215 -> 128 And then, the sequence continues as follows: 128 - 517 - 349 - 748 - 535 - 139 - 739 - 745 - 148 - 529 - 738 - 528 - 521 -> 10 = a good place to stop.
I tried looking for a vaguely similar but possibly more interesting or flexible alternative and I don't feel I have yet found one found one; however, 267 .. 2^1 + 6^3 + 7^2 = 2 + 216 + 49 = 267 .. which seems to be the only such 3 digit solution
First reaction is, why do the powers in that manner (left to right)? This forces knowing in advance the number of digits of the source number. Doing it right to left would create one single function that works for source numbers regardless of their number of digits... Weird
Because you would never get any solution besides single digit numbers. Notice that 10^n*x_n + ... + 10*x_1 + x_0 > x_n^(n+1) + ... + x_1^2 + x_0 for all n > 1 since x_0,...,x_n are digits and therefore smaller than 10
@neomorphicduck idiocy is yours... I probably have more math degrees than you do. One of math's interest is purpose, this specific problem is rather useless
Here's the list of solutions for digit count up to 8 (interestingly, there are no solutions for 5, 6, or 8 digit numbers): 89 135 175 518 598 1306 1676 2427 2646798 It would be interesting to categorize which digit counts do have solutions.
@@IanXMiller 😮 Have you worked on this problem before? How to find general solutions? I replied to another comment earlier with a rough sketch outlining how id go about formulating the general sum over N, but it's rather too long for me to reiterate here. I'll try to find it and cut & paste it as another reply here, and hopefully you can let me know if I'm on the right track? Id keep cracking at it tonight if it weren't already into the early morning hours and I've yet to go to bed.
@@graf_paper Here's a pasted version of my reply to another comment on this video: ************* I like the idea of developing a formalized version of the problem, where for any m digit number, we take the sum: leftmost non-zero digit raised to the first power, plus second from the left digit raised to the 2nd power, and so on... until the final term in the sum is the r-th power of the r-th digit, i.e., the rightmost digit. If r is the number of digits of the positive integer whose leftmost digit is non-zero, then for any number m, r is the number such that Floor[log_10(m)] = Floor[ln(m) / ln(10)] = r. The leftmost digit is then a_1 = Floor[m / 10^r]. Now let b_0 = m and and define also b_1 = b_0 - (a_1 * 10^r) The second digit in m would then be a_2 = Floor[log_10(b_1)]. To find the third digit (assuming r≥2 here), we would first take b_2 = b_1 - (a_2 * 10^(r-2)) and then have a_3 = Floor[log_10(b_2)] and so on until the final term is found from a_r = Floor[log_10(b_(r-1))] where b_(r-1) = b_(r-2) - a_(r-1) * 10^1. Then the sum for f(m) would be given by Sum from k=1 to k=r of (a_k)^k. This a lot to unpack and it's super late right now, but I imagine (or hope) there might be some kind of functional equation that could be found to determine a_k without the need to compute each b_k, but since the Floor function is involved in the process, it might be either to messy to be practical, or perhaps not even possible. But I love the idea of the generalized version of the function with a domain of N (all of the natural numbers) and what properties and limits we could determine about it. ***************
@@knivesoutcatchdamouse2137 considering how big the number is, he most likely found the number using a program written in like C or Python, but I might be wrong though.
there has to be a better approach than just listing cases. I do think it'll boil down to listing cases, but i think we should be able to do some analysis to reduce the no. of cases to check.
If the LHS is multiplication, and A, B, and C are digits with A != 0, then the problem have no solutions, I believe. But if the LHS is not multiplication, but the concatenation of digits, then isn't the problem trivial?
Proof for no solutions, assuming A,B,C are digits, and no leading zero. BC ≤ 81 ABC ≤ 81A < 100A + 10B + C otherwise, if leading zero(s) are allowed, then 0 is (trivially) the only valid solution.
