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 09:02, 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
![$r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0$](http://latex.artofproblemsolving.com/f/6/4/f6491a5fef34ba26ef6657aae299c94700b40a0e.png)
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
![$r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}$](http://latex.artofproblemsolving.com/5/6/a/56aa9ca7c641491fbd454fff558b7a8fb8918ced.png)
to denote the base expansion of
. There are only finitely many integers
that have four-digit expansions
![$k=(a_3a_2a_1a_0)_{-3+i^{}_{}}~~~~a_3\ne 0.$](http://latex.artofproblemsolving.com/6/0/f/60f64fe15a0ae3ffb0c11f59e745a2bd2b665a3d.png)
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 |