Problem:
There exist r unique nonnegative integers n1>n2>⋯>nr and r unique integers ak(1≤k≤r) with each ak either 1 or −1 such that
a13n1+a23n2+⋯+ar3nr=2008
Find n1+n2+⋯+nr.
Solution:
Every positive integer has a unique base-3 representation, which for 2008 is 22021013. Note that 2⋅3k=(3+(−1))⋅3k=3k+1+(−1)⋅3k, so that
===220210132⋅36+2⋅35+2⋅33+1⋅32+1⋅30(37+(−1)⋅36)+(36+(−1)⋅35)+(34+(−1)⋅33)+1⋅32+1⋅301⋅37+(−1)⋅35+1⋅34+(−1)⋅33+1⋅32+1⋅30,
and n1+n2+⋯+nr=7+5+4+3+2+0=21.
The problems on this page are the property of the MAA's American Mathematics Competitions