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,…,n2 as "digits." That is, the equation
r+si=am​(−n+i)m+am−1​(−n+i)m−1+⋯+a1​(−n+i)+a0​
is true for a unique choice of non-negative integer m and digits a0​,a1​,…,am​ chosen from the set {0,1,2,…,n2}, with amâ€‹î€ =0. We then write
r+si=(am​am−1​…a1​a0​)−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=(a3​a2​a1​a0​)−3+i​a3â€‹î€ =0.
Find the sum of all such k.
Solution:
If
k=(a3​a2​a1​a0​)−3+i​​=a3​(−3+i)3+a2​(−3+i)2+a1​(−3+i)+a0​=a3​(−18+26i)+a2​(8−6i)+a1​(−3+i)+a0​=(−18a3​+8a2​−3a1​+a0​)+(26a3​−6a2​+a1​)i​
is a real integer, then its imaginary part must vanish. Thus
26a3​−6a2​+a1​=0(1)
Since a3​,a2​,a1​∈{0,1,…,9} and a3â€‹î€ =0, we see that (1) can hold only if a3​=1 or a3​=2.
If a3​=1, then 6a2​−a1​=26 and the restrictions on a2​ and a1​ force a2​=5 and a1​=4. In this case, we have
k=(a3​a2​a1​a0​)−3+i​=−18a3​+8a2​−3a1​+a0​=10+a0​
Since a0​∈{0,1,2,…,9}, we see that k can be any one of 10,11,…,19.
If a3​=2, then 6a2​−a1​=52, leading to a2​=9 and a1​=2. It follows that k=30+a0​ and k can be any one of 30,31,…,39. Adding the possibilities from the two cases gives the answer
(10+11+⋯+19)+(30+31+⋯+39)=490​
The problems on this page are the property of the MAA's American Mathematics Competitions