Difference between revisions of "1997 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | Notice that all odd numbers can be obtained by using <math>(a+1)^2-a^2=2a+1,</math> where <math>a</math> is a nonnegative integer. All multiples of <math>4</math> can be obtained by using <math>(b+1)^2-(b-1)^2 = 4b</math>, where <math>b</math> is a positive integer. Numbers congruent to <math>2 \pmod 4</math> cannot be obtained because squares are <math> | + | Notice that all odd numbers can be obtained by using <math>(a+1)^2-a^2=2a+1,</math> where <math>a</math> is a nonnegative integer. All multiples of <math>4</math> can be obtained by using <math>(b+1)^2-(b-1)^2 = 4b</math>, where <math>b</math> is a positive integer. Numbers congruent to <math>2 \pmod 4</math> cannot be obtained because squares are <math> 0, 1 \pmod 4.</math> Thus, the answer is <math>500+250 = \boxed{750}.</math> |
== See also == | == See also == | ||
Latest revision as of 15:52, 2 March 2020
Problem
How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?
Solution
Notice that all odd numbers can be obtained by using 
where 
is a nonnegative integer. All multiples of 
can be obtained by using 
, where 
is a positive integer. Numbers congruent to 
cannot be obtained because squares are 
Thus, the answer is 
See also
| 1997 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. 
