The Legend of Four












4












$begingroup$


As far as I know, all numbers have a root of 4. What I mean by this is as follows:



Starting with any number, for example 384, I take the number of letters in that number. Then I repeat this process until it infinitely repeats the number 4.



Three Hundred Eighty Four = 22



Twenty Two = 9



Nine = 4



Four = 4



So I challenge you to find any number which does not follow this rule.






mathematics number-sequence number-theory







share|improve this question







New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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  • $begingroup$
    I believe this is impossible. However, it is so strange that I know it is possible to not be :).
    $endgroup$
    – Rigidity
    2 hours ago
















4












$begingroup$


As far as I know, all numbers have a root of 4. What I mean by this is as follows:



Starting with any number, for example 384, I take the number of letters in that number. Then I repeat this process until it infinitely repeats the number 4.



Three Hundred Eighty Four = 22



Twenty Two = 9



Nine = 4



Four = 4



So I challenge you to find any number which does not follow this rule.






mathematics number-sequence number-theory







share|improve this question







New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I believe this is impossible. However, it is so strange that I know it is possible to not be :).
    $endgroup$
    – Rigidity
    2 hours ago














4












4








4





$begingroup$


As far as I know, all numbers have a root of 4. What I mean by this is as follows:



Starting with any number, for example 384, I take the number of letters in that number. Then I repeat this process until it infinitely repeats the number 4.



Three Hundred Eighty Four = 22



Twenty Two = 9



Nine = 4



Four = 4



So I challenge you to find any number which does not follow this rule.






mathematics number-sequence number-theory







share|improve this question







New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




As far as I know, all numbers have a root of 4. What I mean by this is as follows:



Starting with any number, for example 384, I take the number of letters in that number. Then I repeat this process until it infinitely repeats the number 4.



Three Hundred Eighty Four = 22



Twenty Two = 9



Nine = 4



Four = 4



So I challenge you to find any number which does not follow this rule.






mathematics number-sequence number-theory




mathematics number-sequence number-theory






share|improve this question







New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









RigidityRigidity

211




211




New contributor




Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I believe this is impossible. However, it is so strange that I know it is possible to not be :).
    $endgroup$
    – Rigidity
    2 hours ago


















  • $begingroup$
    I believe this is impossible. However, it is so strange that I know it is possible to not be :).
    $endgroup$
    – Rigidity
    2 hours ago
















$begingroup$
I believe this is impossible. However, it is so strange that I know it is possible to not be :).
$endgroup$
– Rigidity
2 hours ago




$begingroup$
I believe this is impossible. However, it is so strange that I know it is possible to not be :).
$endgroup$
– Rigidity
2 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

There are some trick solutions




For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.




With that in mind




Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.




So, for the final step




One, two, six and ten lead to three.
Three, seven, eight lead to five.
Zero, four, five and nine lead to four.
This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.




To answer your question




There is no number that does not follow this rule.







share|improve this answer









$endgroup$





















    1












    $begingroup$


    There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.







    share|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)




      Both these cannot be counted (till infinity??)



      So technically they do not end with four??







      share|improve this answer









      $endgroup$













      • $begingroup$
        You could argue that infinity has eight letters.
        $endgroup$
        – ZanyG
        22 mins ago










      • $begingroup$
        Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
        $endgroup$
        – DEEM
        16 mins ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      There are some trick solutions




      For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.




      With that in mind




      Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.




      So, for the final step




      One, two, six and ten lead to three.
      Three, seven, eight lead to five.
      Zero, four, five and nine lead to four.
      This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.




      To answer your question




      There is no number that does not follow this rule.







      share|improve this answer









      $endgroup$


















        6












        $begingroup$

        There are some trick solutions




        For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.




        With that in mind




        Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.




        So, for the final step




        One, two, six and ten lead to three.
        Three, seven, eight lead to five.
        Zero, four, five and nine lead to four.
        This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.




        To answer your question




        There is no number that does not follow this rule.







        share|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          There are some trick solutions




          For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.




          With that in mind




          Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.




          So, for the final step




          One, two, six and ten lead to three.
          Three, seven, eight lead to five.
          Zero, four, five and nine lead to four.
          This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.




          To answer your question




          There is no number that does not follow this rule.







          share|improve this answer









          $endgroup$



          There are some trick solutions




          For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.




          With that in mind




          Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.




          So, for the final step




          One, two, six and ten lead to three.
          Three, seven, eight lead to five.
          Zero, four, five and nine lead to four.
          This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.




          To answer your question




          There is no number that does not follow this rule.








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 hours ago









          ZanyGZanyG

          599216




          599216























              1












              $begingroup$


              There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.







              share|improve this answer











              $endgroup$


















                1












                $begingroup$


                There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.







                share|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.







                  share|improve this answer











                  $endgroup$




                  There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 19 mins ago

























                  answered 1 hour ago









                  NautilusNautilus

                  3,968525




                  3,968525























                      0












                      $begingroup$

                      Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)




                      Both these cannot be counted (till infinity??)



                      So technically they do not end with four??







                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        You could argue that infinity has eight letters.
                        $endgroup$
                        – ZanyG
                        22 mins ago










                      • $begingroup$
                        Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                        $endgroup$
                        – DEEM
                        16 mins ago
















                      0












                      $begingroup$

                      Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)




                      Both these cannot be counted (till infinity??)



                      So technically they do not end with four??







                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        You could argue that infinity has eight letters.
                        $endgroup$
                        – ZanyG
                        22 mins ago










                      • $begingroup$
                        Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                        $endgroup$
                        – DEEM
                        16 mins ago














                      0












                      0








                      0





                      $begingroup$

                      Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)




                      Both these cannot be counted (till infinity??)



                      So technically they do not end with four??







                      share|improve this answer









                      $endgroup$



                      Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)




                      Both these cannot be counted (till infinity??)



                      So technically they do not end with four??








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 54 mins ago









                      DEEMDEEM

                      5,953119106




                      5,953119106












                      • $begingroup$
                        You could argue that infinity has eight letters.
                        $endgroup$
                        – ZanyG
                        22 mins ago










                      • $begingroup$
                        Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                        $endgroup$
                        – DEEM
                        16 mins ago


















                      • $begingroup$
                        You could argue that infinity has eight letters.
                        $endgroup$
                        – ZanyG
                        22 mins ago










                      • $begingroup$
                        Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                        $endgroup$
                        – DEEM
                        16 mins ago
















                      $begingroup$
                      You could argue that infinity has eight letters.
                      $endgroup$
                      – ZanyG
                      22 mins ago




                      $begingroup$
                      You could argue that infinity has eight letters.
                      $endgroup$
                      – ZanyG
                      22 mins ago












                      $begingroup$
                      Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                      $endgroup$
                      – DEEM
                      16 mins ago




                      $begingroup$
                      Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end
                      $endgroup$
                      – DEEM
                      16 mins ago










                      Rigidity is a new contributor. Be nice, and check out our Code of Conduct.










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                      Rigidity is a new contributor. Be nice, and check out our Code of Conduct.












                      Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
















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