1994 AIME Problems/Problem 1
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 be congruent to . It will be the th such term, so . The value of is .
~minor edit by Yiyj1
See also
1994 AIME (Problems • Answer Key • Resources) | ||
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