The first four digits of an eight-digit perfect square are 2012. Find its square root?

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Answer 1 · 2017-04-28T13:40:49

#+-2sqrt503#

#2012=2^2*503#
And 503 is a prime number
Because #22^2<503<23^2#
So the square root of 2012 is
#+-sqrt2012=+-2sqrt503#

Answer 2 · 2017-04-28T17:00:09

See below.

We have

#sqrt(20120000) approx 4485.53#

This square root can be extracted manually

and

#4485^2 = 20115225#

so the number is

#4486->4486^2= 20124196#