Discrete mathematics, number of solutions.

[stark narwhal]

Let $x=n+k$ where $n$ is an integer and $0 \leq k<1$

[potent trout]

Civil Service Pigeon

[stark narwhal]

You can find $\lfloor x \rfloor$ and ${x}$ in terms of $n$ and $k$

[potent trout]

Civil Service Pigeon

[stark narwhal]

As a further hint, find ||the range of the left hand side for [n, n+1)||

[loud vortex]

writing $x=n+k$. I was able to find that $$n<(n+k)nk+n+k\le n(n+1)n+n+1=n^2+2n+1$$

[potent trout]

Estbs

[stark narwhal]

It should be [n,(n+1)^2) but yeah

[loud vortex]

$n\le a\le (n+1)^2$

[potent trout]

Estbs

[loud vortex]

I can't understand how I could continue from here 😦

[stark narwhal]

Hint: two cases, a is or isn’t an integer

It light help to consider (kind of large) a so that the ideas become more obvious

[stark narwhal]

Also (n+1)^2 is excluded since it’s achieved at n+1

[loud vortex]

😭

[loud vortex]

why if $n$ satisfaces $n\leq a \le (n+1)^2$ then there exist a $k$ such that $x=n+k$?

[potent trout]

Estbs

[stark narwhal]

k isn't 1

[loud vortex]

the answer is $2(floor(a)-floor(\sqrt(a)-1))$?

[potent trout]

Estbs