1989 AIME Problem 14

Question

Given a positive integer

n,

it can be shown that every complex number of the form

r+si,

where

r

and

s

are integers, can be uniquely expressed in the base

-n+i

using the integers

0,

1,

\ldots,

n^2

as digits. That is, the equation

\begin{aligned}r+si={}&a_m(-n+i)^m\\&+a_{m-1}(-n+i)^{m-1}\\&+\cdots+a_1(-n+i)\\&+a_0\end{aligned}

is true for a unique choice of nonnegative integer

m

and digits

a_0,

a_1,

\ldots,

a_m

chosen from the set

\{0,1,2,\ldots,n^2\},

with

a_m\ne0.

We write

r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}

to denote the base

-n+i

expansion of

r+si.

There are only finitely many integers

k+0i

that have four-digit expansions

k=(a_3a_2a_1a_0)_{-3+i}\qquad a_3\ne0.

Find the sum of all such

k.

Answer

Answer: 490

Concepts:complex number number base system of equations

Difficulty rating: 2840

Small hint:

Compute the second and third powers of

-3+i

and set the imaginary part of the expansion equal to zero

Big hint:

The digit bounds leave only two possible triples

(a_3,a_2,a_1)

; then let

a_0

range over all digits

Solution:

Let

b=-3+i.

Then

b^2=8-6i

and

b^3=-18+26i.

The imaginary part of

a_3b^3+a_2b^2+a_1b+a_0

is

26a_3-6a_2+a_1.

Thus

a_1=6a_2-26a_3.

With

1\leq a_3\leq9

and

0\leq a_1,a_2\leq9,

the only possibilities are

\begin{gathered}(a_3,a_2,a_1)=(1,5,4),\\(a_3,a_2,a_1)=(2,9,2).\end{gathered}

The corresponding real parts are

10+a_0

and

30+a_0,

respectively. As

a_0

ranges from

0

through

9,

the required sum is

\begin{aligned}&(10+11+\cdots+19)\\&\quad+(30+31+\cdots+39)\\&=145+345=490.\end{aligned}