## Calculate the square root of this number (without a calculator): 2000*2001*2002*2003+1

Answer The trick is to realize that the product of four consecutive integers is always less than 1 from a perfect square! This can be seen from the following identity : (n-1)n(n+1)(n+2) +1 = (n^2 + n – 1)^2 In your ‘specific’ case n = 2001, Thus, we have, 2000*2001*2002*2003+1 = ( 2001^2 +...

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