Difference between revisions of "2021 AIME I Problems/Problem 3"
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<math>\binom{11}{2} = 55</math> We need to subtract 5 since <math>2^{10} - 2^0, 2^1, 2^2, 2^3, 2^4</math> don't work. <math>55-5=\boxed{050}</math> ~hansenhe | <math>\binom{11}{2} = 55</math> We need to subtract 5 since <math>2^{10} - 2^0, 2^1, 2^2, 2^3, 2^4</math> don't work. <math>55-5=\boxed{050}</math> ~hansenhe | ||
| − | ==Solution 2 (More | + | ==Solution 2 (More Detailed Explaination)== |
All of the powers of <math>2</math> subtracted by another power of <math>2</math> that can result within 1000 is listed: | All of the powers of <math>2</math> subtracted by another power of <math>2</math> that can result within 1000 is listed: | ||
<math>0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024</math> since <math>2048-1024>1000</math>. None of the numbers when chosen two numbers will be the same because the difference of powers of <math>2</math> can be written as a power of two times a non-power of two. | <math>0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024</math> since <math>2048-1024>1000</math>. None of the numbers when chosen two numbers will be the same because the difference of powers of <math>2</math> can be written as a power of two times a non-power of two. | ||
Revision as of 18:25, 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 is listed:
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. 
