Question

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

Answer 1

Express 1800 as a product of its prime factors:1800=2*2*2*3*3*5*5
The three numbers of 2 can be distributed to three integers in
(3+(3-1)) =(5)
( 3-1 ) (2) ways and by permutation we have:5!/2!(5-2)! This gives:5*4*3!/2*1*3!=10ways.
The two numbers of 3 can be written to 3 integers in:(2+(3-1)) =(4)
(3-1 ) ( 2) and by permutation we have:4*3*2*1/2*2=6ways
The ways in which 2 numbers of 5 can be written to 3 intergers
(2+(3-1))=(4)
(3-1 ) (2) and by permutation we have 4*3*2*1/2*2=6ways.
To get the total number of ways 1800 can be written as a product of 3 integers we get the product of ways it's prime factors can be written. That is:10*6*6=360

Answer 2

ans is 360

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