 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Higher Powers of Eleven (Posted on 2017-05-24) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

 Search: Search body:
Forums (2)