If a number N is divisible by 18 and has exactly 10 divisors (including 1 and N).Find N.

• @ ALL :
  
  A LITTLE CORRECTION :
  
  In the posted solution a little correction regarding the notation is needed.
  
  Please refer to the 4th line of the solution ,i.e , the following statement :
  
  " where 2 ,3,p1,p2,p3,p4,p5,....,pn are the PRIME factors of N and (a1)+1 ,(a2)+2 ,a3 , a4 ,a5 ,....,an are the powers of them respectively and so each of (a1),(a2),a3 , a4 ,a5 ,....,an is ofcourse non-negative +ve integer."
  
  ----Here in the 1st line p1 and p2 have been typed by mistake though these two
  
  notations are absent in the whole solution and therefore p1 and p2 should be
  
  deleted from the above statement,consequently the statement should be read as :
  
  " where 2,3,p3,p4,p5,....,pn are the PRIME factors of N and (a1)+1 ,(a2)+2 ,a3 , a4 ,a5 ,....,an are the powers of them respectively and so each of (a1),(a2),a3 , a4 ,a5 ,....,an is ofcourse non-negative +ve integer."
  
  If u have any confusion or complaint regarding the posted solution along with
  
  the correction, and also any difficulty regarding the understanding of the
  
  solution then u are earnestly requested to communicate it with me..
  
  Your feedback is highly solicited.
  
  Thanks to all:-) :-)
  
• 12 years agoHelpfull: Yes(12) No(1)
• According to the general prime factorization of a natural number We can factor the given integer N as follows.
  
  Let, N = 18*{2^(a1)}*{3^(a2)}*{(p3)^(a3)}*{(p4)^(a4)}*{(p5)^(a5)}....{(pn)^(an)}
  
  =>N={2^[(a1)+1]}*{3^[(a2)+2]}*{(p3)^(a3)}*{(p4)^(a4)}*{(p5)^(a5)}...{(pn)^(an)}
  
  where 2 ,3,p1,p2,p3,p4,p5,....,pn are the PRIME factors of N and (a1)+1 ,(a2)+2 ,a3 , a4 ,a5 ,....,an are the powers of them respectively and so each of (a1),(a2),a3 , a4 ,a5 ,....,an is ofcourse non-negative +ve integer.
  
  So, from above we can write that
  
  (the number of divisors of N) = {(a1)+2}*{(a2)+3}*(a3 +1)*(a4 +1)*....(an +1)
  
  ==> 10 = {(a1)+2}*{(a2)+3}*(a3 +1)*(a4 +1)*....(an +1)
  
  ==> 2 *5 = {(a1)+2}*{(a2)+3}*(a3 +1)*(a4 +1)*....(an +1)......[(eqn1)]
  
  Now, since each of (a1),(a2),a3 , a4 ,a5 ,....,an is ofcourse non-negative +ve integer , so , clearly
  
  {(a1)+2} > or, = 2;
  and
  {(a2)+3} > or, = 3;
  
  Now, at this situation it clearly implies that
  a3=0; a4=0; a5=0; .... ; an=0.
  and
  {(a1)+2} = 2 ; ==> a1 = 0 ;
  
  {(a2)+3} = 5 ; ==> a2 = 2 ;
  
  So, the integer
  N = {2^[(a1)+1]}*{3^[(a2)+2]}*{(p3)^(a3)}*{(p4)^(a4)}*{(p5)^(a5)}...{(pn)^(an)}
  
  => N = (2^1)*(3^4);
  =2*81
  =162.
  
  So, N=162..
  
  
  
  
  
  
• 12 years agoHelpfull: Yes(11) No(0)