An easy method readily springs to mind: write a program to test the cases. There's only 900, so you needn't even spend time looking for a subtle way to save time. Programming is skill not valued highly enough by mathematicians. If the person who composed the question didn't intend to test the pupils' proficiency in that skill, they shouldn't have composed a question like that.
for a in range(100,1000):
x = (a//100)
y = (a - x*100)//10
z = (a - x*100 - y*10)
if a == (x + y**2 + z**3):
print(a)
Would have gone with
for a in range(1,10):
for b in range(0,10):
for c in range(0,10):
if a*100+b*10+c == a + b**2 + c**3:
print(f'{a}{b}{c} = {a} + {b}^2 + {c}^3')
But your solution is fine too :)
100,999
The only other solutions seem to be 518 and 598.
Your pfp really is the best character in the game 😊 good choice!
135
@@EHMM this is one of the solutions already found.
The three digit problem has four 'proper' solutions (135, 175, 518, 598) along with three improper solutions (1, 43, 63).
Extending the problem to four digits with the sum being f(n=abcd) = a+b^2+c^3+c^4
gives three proper solutions (1306, 1676, 2427) along with the trivial improper solution of 1.
How many solutions might there be for the equivalent five digit problem?
Just for fun, I employed Mathematica to find higher numbers that satisfy this pattern.
Surprisingly, no 5-digit 6-digit numbers with this property, but I found a single 7-digit number: 2,646,798 = 2+6^2+4^3+6^4+7^5+9^6+8^7. I wonder if there is infinitely many numbers that satisfy this.
I find it more logical to make the ones carry power of 1 and increase the power with the decimal place to the left, so like f(123) = 1^3 + 2^2 + 3. I found no numbers that would satisfy f(n) = n with this definition, up to 10,000,000 (except 1 through 9 of course). Is it possible that no such number exist for n > 9?
I like the idea of developing a formalized version of the problem, where for any m digit number, we take the sum: leftmost non-zero digit raised to the first power, plus second from the left digit raised to the 2nd power, and so on... until the final term in the sum is the r-th power of the r-th digit, i.e., the rightmost digit.
If r is the number of digits of the positive integer whose leftmost digit is non-zero, then for any number m, r is the number such that
Floor[log_10(m)] = Floor[ln(m) / ln(10)] = r.
The leftmost digit is then
a_1 = Floor[m / 10^r].
Now let b_0 = m and and define also
b_1 = b_0 - (a_1 * 10^r)
The second digit in m would then be
a_2 = Floor[log_10(b_1)].
To find the third digit (assuming r≥2 here), we would first take
b_2 = b_1 - (a_2 * 10^(r-2))
and then have
a_3 = Floor[log_10(b_2)]
and so on until the final term is found from
a_r = Floor[log_10(b_(r-1))]
where
b_(r-1) = b_(r-2) - a_(r-1) * 10^1.
Then the sum for f(m) would be given by
Sum from k=1 to k=r of (a_k)^k.
This a lot to unpack and it's super late right now, but I imagine (or hope) there might be some kind of functional equation that could be found to determine a_k without the need to compute each b_k, but since the Floor function is involved in the process, it might be either to messy to be practical, or perhaps not even possible.
But I love the idea of the generalized version of the function with a domain of N (all of the natural numbers) and what properties and limits we could determine about it.
@@Kapomafioso Excellent! I also though of running the powers in the same direction as the powers of the base for the number - the attraction is it is slightly more intuitive to program ... but think of the mathematics for a second. It simply can't have anything other than the unit digit solutions.
Take abcd = a(10^3)+b(10^2)+c(10)+d
f(abcd) = a^4+b^3+c^2+d
For abcd=f(abcd):
a(10^3)+b(10^2)+c(10)+d = a^4+b^3+c^2+d
=a(a^3)+b(b^2)+c(c)+d
But 10^3>a^3, 10^2>b^2, 10>c, so for all abcd>9, f(abcd)>abcd.
@@Hiltok ohhh that's right! Okay, I looked into what if we shift the powers up by one (start with the second power for 1s, third power for 10s etc..)
And I found a spooky coincidence!
1676 = 1^1 + 6^2 + 7^3 + 6^4 (just like in the video), but also!
1676 = 1^5 + 6^4 + 7^3 + 6^2
Whoah.
Another one is 4,975,929 = 4^8+9^7+7^6+5^5+9^4+2^3+9^2
Again, wonder if there's infinitely many of these.
