# 1980 AHSME Problems/Problem 25

## Problem

In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$, and $d$ such that for all positive integers $n$, $a_n=b\lfloor \sqrt{n+c} \rfloor +d$, where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. The sum $b+c+d$ equals $\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$

## Solution

Solution by e_power_pi_times_i

Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$. However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$, $a = 1$. Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$, and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{\text{(C)}\ 2}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 