⌊ 2 n ⌋ + ⌊ 3 n ⌋ + ⌊ 5 n ⌋ = 2 n + 3 n + 5 n
If n is a natural number and 1 ⩽ n ⩽ 1 0 0 , then find the number of solutions to the equation above.
Notation : ⌊ ⋅ ⌋ denotes the floor function .
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I strongly suspect that it is incorrect to say that if 2 n + 3 n + 5 n is an integer then 2 n , 3 n and 5 n must be integers. You should rather simplify it and then prove that it is divisible by 3 0 . 2 n + 3 n + 6 n = 3 0 3 1 n Hence we have 3 0 ∣ n
@Svatejas Shivakumar @Shanthanu Rai
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Well if n/3,n/5,n/2 are not integers then LHS will be smaller than RHS because floor of for eg. n/3 will be less than n/3 and same goes for other numbers. So they have to be integers.
I quite agree!
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We must have 2 n , 3 n , 5 n as an integer. So n has to be an integer which is a multiple of 3 0 .
There are 3 multiples of 3 0 from 1 to 1 0 0 thus are answer is 3 .