document.write( "Question 999014: Qus1-if a digit is written 6 time always divisible by?
\n" ); document.write( "Qus2-Number divisible by 99 is: \n" ); document.write( "

Algebra.Com's Answer #616785 by KMST(5328) $\"\"$ $\"About$

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Question 1:
\n" ); document.write( "A number made of a digit x repeated 6 times is x(111111),
\n" ); document.write( "so it is divisible by 111111 and its factors, such as
\n" ); document.write( "3, 7, 11, 13, 37, and products of those factors.
\n" ); document.write( "Since $\"111111=3%2A7%2A11%2A13%2A37\"$ , there are
\n" ); document.write( " $\"2%5E5=32\"$ factors of 111111, including 1, and 111111.\r
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\n" ); document.write( "\n" ); document.write( "Question 2:
\n" ); document.write( "A number divisible by 99 is divisible by 9 and by 11.
\n" ); document.write( "Adding up all its digits, and adding up the digits of the sum,
\n" ); document.write( "and repeating the adding process as needed, the result is 9.
\n" ); document.write( "The difference between the sum of the digits in the first, third, fifth, etc place,
\n" ); document.write( "and the sum of the other digits is 0, or 11, or 22, etc. \n" ); document.write( "

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