document.write( "Question 701448: Are there more seven digit numbers with a 1 in them than without a 1 in them? Explain. \n" ); document.write( "
Algebra.Com's Answer #432456 by KMST(5328)\"\" \"About 
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A more general question would be:
\n" ); document.write( "Are there more n-digit numbers with a 1 in them than without a 1 in them?
\n" ); document.write( "The fraction of n-digit numbers without a 1 in them
\n" ); document.write( "(the probability that 1 would not be any of the n digits) is
\n" ); document.write( "\"b%5Bn%5D=%288%2F9%29%289%2F10%29%5E%28n-1%29\"
\n" ); document.write( "because the first digit cannot be zero,
\n" ); document.write( "and it would not be 1 for \"8%2F9\" of the possible digits,
\n" ); document.write( "while the other \"%28n-1%29\" digits have no restriction,
\n" ); document.write( "and would not be 1 in \"9%2F10\" of the cases.
\n" ); document.write( "If you were taught the name you would recognize that as a
\n" ); document.write( "\"geometric sequence\" or \"geometric progression\".
\n" ); document.write( "Even if you did not know of a name for them,
\n" ); document.write( "it is clear that the fraction of n-digit numbers without a 1 in them
\n" ); document.write( "(or the probability of an n-digit number not having a 1)
\n" ); document.write( "decreases as towards zero as \"n\" increases.
\n" ); document.write( "It is \"8%2F9\" for 1-digit numbers,
\n" ); document.write( "\"%288%2F9%29%289%2F10%29=8%2F10\" for 2-digit numbers,
\n" ); document.write( "and keeps getting smaller as we keep including more \"%289%2F10%29%3C1\" factors.
\n" ); document.write( "For 7-digit numbers the fraction of them without a 1 is
\n" ); document.write( "\"%288%2F9%29%289%2F10%29%5E6=0.472392%3C1%2F2\" ,
\n" ); document.write( "so there are \"highlight%28more%29\" 7-digit numbers with a 1 in them than without a 1 in them.
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\n" ); document.write( "EXTRAS:
\n" ); document.write( "Calculating \"b%5B6%5D\", I would find that \"b%5B6%5D%3E1%2F2\",
\n" ); document.write( "but how would I find the value of \"n\" where \"b%5Bn%5D\" becomes smaller than \"1%2F2\" without calculating and tabulating \"b%5Bn%5D\" values for n=1, 2, 3, ...?
\n" ); document.write( "That's where you use logarithms.
\n" ); document.write( "Sorry, I'm getting carried away.\r
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