Determine all possible positive integer(s) N < 10,000, such that:

2^N = 88 (mod 167), and:

2^N = 70 (mod 83)

**** For an extra challenge, solve this problem without using a computer program.

lhs=1; prev=0
for n= 1 to 99999
  {
  lhs=lhs*2
  if lhs mod 167 == 88 and lhs mod 83 ==70
    {
    l=n*log[2]
    diff=n-prev
    prev=n
    println ["$n $l $diff"]
    }
  }
 
finds only two that are below 10,000: 2004 and 8810, leading to a 604-digit number and to a 2653-digit number as their respective powers of 2.

Those two values of n are separated by 6806, and apparently every 6806th value of n thereafter satisfies the conditions as given in this table:
 

  n       log(2^n)             dn
 2004   603.26411131061822422 2004
 8810  2652.0742617996739298  6806
15616  4700.8844122887296354  6806
22422  6749.694562777785341   6806
29228  8798.5047132668410466  6806
36034 10847.314863755896752   6806
42840 12896.125014244952458   6806
49646 14944.935164734008163   6806
56452 16993.745315223063869   6806
63258 19042.555465712119575   6806
70064 21091.36561620117528    6806
76870 23140.175766690230986   6806
83676 25188.985917179286692   6806
90482 27237.796067668342397   6806
97288 29286.606218157398103   6806  

the ceiling of the common log can be used to show the number of digits in the power of 2, and the fractional part allowing view of the beginning of the number by taking the antilog.

Posted by Charlie on 2011-04-24 14:28:41