find no of ways in 32100 can be expressed as a product of two coprime factors?

• 32100= 3*107*2*2*5*5 i.e the numbers of prime factors are 4(107,3,5,2) and according to the formula 2^(n-1)
  we have 2^(4-1)=2^3= 8 this is the answer.
  Ans=8.
• 10 years agoHelpfull: Yes(26) No(2)
• ans is 8 bro i write formula wrong,its 2^n-1 so we have 4 distinct prime num so 2^4-1=2^3=8 ,32100=2*2*3*5*5*107
• 10 years agoHelpfull: Yes(4) No(3)
• @ Sonu,can you tell me how the formula is formulated?
• 10 years agoHelpfull: Yes(2) No(2)
• (107) * (3*4*25)
  (107*3) * (4*25)
  (107*4) * (3*25)
  (107*25) * (4*3)
  (107*3*4) * (25)
  (107*3*25) * (4)
  (107*25*4) * (3)
  ans 7
• 10 years agoHelpfull: Yes(2) No(2)
• options are 8,4,2,1
• 10 years agoHelpfull: Yes(1) No(1)
• 8 is the answercz
  32100=4*25*3*107
  thus on forming all possible pairs as:we get 8 possible pairs
• 10 years agoHelpfull: Yes(1) No(1)
• 32100=2*2*3*5*5*107 ,4 typ of prime factor,according to formula 2^(n-1)-1= 2^(4-1)-1=7 is right ans
• 10 years agoHelpfull: Yes(0) No(3)
• 32100=3*107*2^2*5^2
  we have 4 distinct primes.
  count the number of possibilities such that we have at least one prime on either factors. so totally we have 4c2+4c1.
  correct me if i am wrong.
• 10 years agoHelpfull: Yes(0) No(7)
• @ NAGA MALLESWARA RAO YAGANTI can you please mention the all pairs.
• 10 years agoHelpfull: Yes(0) No(1)
• 32100=3*107*5*5*2*2
  so accn to the formulla 2^(n-1)
  2^(4-1)=8
• 8 years agoHelpfull: Yes(0) No(0)