CAT
Exam
What is the smallest value of a for which (2100)*(a) is a perfect cube ?
Read Solution (Total 4)
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- 4410
2100*a= 21*10^2*a
for perfect cube , smallest integer value is 10*21^2= 4410
If a can have fraction value, then
smallest value of a = 1/2100 - 11 years agoHelpfull: Yes(6) No(1)
- I think 1/2100 is smallest number .
as 1/2100 < 10/21 - 11 years agoHelpfull: Yes(1) No(2)
- prime factor of 2100 is 2*2*3*5*5*7
to make it perfect cube we need 2*3*3*5*7*7=4410 (all must be 3 time) - 11 years agoHelpfull: Yes(1) No(0)
- Smallest value of a for which (2100)*(a) is a perfect cube = 10/21.
- 11 years agoHelpfull: Yes(0) No(3)
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