Difference between revisions of "1994 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | 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>\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>. | + | 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>\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>\boxed{063}</math>. |
== See also == | == See also == | ||
Revision as of 20:31, 11 January 2022
Problem
The increasing sequence 
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 
. Either 
or 
must be divisible by 3. This is true when 
. Since 1994 is even, 
must 
. It will be the 
th such term, so 
. The value of 
is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
