Difference between revisions of "2000 SMT/Algebra Problems/Problem 2"
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Solution 1 - SUBMITTED BY HOWDOI_YT | 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; | 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 | 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 | ||
| + | (oml i just realized how bad this looks someone help me edit it lmao) | ||
Revision as of 13:40, 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 (oml i just realized how bad this looks someone help me edit it lmao)
