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[x/2]+[x/3]+[x/5]=31x/30, where [x] is the greatest integer less than or equal to x, the number of possible values of x is,

[x/2]+[x/3]+[x/5]=31x/30, where [x] is the greatest integer less than or equal to x, the number of possible values of x is,

Grade:12th Pass

1 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
see right part has to be integer s x is multiple of 30
no since 30 is factor of 2,3,5
so left part becomes
31[x]/30
=> n30 where n belongs to integer

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