Difference between revisions of "1997 AIME Problems/Problem 1"

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== Problem ==
 
== Problem ==
How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?
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How many of the integers between 1 and 1000, inclusive, can be expressed as the [[difference of squares|difference of the squares]] of two nonnegative integers?
  
 
== Solution ==
 
== Solution ==
{{solution}}
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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>
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== See also ==
 
== See also ==
* [[1997 AIME Problems]]
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{{AIME box|year=1997|before=First Question|num-a=2}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 14:52, 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 $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$, where $b$ is a positive integer. Numbers congruent to $2 \pmod 4$ cannot be obtained because squares are $0, 1 \pmod 4.$ Thus, the answer is $500+250 = \boxed{750}.$

See also

1997 AIME (ProblemsAnswer KeyResources)
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

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