@@Kapomafioso There won't be infinite solutions. The right hand side (the digit power sum) grows slower than required. For instance you could show a bound of 23 digits: Even if it were all 9s, the f-value is upper bounded by 9^23 + 9^22 + ... which is smaller than the smallest 23 digit number (10^22).
Intuitively, I'd expect the solutions in higher digits to be rarer, wouldn't be surprised if the 7-digit one is the last one.
9:37 you have to make the same argument for c=1!!!
a + 1 = 100a + 1
a = 100a, a = 0
If we don't limit ourselves to 3 digit nos (i.e., require a>0 in Penn's notation), then 1, 43, and 63 are also solutions.
this is really against the spirit of the problem though.
Michael kind of proves that 000 works as well
9:13 Missed the case that c-1 = 0 here. That also doesn't work because it leads to N=001.
8:38 the easier way to show that one of those terms is divisible by 11 is to notice that 11 is prime, and apply the Fundamental Theorem of Arithmetic.
Possible solutions: 000, 001, 043, 063, 135, 175, 518, 598
001, 043, and 063 are really all against the spirit of the problem
135,175,518,598 fit into the solution of this problem
When you wrote 99a+b(10-b)=(c-1)c(c+1) you can make the argument that 99a and (c-1)c(c+1) both are divisible by 3, the latter a product of three consecutive numbers, so b(10-b) has to be divisible by three also. Checking all cases can we get b=1,3,4,6,7,9. From there we have 3 cases, because we b(10-b) is the same for b=3,7 , b=1,9 and b=4,6. From there you can also do cases, use divisibility and inequalities
Since squares and digits are nowhere near a 3 digit answer, i suggest we check just the cubes only, given that the cubic can extend high enough to get in reach of a 3 digit sum.
19x (81+1) and then a cube, requires 19x-82 = 110~ cube root = 4.~
So we know we need at least a 5 at the end of the number to reach into the relevant 3 digit range.
5^3 = 125. We can use the other two digits to fill in the neccesary number, 1,3 = 10, therefore 135 is valid.
6^3 = 216. We need another +10. 2+what square = ends in a 0? None, so it doesn't exist, because 216+81 7 we need a +4. 3+what square ends in a 4?. 1^2,9^2 317 is too small,. 397 results in a number too big.
Repeat question with 8 and 9 but with ending in a 3. 4+what square (8 or 9 only) ends in a 3? none.
8^3 = 512, 518, easy as pie.
9^3 = 729, no such number works as 7+square ends in 0. maybe 8 + square = 10? no. No number exists with 9^3
Finally, go back and check we didn't miss any duplicates, by checking the general formula now that we understand it better.
1+7^2 = 10, 175, 50+125=175, also works.
5+square = 6, 1^2, 9^2, 598. 512+81+5=598
So we have 135,175, 518,598.
9:45, what??!
Did he say "B times 10 B minus 1"??!
there are only 2 more numbers that work btw(for those who are lazy and not like me), it's 598 and 518
You can do it a bit less rigorously just by listing cases. Solutions are 135, 175, 518 and 598
a + b^2 + c^3 = 100a + 10b + c a=1 to 9; b,c = 0 to 9
99a + b(10 - b) + c(1 - c^2) = 0
b 0 1 2 3 4 5 6 7 8 9
b(10-b) 0 9 16 21 24 25 24 21 16 9
c (c-1)c(c+1) = 99a + b(10-b)
0 0
1 0
2 6
3 24
4 60
5 120 = 99 + 21, abc = 135, 175
6 210 = 99 + 111, a=1, b(10-b)=111=3x37, no solution
= 198 + 12, a=2, b(10-b)=12, no solution
7 336 = 99 + 237, no solution
= 198 + 134, no sol
= 297 + 39, no sol
8 504= 495 + 9, abc = 518, 598
9 720 = 693 + 27, no sol
A python one-liner: print([n for n in range(100, 1000) if n == n // 100 + (n // 10 % 10) ** 2 + (n % 10) ** 3])
Or generating n from its digits rather than extracting digits from n (I like this less):
print([n for a in range(1, 10) for b in range(0, 10) for c in range(0, 10) if (n:=100*a + 10*b + c) == a + b*b + c*c*c])
But also 336 & 210 can be proposed to have solutions at a break off
39=333/πe then
336= 99.a+39
a=3
210=99+39πe/3
Or saying the equation becomes 99a+39(πe/3)^n
For 210 a=1 & n=1
For 336 a=3 & n=0
It's as if the equation mutates as it goes on for 720= [(39+99/2)
+39(π×e/3)^1]×4 That's why this is conjectural or you take the 2-6 solution & multiply it by the 5-120 solution . Thanks for your valuable input.
