Difference between revisions of "2000 SMT/Algebra Problems/Problem 2"

(Created page with "Problem 2 Evaluate 2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3")
 
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Evaluate 2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3
 
Evaluate 2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3
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Solution 1 - SUBMITTED BY HOWDOI_YT
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I can write the first 1999 as 2000-1, and the 3rd 2000 as 1999+1;
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2000^3 - 2000^2(2000-1) - 1999^2(1999+1) + 1999^3 = 2000^3 - 2000^3 + 2000^2 - 1999^3 - 1999^2 + 1999^3 = 2000^2 - 1999^2 = (2000 - 1999)(2000 + 1999) = **3999** \qed

Revision as of 14:39, 18 April 2023

Problem 2

Evaluate 2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3

Solution 1 - SUBMITTED BY HOWDOI_YT I can write the first 1999 as 2000-1, and the 3rd 2000 as 1999+1; 2000^3 - 2000^2(2000-1) - 1999^2(1999+1) + 1999^3 = 2000^3 - 2000^3 + 2000^2 - 1999^3 - 1999^2 + 1999^3 = 2000^2 - 1999^2 = (2000 - 1999)(2000 + 1999) = **3999** \qed