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Minimum Integer from Last 2 Digits Puzzle (Posted on 2023-01-20) Difficulty: 3 of 5
Determine the minimum value of a positive integer n such that when 3n is represented in base 143, the last two digits (reading left to right) is 01 (in this order).
Provide the required smallest value of n in base 10.

  Submitted by K Sengupta    
Rating: 5.0000 (1 votes)
Solution: (Hide)
Charlie has furnished a computer program assisted solution here.

My analytic solution is furnished hereunder as follows:

The least positive integer satisfying all the given conditions is 195.

EXPLANATION:-

By the given conditions, we have:
3^n == (mod 143^2)
Since gcd(11,13)=1, we must have:

3^n == 01 (mod 11^2)
3^n == 01 (mod 13^2)
Now, 3^5== 1 (mod 11^2)
So, n=5 is the minimum solution to 3^n ==1 (mod 11^2)
We now consider:
3^n == 01 (mod 13^2)
Now, 3^3==1 (mod 13)
=> 3^(3k) == 1 mod (13)
Then, we will solve for 3^(3k) == 1 mod (13^2)
Now,
3^3k =27^k
= (13*2+1)^k
= (13*2)^0 *1^k + k*13* 2* (1)^(k-1) (mod 1 3^2)
= 1+ (13)*(2k) (mod 13^2)
Accordingly,
3^(3k) == 1 (mod 13^2) gives:
13*2k ==0(mod 13^2)
=> 2k == 0(mod 13)
=> k== 0(mod13)
Accordingly, the minimum positive integer value of k is 13.
But, n= 3k, and accordingly, n = 3*13=39
So, 3^39 == 0(mod 13^2)
So, n=39 is the minimum positive integer solution to: 3^39 == 0(mod 13^2)
However, we observe that: lcm(5, 39)=195
Consequently, it follows that the least positive integer satisfying all the given conditions is 195

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutioncomputer solutionCharlie2023-01-20 14:13:12
SolutionSolution by Two MethodsLarry2023-01-20 08:56:26
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