Thursday, April 15, 2010

Digital root

#Problem 16
Prove any perfect number ,except 6, has digital root 1.  
Solution scheme and approach:
We already know that what is digital root(look under concept of number theory) and we also learned that dividing any number by 9 we get digital root. 
Let, the number is  
dkdk-1dk-2…….d2 d1                                                                                                                          = dk *10^(k-1)+ dk-1*10^(k-2)+dk-2*10^(k-3)…….d2*10 + d1(mod 9)                                            =dk +dk-1+dk-2…….+d2+d1(mod 9) [As 10^k=1(mod9)]                                                                                                                                       All perfect numbers can be written as 2p-1(2p-1) [Where p is a prime number]                  for p=3, 2p-1(2p-1)=28 which is a perfect square and 28=1(mod9)..(i)                                                                                                                          for p>3, we can write p=6k+1 or p=6k+5                                                                      Case 1: p=6k+1
As 26k=1(mod 9) and 26k+1-1=1(mod 9) we can write,                                                              
26k(26k+1-1) =1(mod 9)….(ii)
Case 2: p=6k+5
As 26k+4= 24(mod 9)=7(mod 9) and 26k+5-1=25-1(mod 9)= 4 we can write,
26k+4(26k+5-1) =7*4(mod 9)=1(mod 9)……….(iii)
From (i),(ii) and (iii) we can conclude any perfect number ,without 6, will be having digital root 1.(Q.E.D.)

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