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        riddles >> medium >> Deleting the first digit
        
(Message started by: fatball on Jan 22^nd, 2006, 9:02pm)

Title: Deleting the first digit
Post by fatball on Jan 22^nd, 2006, 9:02pm

Find all powers of 2 such that, after deleting the first digit, another power of 2 remains. (For example, 2⁵ = 32. On deleting the initial 3, we are left with 2 = 2¹.) Numbers are written in standard decimal notation, with no leading zeroes.

Title: Re: Deleting the first digit
Post by Eigenray on Jan 23^rd, 2006, 2:50am

[hideb]Note that if the first digit k is even, then dividing the number by 2 gives another solution. So let's assume the first digit k is odd. Say
2^x = k10^y + 2^z.
Since x>z, the largest power of 2 dividing k10^y=2^x-2^z is y=z. Thus
2^x-z = k5^z + 1.
Now, if we're not allowing the second digit to be 0, that is, 2^x has just one more digit than 2^z, then 0<x-z<7, and we find only the solution k=3, z=1. Thus the only such number with odd first digit is 32, and the only other number is then 64.

But we can handle the more general case. As explained [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1137859995;start=0#1]before[/link], 2^x=2^z mod 5^z implies that x=z mod 4*5^z-1, so that x-z > 4*5^z-1. But it should not take much convincing that
2^{4*5^(z-1)} < 10 * 5^z + 1
is only possible for z=1.[/hideb]