Consider N=2004^2004.
1) What are the first 3 digits of N?
2) What are the last 3 digits of N?
log(2004)=3.301897717195208
Multiply by 2004 to get log(2004^2004) = 6617.003025259197.
Ignoring the characteristic, the antilog of the mantissa is 1.006990235161036 (but the last 3 or 4 digits would be spurious due to the loss of 4 significant figures by subtracting the 6617 characteristic). The accuracy is certainly enough to say the first three digits are 100. In fact the first several digits of this 6618digit number must be 10069902351. For the first 3 digits only 8significantdigit accuracy is needed in the calculator.
2004 mod 1000 is 4.
2004^5 mod 1000 is 24. (because 4^5 is 1024)
2004^50 mod 1000 is 376. (because 24^10 is 63403380965376)
2004^200 mod 1000 is 376. (because 376^4 is 19987173376)
and 376 to any power mod 1000 is 376, so
2004^2000 mod 1000 is 376.
Then 376 * 4^4 = 256 mod 1000.
So the last three digits of 2004^2004 are 256.
For a calculator with fewer digits of accuracy, more steps would be needed to keep the integers within the limits of accuracy, such as 24^5=7962624, then 624^2=389376, to arrive at the 2004^50 mod 1000 = 376.
Edited on November 23, 2004, 3:21 pm

Posted by Charlie
on 20041123 15:11:39 