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

1989 AIME Problem 14

Answer: 490

Difficulty rating: 2840

Small Hint:

Big Hint:

Solution:

Problem 14 in Other Years