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?

x=11^(-n)/9 (100^(11^(n - 1)) - 1) gives all solutions larger than 1.

e.g.

for n=1, x=1: 1*11^2=121

for n=2, x= 9182736455463728191 ( x(n=2) an interesting number in itself); then

x*11 = 101010101010101010101,
x*11^2 = 1111111111111111111111,
x*11^3 = 12222222222222222222221, working the digits into the desired form.

for n=3, x=8347942232239752900909925703314133066199181901661240504215710827281075214959512480173637198430586860338926454628933967776942983554553802487686785207446364471157859587611653727356206695049670256281826529760414057934719091743885132314884381*
x*11 = 91827364554637281910009182736455463728191000918273645546372819100091827364554637281910009182736455463728191000918273645546372819100091827364554637281910009182736455463728191000918273645546372819100091827364554637281910009182736455463728191 (note the digit repetition, 11 copies of x(n=2) with breaks)
x*11^2 = 1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
x*11^3 = 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
x*11^4 = 122222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222221

from which pattern the solution readily emerges.

*With a little deductive effort, it seems this number has the nice form 90909090909090909090918181818181818181818182727272727272727272727363636363636363636363645454545454545454545455454545454545454545454636363636363636363636372727272727272727272728181818181818181818181090909090909090909091*x(n=2)

 

Edited on May 26, 2017, 1:41 am

Posted by broll on 2017-05-26 00:06:45