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=7632 mod100 (since 76^k = 76 mod100). So 31x=44 mod100 > x=24 mod100 > last two digits are 24.

Posted by Dennis
on 20061230 17:37:13 