+0

A bit stuck.

+2
205
6

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?

Jan 15, 2019

#1
+2

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

but anyway,

HOPE THIS HELPED!

Jan 15, 2019
#2
+3

The answer is 195.  I'm not sure how to show this.  I'm going to post this over at MHF.

Jan 15, 2019
#3
+3

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.  ??

Melody  Jan 15, 2019
#4
+2

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.

Jan 15, 2019
#6
+2

I bashed numbers ending in 1 for mod(143) and got 195 as an answer. Not the elegant solution I hoped for :l

Jan 16, 2019