Find the least positive integer \(n\) such that when \(3^n\) is written in base 143, its two right-most digits in base 143 are 01.
Unsure of how to approach. Hints?
it leaves a remainder of 1 when divided by 143?
so the remainders would be 3, 9, 27, 81, 100, 14, 42, 126, 92, 133, 113, 53, 16, 48, 1 and thats the cycle. the last one is 1 and that is the 15th term, but then since it is (143*143*x)+1 and the very end, then it leaves remainder one when divided by 143*143 too, and a bunch of complicated stuff. you could maybe square everything and then it would become 225, so maybe it is 225? idk man
HOPE THIS HELPED!
The answer is 195. I'm not sure how to show this. I'm going to post this over at MHF.
Thanks Rom, much appreciated.
I thought about it but I am not good with mod questions.
I assume the fact that 143=11*13 can somehow be used. ??
The response I received.
If you convert 3^195 to base 143, Mathematica 11 gives the following result: Note that every group separated by colon(:) is considered as "one digit":
3^195 in base 143 =2:40:141:29:79:52:61:74:79:122:117:123:66:25:41:52:128:114:75:11:142:18:135:75:35:31:79:89:25:73:123:36:96:4:71:10:124:75:9:30:118:113:0:1_143 (44 digits).
As you can see, it ends in 01.