Subj:.....A Power-Full Problem  (S644)           From: MathNexus.wwu.edu           on 5/13/2008 Drawing from CarsCarsCars...

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Source: http://mathnexus.wwu.edu/Archive/problem/detail.asp?ID=164 At a mathematics conference recently, I picked up an interesting problems from one of the speakers (C.B., CWU). Problem: Evaluate 100² - 99² + 98² - 97² + 96² - 95² +...+ 2² - 1² The problem brought back memories of when I once asked students to solve the following: (A-X)(B-X)(C-X)(D-X)....(Y-X)(Z-X)! Hint: What algebraic technique begs to be applied? Don't use a calculator. . . Finger pointing down from darrell94590 on 1/2/2006 .       Drawing from Ripleys-Believe It Or Not ..               THE SOLUTION   100² - 99² can be factored into (100 - 99) (100 + 99) so the problem can be completely factored into (100-99)(100+99) + (98-97)(98+97) + (96-95)(96+95) +...    + (6-5)(6+5) + (4-3)(4+3) + (2-1)(2+1) Note that 100-99 and 98-97 and 96-97 all equal 1. So this problem reduces to (100+99) + (98+97) + (96+95) +...+ (6+5) + (4+3) + (2+1) which is 199 + 195 + 191 +...+ 11 + 7 + 3 Notice that the first term, 199, and the last term, 3, add to 202.    So I could replace 199 and 3 with any two numbers that add to    202.  I will choose to replace the both them with two 101s. Similarly 195 and 7 add 202 and 191 and 11 add to 202.  So I can    replace these pairs all with 101s.  Continuing the pattern,    all numbers can be replaced with 101s. The problem now looks like 101 + 101 + 101 +...+ 101 + 101 + 101 How many 101s do we have here?  If you count from 1 to 100 that's a hundred numbers.  If you group them is pairs, that is fifty pairs so there are fifty 101s.  Therefore 101 + 101 + 101 +...+ 101 + 101 + 101 = 50(101) which is 5050, completely determined with out a calculator.

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