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(#M40039102) ITC Numerical Ability question numerical Keep an EYE

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

Asked In ITC jayanth (10 years ago)

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parth parashar  (10 years ago)

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

Is this solution Helpfull? Yes (7) | No (4)  

abhijit  (7 years ago)

the total number of ways to write it as a product of 3 positive integers is 10⋅15⋅3=45010⋅15⋅3=450

Is this solution Helpfull? Yes (2) | No   

Abhas  (10 years ago)

Ans is 23 ways

Is this solution Helpfull? Yes (1) | No (8)  

Aphin  (5 years ago)

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.

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vinod kumar  (10 years ago)

ans is 12 ways

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pritesh  (9 years ago)

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

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