# 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 + 2001 -1 ) ^2

Thus SQRT [ 2000*2001*2002*2003+1 ]

= SQRT [ 2001^2 + 2001 -1 ) ^2 ]

= 2001^2 + 2001 -1

= 2001^2 + 2000

= ( 2000 + 1 ) * ( 2001 ) + 2000

= (2000*2001) + 2001 + 2000 [ you can easily calculate these without a calculator ]

= (1000*2*2001) + 2001 + 2000

= (1000*4002) + 2001 + 2000

= 4002000 + 4001

= 4006001

Done.