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

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
7 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.
7 years agoHelpfull: Yes(7) No(25)
Break down the no in to its prime powers.
Ans-6
7 years agoHelpfull: Yes(0) No(14)
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
3 Months agoHelpfull: Yes(0) No(0)

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