From: MathNexus.wwu.edu
          on 3/15/2008

But Pepso did leave the cryptic message: "To brother Stu who 
acts so wise!  The new combination is the last three digits 
of 

Stu Dent asks for your help...What is the new combination? 

Source: Adapted from D. Piele's 7th International Computer 
Problem Solving Contest, 1987 

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Hint: You could try every one of the 10x10x10 possible 
combinations, but that would be no fun and take a lot 
of time. 

Look for possible patterns in powers of 9... 

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Finger pointing down from darrell94590 on 1/2/2006

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Drawing from Ripleys-Believe It Or Not

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              THE SOLUTION

Nine to the power of one is equal to ...........9.
Nine to the power of two is equal to ..........81.
Nine to the power of three is equal to .......729.
Nine to the power of four is equal to .......6561.
Nine to the power of five is equal to ......59049.
Nine to the power of six is equal to ......531441.
Nine to the power of seven is equal to ...4782969.
Nine to the power of eight is equal to ..43046721.
Nine to the power of nine is equal to ..387420489.
Nine to the power of ten is equal to ..3486784401.

Solution Commentary:

Some observed patterns in 9 to the nth power:

If n is even, the 1's digit is 1, otherwise it is a 9.

The 10's digits cycle through the sequence 0-8-2-6-4-4-6-2-8-0...
   some nice symmetry!

No nice pattern seems evident in the 100's digits.

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Thus, since 6! = 720, we can conclude that the combination is X-0-1,
as 720 is even and a multiple of ten.  Now, Stu would have to only
test 10 combinations, 0-0-1, 1-0-1, ..., 9-0-1.

But that's no fun.  Ignoring all of the "uninvolved" digits, focus
on the patterns of the last three digits of 9n for n a multiple of 10:
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BINGO! There was a pattern lurking in the 100's digits!  Now, since
720 = 14*50 + 20, we know that the combination will be 8-0-1.