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>0,1,2,\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 | ||
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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}~~</math></center> <p> <center> <math>~~a_3\ne 0.</math> </center> | <center><math>k=(a_3a_2a_1a_0)_{-3+i}~~</math></center> <p> <center> <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>, |
== Solution == | == Solution == | ||
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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 | + | So we 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>1\leq a_3 \leq 2</math> since <math>a_3</math> can't equal 0, so we try cases: |
*Case 1: <math>a_3=2</math> | *Case 1: <math>a_3=2</math> | ||
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*Case 2: <math>a_3=1</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>. | :The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>. | ||
− | |||
− | |||
− | So we have | + | So we have 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>: | ||
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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>. | 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>. | ||
+ | ~minor edit by Yiyj1 | ||
== See also == | == See also == |
Latest revision as of 17:28, 16 September 2023
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
. 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/f/8/1/f81d425eb93665b6c8598c7cf275730089c4df4d.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 solve the diophantine equation .
The minimum the left-hand side can go is -54, so since
can't equal 0, so we try cases:
- Case 1:
- The only solution to that is
.
- Case 2:
- The only solution to that is
.
So we have 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
.
~minor edit by Yiyj1
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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