document.write( "Question 1108845: Let n be a three digit number and let m be the number obtained by reversing the order of the digits in n. Suppose that m does not equal n and that n+m and n-m are both divisible by 7. Find all such pairs n and m.\r
\n" ); document.write( "\n" ); document.write( "I have found
\n" ); document.write( "168 and 861
\n" ); document.write( "259 and 952\r
\n" ); document.write( "\n" ); document.write( "Are there any more?
\n" ); document.write( "

Algebra.Com's Answer #723872 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
There are no other solutions. My proof is below.
\n" ); document.write( "The kids at artofproblemsolving would read it and laugh.
\n" ); document.write( "They would probably say that it was obvious based on the Ruszcyk theorem,
\n" ); document.write( "or some such thing.
\n" ); document.write( "(I just made that theorem up, but they probably have a few modular arithmetic theorems).
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "Let the numbers be made of the digits \"a\" , \"b\" , and \"c\" .
\n" ); document.write( "\"n=100a%2B10b%2Bc\" and \"m=100c%2B10b%2Ba\" are 3-digit numbers, so \"a%3C%3E0\" and \"c%3C%3E0\".
\n" ); document.write( "To have \"n%3C%3Em\" , it must be \"a%3C%3Ec\" ,
\n" ); document.write( "and to have \"n-m%3E0\" , it must be \"a%3Ec\" .
\n" ); document.write( "
\n" ); document.write( "For any two numbers x and y,
\n" ); document.write( "if x-y and x+7 are both multiples of 7,
\n" ); document.write( "x and y are both multiples of 7,
\n" ); document.write( "and vice versa.
\n" ); document.write( "So, m and n must be multiples of 7.
\n" ); document.write( "For \"n=100a%2B10b%2Bc=%287%2A14%2B2%29a%2B%287%2B3%29b%2Bc=7%2814a%2Bb%29%2B%282a%2B3b%2Bc%29\" to be a multiple of 7,
\n" ); document.write( "\"2a%2B3b%2Bc\" must be a multiple of 7.
\n" ); document.write( "Similarly, for \"m=100c%2B10b%2Ba\" to be a multiple of 7,
\n" ); document.write( "\"2c%2B3b%2Ba\" must be a multiple of 7.
\n" ); document.write( "If \"2a%2B3b%2Bc\" and \"2c%2B3b%2Ba\" are both multiples of 7,
\n" ); document.write( "\"%282a%2B3b%2Bc%29-%282c%2B3b%2Ba%29=a-c\" must be a multiple of 7, and
\n" ); document.write( "\"%282a%2B3b%2Bc%29%2B%282c%2B3b%2Ba%29=3%28a%2Bc%29%2B6b=3%28a%2Bc%2B2b%29\" must be a multiple of 7,
\n" ); document.write( "which means \"a%2Bc%2B2b\" must be a multiple of 7.
\n" ); document.write( "
\n" ); document.write( "For \"a-c\" to be a multiple of 7, the only choices are
\n" ); document.write( "\"system%28c=1%2Ca=1%2B7=8%29\" and \"system%28c=2%2Ca=2%2B7=9%29\" .
\n" ); document.write( "
\n" ); document.write( "With \"system%28c=1%2Ca=8%29\" , \"a%2Bc=9\" ,
\n" ); document.write( "and then for \"a%2Bc%2B2b=9%2B2b\" to be a multiple of 7,
\n" ); document.write( "it must be \"9%2B2b=21\" , \"b=6\" .
\n" ); document.write( "So one solution is \"system%28a=8%2Cb=6%2Cc=1%29\" or \"system%28n=861%2Cm=168%29\" .
\n" ); document.write( "
\n" ); document.write( "With \"system%28c=2%2Ca=9%29\" , \"a%2Bc=11\" ,
\n" ); document.write( "and then for \"a%2Bc%2B2b=11%2B2b\" to be a multiple of 7,
\n" ); document.write( "it must be \"11%2B2b=21\" , \"b=5\" .
\n" ); document.write( "So the other solution is \"system%28a=9%2Cb=5%2Cc=2%29\" or \"system%28n=952%2Cm=259%29\" .
\n" ); document.write( "
\n" );