In how many possible ways can you write 1800 as a product of 3 positive integers a,b and c.

• For this question remember 1 formula (n+r-1) C (r-1)
  where n=power
  r=no of integers
  1800=3^2*5^2*2^3
  hence (2+3-1) C (3-1)=6
  (2+3-1) C (3-1)=6
  (3+3-1) C (3-1)=10
  Ans=6*6*10=360
• 6 years agoHelpfull: Yes(45) No(11)
• first prime factorise 1800-2^3*3^2*5^2
  which is nothing but -2*2*2*3*3*5*5
  First we need to distribute 2's we can put _2 _2 _2_ i need to inser two * inbetween to get product of three numbers = 5C2=10
  similarly for 3 we have 2 three's and we need to insert two * = 4C2=6
  Similarly for 5 we have 2 five's we have 4C2=6
  So answer will be = 10*6*6=360
• 6 years agoHelpfull: Yes(14) No(16)
• Ans: 12
  Solution:
  Total number of factors of 1800 is 36.
  So total factors /number of integers to be expressed as product.
  i.e.,36/3=12.
• 6 years agoHelpfull: Yes(7) No(25)
• Break down the no in to its prime powers.
  Ans-6
• 6 years agoHelpfull: Yes(0) No(14)