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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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Solution computer solution | Comment 1 of 4
121 (a single 2)  is enough for division by 11^2
12222222222222222222221 (21 2's) is enough for 11^3
241 2's works for 11^4

We overflow the power of UBASIC trying for 11^5

   10   N=121:Tst=N:Pwr=0:Twos=1
   15    repeat
   16     Tst=N:Pwr=0
   20     repeat
   25        Done=0
   40        if Tst@11=0 then Pwr=Pwr+1:else:Done=1
   50        Tst=Tst\11:if Tst=0 then Done=1
   60     until Done
   70     if Pwr>Maxpwr then Maxpwr=Pwr:print N,Maxpwr,Twos:Ct=Ct+1
   80     N=int(N/10)*100+21:Twos=Twos+1
   90   until Ct>22

which crashes in line 80 when there are already 2591 2's.

A better approach finds 2661 2's works for 11^5:

   10   N=121:Tst=N:Pwr=0:Twos=1
   11   Eleven5=11^5
   15    repeat
   20    ' print N@Eleven5
   30    if N@Eleven5=96448 then print Twos:Ct=Ct+1
   50    if N@Eleven5=0 then print Twos;"*":Ct=Ct+1
   80     N=((N-1)*10+21)@Eleven5:Twos=Twos+1
   90   until Ct>19

The reference to 96448 was made to verify against a known value from the first run, which did not get as far as 2661 2's.

  Posted by Charlie on 2017-05-24 11:32:24
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