document.write( "Question 694287: There is a 3digit number in which the 2nd digit is the sum of the 1st and 3rd digit. Can you prove that the number is divisible by 11? \n" ); document.write( "
Algebra.Com's Answer #427826 by KMST(5328)\"\" \"About 
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Yes.
\n" ); document.write( "As a matter of fact, that is the way to multiply a two digit number times 11.
\n" ); document.write( "It is very easy, as long as the sum of those digits is less than 10.
\n" ); document.write( "You just write the two digits, leaving some space in the middle, and then write the sum in that middle space.
\n" ); document.write( "\"a\" = first (hundreds) digit
\n" ); document.write( "\"b\" = third (ones) digit
\n" ); document.write( "Then, second (tens) digit = \"a%2Bb%3C=9\",
\n" ); document.write( "and the value of the 3-digit number is
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\n" ); document.write( "\"10a%2Bb\" is the 2-digit number with \"a\" as a first (tens) digit and \"b\" as a second (ones) digit.
\n" ); document.write( "The 3-digit number in the problem is \"11\" times the two digit number ab.
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