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

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== Problem ==
 
== Problem ==
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?
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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 ==
{{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 <math>\displaystyle \equiv 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>063</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|before=First question|num-a=2}}
 
{{AIME box|year=1994|before=First question|num-a=2}}

Revision as of 11:27, 5 April 2007

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 $\displaystyle \equiv 1 \pmod{3}$. 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 $063$.

See also

1994 AIME (ProblemsAnswer KeyResources)
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