The meaning of problems
practice
Conclusion. 1 : \ (R & lt (A) | R & lt (AM) (A, m \ GE. 1) \)
\[\frac{R(am)}{R(a)}=\frac{\frac{10^{am}-1}{9}}{R(a)}=\frac{\frac{(9+1)^{am}-1}{9}}{R(a)}\]
Corollary. 1 : \ (P | R & lt (10 m ^) \) , then \ (p | R (10 ^ n) (n \ ge m) \)
Set \ (A (p) \) is minimum when the \ (P | R & lt (A (P)), P \ in Prime \) , the other \ (K | A (P) \) , if the \ (k | 10 n-^ \) , then \ (p | R (10 ^
n) \) is now a requirement \ (A (p) \)
\ (A (p) \) is not so easy to find, we estimate its scope, \ (A (the p-) \ the p-Le \) , so if \ (A (the p-) | 10 the n-^ \) , then \ (A (p) \) except for the \ (1 \) (in fact, is not \ (1 \) ) necessarily contain only outer \ (2 \) and \ (5 \) two prime factors
\ (^ 10 n-= (2 *. 5) ^ n-\) , \ (A (P) = 2 ^ A5 ^ B \) , \ (A \ Le 16, B \ Le. 8 \) , so \ (n \ le 16 \)
Therefore, we direct that \ (10 ^ {16} \) judge whether \ (p | R (10 ^ {16}) \) to