Removing Floors

Algebra Level 3

n 2 + n 3 + n 5 = n 2 + n 3 + n 5 \large \left \lfloor \dfrac n2 \right \rfloor + \left \lfloor \dfrac n3 \right \rfloor + \left \lfloor \dfrac n5 \right \rfloor = \dfrac n2 + \dfrac n3 + \dfrac n5

If n n is a natural number and 1 n 100 1\leqslant n \leqslant 100 , then find the number of solutions to the equation above.

Notation : \lfloor \cdot \rfloor denotes the floor function .


The answer is 3.

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1 solution

We must have n 2 , n 3 , n 5 \dfrac{n}{2},\dfrac{n}{3},\dfrac{n}{5} as an integer. So n n has to be an integer which is a multiple of 30 30 .

There are 3 3 multiples of 30 30 from 1 1 to 100 100 thus are answer is 3 \boxed{3} .

I strongly suspect that it is incorrect to say that if n 2 + n 3 + n 5 \dfrac n2 + \dfrac n3 + \dfrac n5 is an integer then n 2 \dfrac n2 , n 3 \dfrac n3 and n 5 \dfrac n5 must be integers. You should rather simplify it and then prove that it is divisible by 30 30 . n 2 + n 3 + n 6 = 31 n 30 \dfrac n2 + \dfrac n3 + \dfrac n6 = \dfrac {31n}{30} Hence we have 30 n 30|n

@Svatejas Shivakumar @Shanthanu Rai

milind prabhu - 5 years, 2 months ago

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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.

Kushagra Sahni - 5 years, 2 months ago

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That is a pretty good argument!

milind prabhu - 5 years, 2 months ago

I quite agree!

Adarsh Kumar - 5 years, 2 months ago

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