document.write( "Question 1046143: I didn't ask the type the question right before
\n" ); document.write( "Find the smallest odd positive integer (m) so that the decimal expansion of 1/m has just 4 non recurring digits and has 2 repeating digits.
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Algebra.Com's Answer #661702 by KMST(5345)\"\" \"About 
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\"1%2F99=%220.010101...%22\" , \"2%2F99=%220.020202...%22\" ,
\n" ); document.write( "and for any \"p\" integer that is not a multiple of 11, or 99,
\n" ); document.write( "the decimal part of \"p%2F99\" will be a 2-digit repeating sequence.\r
\n" ); document.write( "\n" ); document.write( "So, we can represent all \"10%5E4%2Fm\" numbers whose decimal expansion is a 2-digit repeating sequence as
\n" ); document.write( "\"p%2F99\" , where \"p\"= an integer that is not a multiple of 11.
\n" ); document.write( "Then, \"10%5E4%2Fm=p%2F99\" , and \"1%2Fm=%281%2F10%5E4%29%28p%2F99%29\"
\n" ); document.write( "has pairs of repeating digits after the 4th decimal place.
\n" ); document.write( "\"1%2Fm=%281%2F10%5E4%29%28p%2F99%29\"--->\"m=99%2A10%5E4%2Fp\"-->\"m=3%5E2%2A11%2A2%5E4%2A5%5E4%2Fp\" .
\n" ); document.write( "For that \"m\" to be an integer,
\n" ); document.write( "\"p\" cannot have any factors not present in \"3%5E2%2A11%2A2%5E4%2A5%5E4\" ,
\n" ); document.write( "and for \"m\" to be an odd integer,
\n" ); document.write( "\"p\" must have \"2%5E4\" as a factor.
\n" ); document.write( "Choosing \"p=3%5E2%2A2%5E4%2A5%5E4\" , we get \"m=3%5E2%2A11%2A2%5E4%2A5%5E4%2F%283%5E2%2A2%5E4%2A5%5E4%29=11\" with \"1%2Fm=%220.09090909...%22\" .
\n" ); document.write( "That is the smallest odd positive integer \"%28m%29\" so that the decimal expansion of \"1%2Fm\" has pairs of repeating digits after the 4th decimal place.
\n" ); document.write( "However, \"m=11\" may not be considered the right answer,
\n" ); document.write( "because the phrase \"just 4 non recurring digits\" probably means that the third decimal places must be different from the 5th one,
\n" ); document.write( "and the and fourth one must be different from the sixth one.
\n" ); document.write( "So, we need to put less factors into \"p\" ,
\n" ); document.write( "leaving \"m\" with more of the factors in \"3%5E2%2A11%2A5%5E4\" to get
\n" ); document.write( "\"m=3%2A11=33\" with \"1%2Fm=1%2F33=%220.03030303...%22\" (not right either),
\n" ); document.write( "\"m=5%2A11=55\" with \"1%2Fm=1%2F33=%220.02020202m=...%22\" (not right either),
\n" ); document.write( "\"m=3%5E2%2A11=99\" with \"1%2Fm=1%2F99=%220.01010101...%22\" , and so on.
\n" ); document.write( "We could keep trying to drop factors from \"p\" ,
\n" ); document.write( "until we get to \"m=11%2A5%5E6=6875\",
\n" ); document.write( "when \"1%2Fm=1%2F6875=%220.0001454545...%22\"
\n" ); document.write( "has a 4th decimal digit \"%281%29\" that is different from the sixth one \"%285%29\" .
\n" ); document.write( "So, \"highlight%28m=6875%29\" is my answer.
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\n" ); document.write( "NOTE: there is an option that does not require trying so many factor combinations to get to \"m\" ,
\n" ); document.write( "but it does require more algebra, so it may not get to the answer faster.
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