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Higher Powers of Eleven (Posted on 2017-05-24) Difficulty: 3 of 5
Let N(x) be the number 122....221 where the digit 2 occurs x times.

Twice in the past we have determined the highest power of 11 that divides N(2001) is 11^3.

What is the smallest x for N(x) to be a multiple of 11^3? What about multiples of 11^4 and 11^5?

See The Solution Submitted by Brian Smith    
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re: Analytic Solution Comment 4 of 4 |
(In reply to Analytic Solution by Brian Smith)

To continue in this line N(322101) is the minimal term for 11^7. 

11   Eleven5=11^7
15    do

50    if N%Eleven5=0 then 
     print Twos;"*"
80     N=((N-1)*10+21)%Eleven5
90   loop until Ct>0

----a mess of a program as it was translated into Mintoris Basic. 

  Posted by Charlie on 2017-05-26 15:45:40
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