Determine the last two and the first two digits of:

32^1+32^2+32^3+...+32^2007

Let x be the number in question. x=(32^2008 - 32)/31. Since the first two digits are unaffected by the -32 term, consider x to be approximately 32^2008/31. Taking logs and simplifying yields log x = 3020.849795 --> x = 7.076 X 10^3020 so the first two digits are 70. Also 31x = (32^4)^502 - 32 --> 31x=1048576^502 - 32 --> 31x=76^502 - 32 mod100 --> 31x=76-32 mod100 (since 76^k = 76 mod100). So 31x=44 mod100 --> x=24 mod100 --> last two digits are 24.

Posted by Dennis on 2006-12-30 17:37:13