Difference between revisions of "1989 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | Given a positive integer <math>n^{}_{}</math>, it can be shown that every complex number of the form <math>r+si^{}_{}</math>, where <math>r^{}_{}</math> and <math>s^{}_{}</math> are integers, can be uniquely expressed in the base <math>-n+i^{}_{}</math> using the integers <math>1,2^{}_{},\ldots,n^2</math> as digits. That is, the equation | + | Given a positive [[integer]] <math>n^{}_{}</math>, it can be shown that every [[complex number]] of the form <math>r+si^{}_{}</math>, where <math>r^{}_{}</math> and <math>s^{}_{}</math> are integers, can be uniquely expressed in the base <math>-n+i^{}_{}</math> using the integers <math>1,2^{}_{},\ldots,n^2</math> as digits. That is, the equation |
<center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center> | <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center> | ||
is true for a unique choice of non-negative integer <math>m^{}_{}</math> and digits <math>a_0,a_1^{},\ldots,a_m</math> chosen from the set <math>\{0^{}_{},1,2,\ldots,n^2\}</math>, with <math>a_m\ne 0^{}^{}</math>. We write | is true for a unique choice of non-negative integer <math>m^{}_{}</math> and digits <math>a_0,a_1^{},\ldots,a_m</math> chosen from the set <math>\{0^{}_{},1,2,\ldots,n^2\}</math>, with <math>a_m\ne 0^{}^{}</math>. We write | ||
<center><math>r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}</math></center> | <center><math>r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}</math></center> | ||
to denote the base <math>-n+i^{}_{}</math> expansion of <math>r+si^{}_{}</math>. There are only finitely many integers <math>k+0i^{}_{}</math> that have four-digit expansions | to denote the base <math>-n+i^{}_{}</math> expansion of <math>r+si^{}_{}</math>. There are only finitely many integers <math>k+0i^{}_{}</math> that have four-digit expansions | ||
− | <center><math>k=(a_3a_2a_1a_0)_{-3+i^{}_{}}~~~~a_3\ne 0.</math></center> | + | <center><math>k=(a_3a_2a_1a_0)_{-3+i^{}_{}}~~</math><math>~~a_3\ne 0.</math></center> |
Find the sum of all such <math>k^{}_{}</math>. | Find the sum of all such <math>k^{}_{}</math>. | ||
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<math>(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i</math> | <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 | + | So we need to solve the [[diophantine equation]] <math>a_1-6a_2+26a_3=0 \Longrightarrow 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: | 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> | + | *Case 1: <math>a_3=2</math> |
− | + | :The only solution to that is <math>(a_1, a_2, a_3)=(2,9,2)</math>. | |
− | The only solution to that is <math>(a_1, a_2, a_3)=(2,9,2)</math>. | + | *Case 2: <math>a_3=1</math> |
− | + | :The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>. | |
− | Case 2: <math>a_3=1</math> | + | *Case 3: <math>a_3=0</math> |
− | + | :<math>a_3</math> cannot be 0, or else we do not have a four digit number. | |
− | The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>. | ||
− | |||
− | Case 3: <math>a_3=0</math> | ||
− | |||
− | <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. | + | 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 <math>k</math> that can be expressed by one of those. |
<math>(292a_0)_{-3+i}</math>: | <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. | + | We plug the first three digits into base 10 to get <math>30+a_0</math>. The sum of the integers <math>k</math> in that form is <math>345</math>. |
<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 of the integers k in that form is 145. | + | We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum of the integers <math>k</math> in that form is <math>145</math>. The answer is <math>345+145=\boxed{490}</math>. |
− | |||
− | 345+145= | ||
== See also == | == See also == | ||
{{AIME box|year=1989|num-b=13|num-a=15}} | {{AIME box|year=1989|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Complex Numbers Problems]] |
Revision as of 14:15, 25 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
![$k=(a_3a_2a_1a_0)_{-3+i^{}_{}}~~$](http://latex.artofproblemsolving.com/9/1/1/9117bb67a6e8a65a1c9475a12d07bf0d76d78341.png)
![$~~a_3\ne 0.$](http://latex.artofproblemsolving.com/0/6/3/063a24c5b1954678bd93bbec326e554e671a4acf.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
that can be expressed by one of those.
:
We plug the first three digits into base 10 to get . The sum of the integers
in that form is
.
:
We plug the first three digits into base 10 to get . The sum of the integers
in that form is
. The answer is
.
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 |