 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Power of an Integer Beginning with an Arbitrary Sequence (Posted on 2008-01-31) L is any integer other than a power of 10. M is any integer. Show that there is an integral power of L that begins with the sequence of digits given in M.

 See The Solution Submitted by FrankM Rating: 3.8000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Might be the solution | Comment 4 of 6 | `The Problem statement is equivalent to:For some positive integers(x,k)M*10^x ≤ L^k < (M+1)*10^xTaking log for the above equationlogM + x ≤ klogL < log(M+1)+x=> (logM+x)/logL ≤ k < (log(M+1)+x)/logLWe can find x such that logM+x = (k-α)logL where α is approximately 0 for some kThen log(M+1)+x = logM*(1+1/M)+xlog(M+1)+x = logM + x + ((1/M)-(1/2M�)+...) (log(1+x) expansion)log(M+1)+x = (k-α)logL + ((1/M)-(1/2M�)+...)As α is very near to 0,((1/M)-(1/2M�)+...) willbe greater than αlogL, which makes(log(M+1)+x)/logL is just greater than k.Hence there exists an integral power of L which startswith M. `

 Posted by Praneeth on 2008-02-01 02:54:00 Please log in:

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