Difference between revisions of "1989 AIME Problems/Problem 14"
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<math>(154a_0)_{-3+i}</math>: | <math>(154a_0)_{-3+i}</math>: | ||
| − | We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum | + | We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum of the integers k in that form is 145. |
345+145=<math>\boxed{490}</math> | 345+145=<math>\boxed{490}</math> | ||
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== See also == | == See also == | ||
{{AIME box|year=1989|num-b=13|num-a=15}} | {{AIME box|year=1989|num-b=13|num-a=15}} | ||
Revision as of 08:02, 12 November 2007
Problem
Given a positive integer 
, it can be shown that every complex number of the form 
, where 
and 
are integers, can be uniquely expressed in the base 
using the integers 
as digits. That is, the equation
is true for a unique choice of non-negative integer 
and digits 
chosen from the set 
, with $a_m\ne 0^{}^{}$ (Error compiling LaTeX. Unknown error_msg). We write
to denote the base expansion of . There are only finitely many integers 
that have four-digit expansions
Find the sum of all such 
.
Solution
First, we find the first three powers of 
:
So we need to solve the diophantine equation 
.
The minimum the left hand side can go is -54, so 
, so we try cases:
Case 1: 
The only solution to that is 
.
Case 2: 
The only solution to that is 
.
Case 3: 
cannot be 0, or else we do not have a four digit number.
So we have the four digit integers 
and 
, and we need to find the sum of all integers k that can be expressed by one of those.
:
We plug the first three digits into base 10 to get 
. The sum of the integers k in that form is 345.
:
We plug the first three digits into base 10 to get 
. The sum of the integers k in that form is 145.
345+145=
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
