Find all possible pairs (a,b) of positive integers such that a^a has b digits and b^b has a digits, where each of a^a and b^b is expressed in base ten.

list
   10   for A=1 to 999
   20     Pwr=A^A
   30     B=len(cutspc(str(Pwr)))
   40     Pwr=B^B
   50     if len(cutspc(str(Pwr)))=A then print A;B,A^A,B^B
   60   next

 a  b      a^a            b^b
 1  1    1               1
 8  8    16777216        16777216
 9  9    387420489       387420489
Overflow in 50
?a,b
 183     415
OK

That when we go higher than this, the lengths become larger than is possible for a solution is shown by:

a     len(a^a)
1       1
2       1
3       2
4       3
5       4
6       5
7       6
8       8
9       9
10      11
11      12
12      13
13      15
14      17
15      18
16      20
17      21
18      23
19      25
20      27
21      28
22      30
23      32
24      34
25      35
26      37
27      39
28      41
29      43
30      45
31      47
32      49
33      51

So that, for example 33^33's length of 51 would lead to 51^51, which is certainly larger than 33 digits long (in fact, larger than 51 digits long as this series progresses).

Posted by Charlie on 2013-04-12 16:07:42