11Suppose 2k=1a1a2....aj
Here k=1+j...
Now folowing is the list of the 1st 2 digits of the coming powers...
2 or 3
4 or 5 6 or 7
8 or 1.... 1... or 1....
1....
Easy to see that these nos contain j+2 digits with leading digit 1....
And it is also quite obvious that there shall be exactly 1 such power, for example suppose there exists another power, 2l, where l=k+a.....then it is easy to visualise the leading digit does not remain 1 anymore, thus proved, that if it is true for k=j+1, it's true for k'=j+2 as well, j=1 satisfies the statement (24=16), thus proved.
1let p(k) denote a power of 2 with k digits with 1 as the leading digit
101<p(2)<2(101)
102<p(3)<2(102)
suppose 10k-1<p(k)<2(10k-1) {if this is proved, uniqueness is self explanatory}
multiplying throughout by 23
(8)10k-1<(23)p(k)<16(10k-1)
case 1
10(10k)<(23)p(k)<16(10k-1)
so (23)p(k) = p(k+1)
case 2
(8)10k-1<(23)p(k)<(10)(10k-1)
multiplying throughout by 2
(16)10k-1<(24)p(k)<(20)(10k-1)
so (24)p(k)=p(k+1)
now it can be seen either of case 1 or 2 is true but never both
uniqueness of the result is as below
1
rahul1993 Duggal
·2009-08-04 02:41:07
please omit the last line 'uniqueness of the result is as below'
