I believe (and I hope I'm right) that if a

two-digit number to the power of seven yields a number

whose last two digits end in x and y (for example),

then any number ending in those two digits will, when

raised to the power of 7, end in x and y. So if 34^7

ends in 44 (52523350144 to be precise), then 3234^7

should end in 44. And it does

(3699815853555305399228544). <br><br>In this way, all we have to do is raise 7

to the power of 7, raise the last two digits to the

pwer of seven, and repeat this till we get a

pattern.<br><br>Let s(x) mean the the last two digits of the number

got when raising 7 to the power of 7 x times. So...

<br>s(1) = last two digits of 7^7 = 43<br>s(2) = last two

digits of (7^7)^7 = last two digits of 43^7 = 07<br>Get

me?<br><br>s(1) = 43<br>s(2) = 07<br>s(3) = 43<br>s(4) =

07<br>etc etc etc. Very boring, ain't it? So for every even

x, s(x) = 07. Answer: s(1000) = 07