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Difference between revisions of "1989 AIME Problems/Problem 14"

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== Solution ==
 
== Solution ==
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{{solution}}
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First, we find the first three powers of <math>-3+i</math>:
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 +
<math>(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i</math>
 +
 
 +
So we need to solve the [[diophantine equation]] <math>a_1-6a_2+26a_3=0</math>.
 +
 
 +
<math>a_1-6a_2=-26a_3</math>
 +
 
 +
The minimum the left hand side can go is -54, so <math>a_3\leq 2</math>, so we try cases:
 +
 
 +
Case 1: <math>a_3=2</math>
 +
 
 +
The only solution to that is <math>(a_1, a_2, a_3)=(2,9,2)</math>.
 +
 
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Case 2: <math>a_3=1</math>
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The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>.
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 +
Case 3: <math>a_3=0</math>
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 +
<math>a_3</math> cannot be 0, or else we do not have a four digit number.
 +
 
 +
So we have the four digit integers <math>(292a_0)_{-3+i}</math> and <math>(154a_0)_{-3+i}</math>, and we need to find the sum of all integers k that can be expressed by one of those.
 +
 
 +
<math>(292a_0)_{-3+i}</math>:
 +
 
 +
We plug the first three digits into base 10 to get <math>30+a_0</math>. The sum of the integers k in that form is 345.
 +
 
 +
<math>(154a_0)_{-3+i}</math>:
 +
 
 +
We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum f the integers k in that form is 145.
 +
 
 +
345+145=<math>\boxed{490}</math>
 +
 
  
 
== 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 08:01, 12 November 2007

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 $1,2^{}_{},\ldots,n^2$ as digits. That is, the equation

$r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0$

is true for a unique choice of non-negative integer $m^{}_{}$ and digits $a_0,a_1^{},\ldots,a_m$ chosen from the set $\{0^{}_{},1,2,\ldots,n^2\}$, with $a_m\ne 0^{}^{}$ (Error compiling LaTeX. Unknown error_msg). We write

$r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}$

to denote the base $-n+i^{}_{}$ expansion of $r+si^{}_{}$. There are only finitely many integers $k+0i^{}_{}$ that have four-digit expansions

$k=(a_3a_2a_1a_0)_{-3+i^{}_{}}~~~~a_3\ne 0.$

Find the sum of all such $k^{}_{}$.

Solution

First, we find the first three powers of $-3+i$:

$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$

So we need to solve the diophantine equation $a_1-6a_2+26a_3=0$.

$a_1-6a_2=-26a_3$

The minimum the left hand side can go is -54, so $a_3\leq 2$, so we try cases:

Case 1: $a_3=2$

The only solution to that is $(a_1, a_2, a_3)=(2,9,2)$.

Case 2: $a_3=1$

The only solution to that is $(a_1, a_2, a_3)=(4,5,1)$.

Case 3: $a_3=0$

$a_3$ cannot be 0, or else we do not have a four digit number.

So we have the four digit integers $(292a_0)_{-3+i}$ and $(154a_0)_{-3+i}$, and we need to find the sum of all integers k that can be expressed by one of those.

$(292a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $30+a_0$. The sum of the integers k in that form is 345.

$(154a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $10+a_0$. The sum f the integers k in that form is 145.

345+145=$\boxed{490}$


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
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