 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Year by Year (Posted on 2004-11-23) Consider N=2004^2004.

1) What are the first 3 digits of N?

2) What are the last 3 digits of N?

 See The Solution Submitted by SilverKnight Rating: 2.3333 (6 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution (spoiler) | Comment 1 of 8

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 6618-digit number must be 10069902351.  For the first 3 digits only 8-significant-digit 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 2004-11-23 15:11:39 Please log in:

 Search: Search body:
Forums (0)