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 βn+i using the integers 0,1,2,β¦,n2 as "digits." That is, the equation
r+si=amβ(βn+i)m+amβ1β(βn+i)mβ1+β―+a1β(βn+i)+a0β
is true for a unique choice of non-negative integer m and digits a0β,a1β,β¦,amβ chosen from the set {0,1,2,β¦,n2}, with amβξ =0. We then write
r+si=(amβamβ1ββ¦a1βa0β)βn+iβ
to denote the base βn+i expansion of r+si. There are only finitely many integers k+0i that have four-digit expansions
k=(a3βa2βa1βa0β)β3+iβa3βξ =0.
Find the sum of all such k.
Solution:
If
k=(a3βa2βa1βa0β)β3+iββ=a3β(β3+i)3+a2β(β3+i)2+a1β(β3+i)+a0β=a3β(β18+26i)+a2β(8β6i)+a1β(β3+i)+a0β=(β18a3β+8a2ββ3a1β+a0β)+(26a3ββ6a2β+a1β)iβ
is a real integer, then its imaginary part must vanish. Thus
26a3ββ6a2β+a1β=0(1)
Since a3β,a2β,a1ββ{0,1,β¦,9} and a3βξ =0, we see that (1) can hold only if a3β=1 or a3β=2.
If a3β=1, then 6a2ββa1β=26 and the restrictions on a2β and a1β force a2β=5 and a1β=4. In this case, we have
k=(a3βa2βa1βa0β)β3+iβ=β18a3β+8a2ββ3a1β+a0β=10+a0β
Since a0ββ{0,1,2,β¦,9}, we see that k can be any one of 10,11,β¦,19.
If a3β=2, then 6a2ββa1β=52, leading to a2β=9 and a1β=2. It follows that k=30+a0β and k can be any one of 30,31,β¦,39. Adding the possibilities from the two cases gives the answer
(10+11+β―+19)+(30+31+β―+39)=490β
The problems on this page are the property of the MAA's American Mathematics Competitions