Call a natural number n faithful, if there exist natural numbers a < b < c such that a divides
b, b divides c and n = a + b + c.
(i) Show that all but a finite number of natural numbers are faithful.
(ii) Find the sum of all natural numbers which are not faithful.

• mka=kb=c( where m and k are any integers >1)
  hence, n=a*(mka+ma+a)
  now, to find min n , put a as 1, m as 2 and k as 2
  Thus, min n=7 (All n less than this are not faithful)
  Thus, (1) is proven and (2) is 1+2+3+4+5+6=21
• 10 years agoHelpfull: Yes(1) No(2)