Question 1150159
<br>
-----------------------------------------------------------------------<br>
(corrected response, thanks to tutor @ikleyn for pointing out the obvious error....)
-----------------------------------------------------------------------<br>
The sum of the digits of 2007 is 9, so the number is divisible by 9:<br>
{{{2007 = 3*3*223}}}<br>
223 is prime, so we have the prime factorization of 2007:<br>
{{{2007 = (3^2)(223^1)}}}<br>
Then<br>
{{{2007^2007 = (3^4014)(223^2007)}}}<br>
A perfect square factor of the number has to be of the form<br>
{{{(3^a)(223^b)}}}<br>
Where a and b are even integers, with 0 <= a <= 4014 and 0 <= b <= 2007.<br>
That gives us 2008 choices for a and {{{cross(2007)}}} 1004 choices for b; the number of perfect square factors of the number is then {{{cross(2008*2007 = 4030056)}}} 2008*1004 = 2016032.<br>