Here’s another problem I’ve been looking at lately:
Find a number 
, such that if 
is any digit from 0 to 9, 
has as its first digit past the decimal point. For instance, 63.18 works for the first power 
and the third power 
, but it doesn’t work for the second 
where the first digit is 7 instead of 2.
Hint: The smallest such number is less than 10.
To prove that the problem is possible to solve quickly, I’ll provide the solution to a more complicated version. Suppose that the first two digits of must match instead of just the first one. This means that 
has “03” just to the right of the decimal point all way to up to 
having “99” just after the decimal point. Then the smallest solution is the following number:
95.01063709417263364386569653065071891390859441647675346849
01552762611728856071543459219055118914594166160224386992990
49592128996983422861693601042135900702684741868297684742429
253949832737577847233354…
The smallest solution’s digits actually never end so I had to cut it off somewhere. I choose the cut off point to be the first place where the remaining digits still worked. In other words, if you truncated the 4 off of the end, the number would be changed too much and would fail to meet all the conditions. Furthermore, I was forced to round up (because anything smaller doesn’t work).
