2021 AIME I Problems/Problem 3
Problem
Find the number of positive integers less than that can be expressed as the difference of two integral powers of
Solution
We need to subtract 5 since don't work. ~hansenhe
Solution 2 (More Detailed Explaination)
All of the powers of subtracted by another power of that can result within 1000 are since . None of the numbers when chosen two numbers will be the same because the difference of powers of can be written as a power of two times a non-power of two.
Case 1: The subtrahend (the second number in a subtraction expression) must be greater than if the minuend is . In this case, the subtrahend can be ranging from to giving total choices.
Case 2: If both numbers are powers of two less than , then we can choose two numbers from that list and order them to form a positive number. The amount of ways to do this is . In total, there are numbers.
~Interstigation
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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