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Difference between revisions of "1994 AIME Problems/Problem 1"

 
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== Problem ==
 
== Problem ==
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The increasing [[sequence]] <math>3, 15, 24, 48, \ldots\,</math> consists of those [[positive]] multiples of 3 that are one less than a [[perfect square]].  What is the [[remainder]] when the 1994th term of the sequence is divided by 1000?
  
 
== Solution ==
 
== Solution ==
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One less than a perfect square can be represented by <math>n^2 - 1 = (n+1)(n-1)</math>. Either <math>n+1</math> or <math>n-1</math> must be divisible by 3. This is true when <math>n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}</math>. Since 1994 is even, <math>n</math> must be congruent to <math>1 \pmod{3}</math>. It will be the <math>\frac{1994}{2} = 997</math>th such term, so <math>n = 4 + (997-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.
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~minor edit by [[User: Yiyj1|Yiyj1]]
  
 
== See also ==
 
== See also ==
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* [[1994 AIME Problems]]
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{{AIME box|year=1994|before=First question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 20:05, 1 September 2023

Problem

The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000?

Solution

One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$. Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}$. Since 1994 is even, $n$ must be congruent to $1 \pmod{3}$. It will be the $\frac{1994}{2} = 997$th such term, so $n = 4 + (997-1) \cdot 3 = 2992$. The value of $n^2 - 1 = 2992^2 - 1 \pmod{1000}$ is $\boxed{063}$.

~minor edit by Yiyj1

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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