# 1989 AIME Problems/Problem 14

## Problem

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 $0,1,2,\ldots,n^2$ as digits. That is, the equation

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

is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\ldots,a_m$ chosen from the set $\{0,1,2,\ldots,n^2\}$, with $a_m\ne 0$. 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

$k=(a_3a_2a_1a_0)_{-3+i}~~$

$~~a_3\ne 0.$

Find the sum of all such $k$.

## Solution

First, we find the first three powers of $-3+i$:

$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$

So we need to solve the diophantine equation $a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3$.

The minimum the left hand side can go is -54, so $a_3\leq 2$, so we try cases:

• Case 1: $a_3=2$
The only solution to that is $(a_1, a_2, a_3)=(2,9,2)$.
• Case 2: $a_3=1$
The only solution to that is $(a_1, a_2, a_3)=(4,5,1)$.
• Case 3: $a_3=0$
$a_3$ cannot be 0, or else we do not have a four digit number.

So we have the four digit integers $(292a_0)_{-3+i}$ and $(154a_0)_{-3+i}$, and we need to find the sum of all integers $k$ that can be expressed by one of those.

$(292a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $30+a_0$. The sum of the integers $k$ in that form is $345$.

$(154a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $10+a_0$. The sum of the integers $k$ in that form is $145$. The answer is $345+145=\boxed{490}$.