If we don't limit the number of digits, are there an infinite number of solutions?
thats an interesting question, we need solutions of this equation
a0*10^n + a1*10^(n-1) +... + a(n-1)*10 + an = a0 + a1^2 + a2^3 +...+ an^(n+1)
to ur question, I think no, because 10^n>9^n, and the difference 10^n - 9^n increases, so for large enough n, we have LHS > RHS
so we may have solutions for higher number of digits but it cant be infinite
@@adityamukhopadhyay6803 Yeah, I should have thought of that. In fact it breaks down at 10^22. Who wants to find all the solutions?
@@xy-st9dzThis has been done! See OEIS A032799
@@xy-st9dz may be we can run a python code or work it out case by case
Extending the problem here. Is there other solutions for n-digits numbers with 8≤n≤22 (max bound)?
1 digit: 1, 2, 3, 4, 5, 6, 7, 8, 9. 2 digits: 89.
3 digits: 135, 175, 518, 598. 4 digits: 1306, 1676, 2427.
5 digits: no solution. 6 digits: no solution. 7 digits: 2646798.
In my opinion is simpler to check which b and c values satisfy b(b-10) mod 99 = c(c^2-1) mod 99, and then compute a
First on-the-trot reaction (busy - don't ask) @ 1:15 just maybe ... maybe an instruction ... an order ... an appeal ... do a few more of these yourself.
Do at least ten and if you notice a pattern emerging note it down and see if it matches pattern we see in video.
Before trying to solve the problem do try to research it.
Build up subconscious knowledge of the problem... gotta dash now ... hugz
thanks
For c = 5, 6, 7, 8, 9, mod(c^3 - c,99) = 21, 12, 41,9, 27. Only 21 and 9 match b(10-b)
Hey, do you mind finding the value of cos 4 degrees?
www.intmath.com/blog/wp-content/images/2011/06/exact-values-sin-degrees.pdf
Hi Michael, I've been watching your channel for a while and I've big fan of your videos!
I have a problem suggestion that I thik could make for a good video idea. My Linear Algebra teacher had a couple of exercises about matrices and linear systems reduced mod p. 3 years later and I've never been able to solve one of these whch asked how many m by n matrices whose entries where reduced mod 2 had rank 1.
I'm not sure if this is how you usually take problem suggestions, or if the problem is actually simple and I'm just being dumb, but I think the whole idea would make for a nice video.
He has a form in the description where you can submit problems
@@stefanalecu9532 Oh thank for telling me!
This is fun for a 5-grader. Better tell us how to find a polynomial function whose values at all 3-digit numbers are equal to your function f(N)=a+b^2+c^3.
Since we found the fixed points of the function let's move on to finding its other periodic points :) Starting from f(f(N))=N
I couldn't find any additional periodic points for 2-digit and 3-digit functions. However, I did find the longest chains:
896 - 305 -> 128
995 - 215 -> 128
And then, the sequence continues as follows: 128 - 517 - 349 - 748 - 535 - 139 - 739 - 745 - 148 - 529 - 738 - 528 - 521 -> 10 = a good place to stop.
