Removing Floors

Algebra Level 3

$\large \left \lfloor \dfrac n2 \right \rfloor + \left \lfloor \dfrac n3 \right \rfloor + \left \lfloor \dfrac n5 \right \rfloor = \dfrac n2 + \dfrac n3 + \dfrac n5$

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

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

The answer is 3.

1 solution

A Former Brilliant Member
Apr 3, 2016

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

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

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

@Svatejas Shivakumar @Shanthanu Rai

milind prabhu - 5 years, 2 months ago

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

That is a pretty good argument!

milind prabhu - 5 years, 2 months ago

I quite agree!

Adarsh Kumar - 5 years, 2 months ago

Removing Floors

The answer is 3.

1 solution

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