For Each positive integer p

Number Theory Level 5

For every positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \dfrac{1}{2}.$

If $S = \displaystyle\sum_{p=1}^{2007} b(p)$ find the remainder when $S$ is divided by 1000.

For example, $b(6) = 2$ and $b(23) = 5$ .

The answer is 955.

1 solution

Patrick Corn
Apr 1, 2015

It's clear that $b(p) = n$ if and only if $(n-1/2)^2 < p < (n+1/2)^2$ , which translates to $n^2-n+1 \le p \le n^2 + n$ . There are $2n$ values of $p$ in this range. When $n =44$ we get that $p$ runs from $1893$ to $1980$ . When $n = 45$ the range runs from $1981$ to $2070$ , so there are $27$ values of $p$ in the given sum for which $b(p) = 45$ .

So the answer is $\sum_{n=1}^{44} n(2n) + 45 \cdot 27 = 59955$ . The answer is $\fbox{955}$ .

@Parth Lohomi beautiful problem ....... enjoyed it :) ... . a nice modification of ceiling and floor function :p

Abhinav Raichur - 6 years, 2 months ago

Although I am very late to reply , but still thanks!

Parth Lohomi - 5 years, 1 month ago

For Each positive integer p

The answer is 955.

1 solution

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