In how many ways can you write 3240 as product of three positive integers a,b,c

• factors of 3240 =2*2*2*3*3*3*3*5*1
  since we have 9 factors so no of ways = 9*8*7 =504
• 9 years agoHelpfull: Yes(7) No(4)
• the total number of ways to write it as a product of 3 positive integers is 10⋅15⋅3=45010⋅15⋅3=450
• 6 years agoHelpfull: Yes(2) No(0)
• Ans is 23 ways
  
• 9 years agoHelpfull: Yes(1) No(8)
• first, we get the prime factorization of 3240 which is 2^3*3^4*5^1 which could be equated as a*b*c.
  Now for each of these three numbers could be written in 3 ways like:
  a = 2^(x1)*3^(y1)*5^(z1)
  b=2^(x2)*3^(y2)*5^(z2)
  c=2^(x3)*3^(y3)*5^(z3)
  hence, a*b*c=2^(x1+x2+x3)*3^(y1+y2+y3)*5^(z1+z2+z3)=3240=2^3*3^4*5^1.
  if we recall the concept of dividing n things among r people it could be done in n+r-1Cr-1 ways.
  for x,
  3 IN 3 WAYS among x1,x2&x3, therefore 5C2=10 ways.
  for y,
  4 IN 3 WAYS, therefore 6C2=15 ways.
  for z,
  1 in 3 WAYS, therefore 3C2=3 ways.
  The product of these numbers is 10*15*3=450 ways.
• 4 years agoHelpfull: Yes(1) No(0)
• ans is 12 ways
• 9 years agoHelpfull: Yes(0) No(3)
• 23⋅34⋅5=3240. Therefore the number of abelian groups of order 3240 is 3⋅4=12.
  
  Is this the entire proof or do we need to do the table showing divisors like this?
  
  Divisors:
  
  23⋅34⋅5
  22⋅2⋅34⋅5
  22⋅2⋅33⋅3⋅5
  2⋅2⋅2⋅33⋅3⋅5
  2⋅2⋅2⋅32⋅3⋅3⋅5
  2⋅2⋅2⋅3⋅3⋅⋅3⋅5
  2⋅2⋅2⋅34⋅5
  22⋅2⋅32⋅3⋅3⋅5
  23⋅3⋅3⋅3⋅3⋅5
  
  
• 8 years agoHelpfull: Yes(0) No(2)