I tried looking for a vaguely similar but possibly more interesting or flexible alternative and I don't feel I have yet found one found one; however, 267 .. 2^1 + 6^3 + 7^2 = 2 + 216 + 49 = 267 .. which seems to be the only such 3 digit solution
Also, 357 .. 3^2 + 5^1 + 7^3 = 9 + 5 + 343 = 357 a slightly different rule again
First reaction is, why do the powers in that manner (left to right)? This forces knowing in advance the number of digits of the source number. Doing it right to left would create one single function that works for source numbers regardless of their number of digits... Weird
Because you would never get any solution besides single digit numbers. Notice that 10^n*x_n + ... + 10*x_1 + x_0 > x_n^(n+1) + ... + x_1^2 + x_0 for all n > 1 since x_0,...,x_n are digits and therefore smaller than 10
@@christophdietrich4240 right but the left to right function is so limited, studying it is quasi useless
maybe studying mathematics isn't your calling 😊
@neomorphicduck idiocy is yours... I probably have more math degrees than you do. One of math's interest is purpose, this specific problem is rather useless
Here's the list of solutions for digit count up to 8 (interestingly, there are no solutions for 5, 6, or 8 digit numbers):
89
135
175
518
598
1306
1676
2427
2646798
It would be interesting to categorize which digit counts do have solutions.
12157692622039623539
@@IanXMillerI just validated that is is a valid solution. Bravo.
@@IanXMiller 😮
Have you worked on this problem before? How to find general solutions? I replied to another comment earlier with a rough sketch outlining how id go about formulating the general sum over N, but it's rather too long for me to reiterate here. I'll try to find it and cut & paste it as another reply here, and hopefully you can let me know if I'm on the right track? Id keep cracking at it tonight if it weren't already into the early morning hours and I've yet to go to bed.
@@graf_paper Here's a pasted version of my reply to another comment on this video:
*************
I like the idea of developing a formalized version of the problem, where for any m digit number, we take the sum: leftmost non-zero digit raised to the first power, plus second from the left digit raised to the 2nd power, and so on... until the final term in the sum is the r-th power of the r-th digit, i.e., the rightmost digit.
If r is the number of digits of the positive integer whose leftmost digit is non-zero, then for any number m, r is the number such that
Floor[log_10(m)] = Floor[ln(m) / ln(10)] = r.
The leftmost digit is then
a_1 = Floor[m / 10^r].
Now let b_0 = m and and define also
b_1 = b_0 - (a_1 * 10^r)
The second digit in m would then be
a_2 = Floor[log_10(b_1)].
To find the third digit (assuming r≥2 here), we would first take
b_2 = b_1 - (a_2 * 10^(r-2))
and then have
a_3 = Floor[log_10(b_2)]
and so on until the final term is found from
a_r = Floor[log_10(b_(r-1))]
where
b_(r-1) = b_(r-2) - a_(r-1) * 10^1.
Then the sum for f(m) would be given by
Sum from k=1 to k=r of (a_k)^k.
This a lot to unpack and it's super late right now, but I imagine (or hope) there might be some kind of functional equation that could be found to determine a_k without the need to compute each b_k, but since the Floor function is involved in the process, it might be either to messy to be practical, or perhaps not even possible.
But I love the idea of the generalized version of the function with a domain of N (all of the natural numbers) and what properties and limits we could determine about it.
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@@knivesoutcatchdamouse2137 considering how big the number is, he most likely found the number using a program written in like C or Python, but I might be wrong though.
there has to be a better approach than just listing cases. I do think it'll boil down to listing cases, but i think we should be able to do some analysis to reduce the no. of cases to check.
What about for number of digits larger than 3?
There is a wonderful thread in the comments section that looks at this!!
Note: 0
You could also simply spend 20 seconds to write a small script to find the solutions 😄
This isn’t a coding channel…
518, 598
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fun spin-off problem: find all fixed points of:
ABC=100A+10B+C
If the LHS is multiplication, and A, B, and C are digits with A != 0, then the problem have no solutions, I believe. But if the LHS is not multiplication, but the concatenation of digits, then isn't the problem trivial?
Proof for no solutions, assuming A,B,C are digits, and no leading zero.
BC ≤ 81
ABC ≤ 81A < 100A + 10B + C
otherwise, if leading zero(s) are allowed, then 0 is (trivially) the only valid solution.
First :*
Womp womp
An easy method readily springs to mind: write a program to test the cases. There's only 900, so you needn't even spend time looking for a subtle way to save time. Programming is skill not valued highly enough by mathematicians. If the person who composed the question didn't intend to test the pupils' proficiency in that skill, they shouldn't have composed a question like that.