Difference between revisions of "2021 AIME I Problems/Problem 3"
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<b><u>Case 1:</u></b> The subtrahend (the second number in a subtraction expression) must be greater than <math>24</math> if the minuend is <math>1024</math>. In this case, the subtrahend can be ranging from <math>32</math> to <math>512</math> giving <math>5</math> total choices. | <b><u>Case 1:</u></b> The subtrahend (the second number in a subtraction expression) must be greater than <math>24</math> if the minuend is <math>1024</math>. In this case, the subtrahend can be ranging from <math>32</math> to <math>512</math> giving <math>5</math> total choices. | ||
| − | <b><u>Case 2:</u></b> If both numbers are powers of two less than <math>1024</math>, 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 <math>10\choose2</math><math>=45</math>. | + | <b><u>Case 2:</u></b> If both numbers are powers of two less than <math>1024</math>, 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 <math>10\choose2</math> <math>=\frac{10\cdot9}{2}=45</math>. |
In total, there are <math>45+5=\boxed{050}</math> numbers. | In total, there are <math>45+5=\boxed{050}</math> numbers. | ||
Revision as of 18:28, 11 March 2021
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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
