ACC's Math Conjecture II
Another one for posterity:
If N=31622777, then N^3 = 31622777796632428411433, and that's the largest N that has such a property.
I have tested this conjecture up to N's 100-digits long, and it holds true. Used the code below.
If you can find a larger N, let me know.
Thanks, ACC.
If N=31622777, then N^3 = 31622777796632428411433, and that's the largest N that has such a property.
I have tested this conjecture up to N's 100-digits long, and it holds true. Used the code below.
If you can find a larger N, let me know.
Thanks, ACC.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Numerics;
namespace SquareStart
{
class Program
{
static void Main(string[] args)
{
int power = Int32.Parse(args[0]);
BigInteger factor = 1;
BigInteger MAX = 1000000;
BigInteger count = 0;
for (BigInteger n = 1; ; n++, count++)
{
if (n % 10 != 0 && (BigInteger.Pow(n, power)).ToString().StartsWith(n.ToString()))
{
Console.WriteLine("{0} => {1}", n, BigInteger.Pow(n, power));
BigInteger temp = n;
n = 10 * n - factor;
if (n <= temp)
{
n = temp;
}
else
{
factor *= 10;
}
count = 0;
}
if (count >= MAX)
{
n *= 10;
count = 0;
Console.Write("{0}-", n.ToString().Length);
}
}
}
}
}
Comments
Well, it's easy to find at least one number with arbitrary length for which this property holds true - any power of 10 :) Here is my napkin-style Python script that prints all such numbers and makes it very obvious that any power of 10 always fits the bill :)
Reply Deletefor i in range(10**1000):
if str(i ** 3).startswith(str(i)):
print(i, i**3)
0 0
1 1
10 1000
32 32768
100 1000000
1000 1000000000
10000 1000000000000
31623 31623446801367
100000 1000000000000000
316228 31622846796684352
1000000 1000000000000000000
3162278 31622786796633508952
10000000 1000000000000000000000
31622777 31622777796632428411433
100000000 1000000000000000000000000
Forgot to mention: except powers of 10 :D
Deletehaha, makes sense :)
DeleteWie wär's mit 316227766017? 31622776601^3=わ31622776601732413360022154029126
Reply Deleteoder 316227766016838^3=わ31622776601683813360022129111349862592192472
Miracle173: I guess the problem's been solved :) Good job!!!
Reply Deleteeinfach die Ziffern von 10^0.5 nehmen. Diese Wurzel kann man z.B. hier (https://www.ttmath.org/online_calculator) berechnen. Dann findet man z.B.
Reply Delete316227766016837933199889354443271853371955513932521682685750485279259443863923822134424810837930029518734728415284005514854885603045388001469051959670015390334492165717925994065915015347411333948412408531692957709047157646104436925787906203780860994182837171154840632855299911859682456420332696160469131433612894979189027^3 =わ 31622776601683793319988935444327185337195551393252168268575048527925944386392382213442481083793002951873472841528400551485488560304538800146905195967001539033449216571792599406591501534741133394841240853169295770904715764610443692578790620378086099418283717115484063285529991185968245642033269616046913143361289497918902769409127746476424372998772236274428390726337300950439377124613306560523609736286431935375164091955633839082543107707994928448405943427119419511840442079309120216733015546947758644146966479379031266044124861476885589992602101810611562999852833023107122345377814217819153038915286931921854519604269125481207165479973860200095144367860768079216739417320940438348635868789195186885141279651353875619216981657040907404475657787785785828175923925174846559126447745292267007331960857567949514398539137781848120134336213196224279280093116123725767418602571473583480804020163991852839938793882772549236048118970510860691652596864884863920503887362683
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