1980 AMC 12 Problem 25

Question

In the nondecreasing sequence of odd integers

\begin{aligned} (a_1,a_2,a_3,\ldots) ={}&(1,3,3,3,\\ &5,5,5,5,5,\ldots) \end{aligned}

each positive odd 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\left\lfloor\sqrt{n+c}\right\rfloor+d,

where

\lfloor x\rfloor

denotes the largest integer not exceeding

x.

The sum

b+c+d

equals

0

1

2

3

4

The first

m

positive odd integers have sum

m^2

Express the least

m

with

m^2\ge n

using a floor of

\sqrt{n-1}

The final occurrence of

2m-1

is in position

1+3+\cdots+(2m-1)=m^2.

Therefore

\begin{aligned} a_n&=2\left\lceil\sqrt n\right\rceil-1\\ &=2\left\lfloor\sqrt{n-1}\right\rfloor+1. \end{aligned}

Thus

b=2,

c=-1,

and

d=1,

so

b+c+d=2.

Therefore, the correct answer is C.

Problem 25 in other years