document.write( "Question 252191: A two digit number has different digits. If the difference between the square of the number and the square of the number whose digits are interchanged is a positive perfect square, what is the two digit number? \n" ); document.write( "

Algebra.Com's Answer #183982 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!

\r\n" );
document.write( "This time drk's solution was shorter than mine.  Essentially he\r\n" );
document.write( "took square roots of both sides of\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "getting\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Ssince the right side is a positive integer, the left side must \r\n" );
document.write( "be also.  That means there must be another factor of 11 under \r\n" );
document.write( "the square root radical, and the only possibility of the sum of \r\n" );
document.write( "two digits being a multiple of 11 is for (t+u) to be 11, and that \r\n" );
document.write( "is only true when t=6 and t=5. \r\n" );
document.write( " \r\n" );
document.write( "Here was my longer solution:\r\n" );
document.write( "-----------------------------\r\n" );
document.write( "

\n" ); document.write( "A two digit number has different digits. If the difference between the square of the number and the square of the number whose digits are interchanged is a positive perfect square, what is the two digit number?\r
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "\r\n" );
document.write( "There must exist integer k so that:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Factor the left side as the difference of two perfect squares:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Simplifying:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Write 99 as a product of primes:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Since the left side is an integer, the right side must be also.\r\n" );
document.write( "So  must contain factors  as well as \r\n" );
document.write( "possibly another perfect square .\r\n" );
document.write( "\r\n" );
document.write( "So  must be a perfect square of the form ,\r\n" );
document.write( "where m is a positive integer. \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "That gives:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The largest possible value of (t-u)(t+u) is \r\n" );
document.write( "when t=9 and u=0, or (9-0)(9+0)=81 and the smallest\r\n" );
document.write( "is when t=1 and u=0, or (1-0)(1+0)=1.  Therefore the\r\n" );
document.write( "right side must satisfy this:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " since 4 is the largest perfect \r\n" );
document.write( "square that does not exceed , and since\r\n" );
document.write( "1 is the smallest largest perfect square that exceeds\r\n" );
document.write( ".\r\n" );
document.write( "Therefore\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So m is either 1 or 2.\r\n" );
document.write( "\r\n" );
document.write( "When m = 2, we have\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The only ways 44 can be written as the product of two\r\n" );
document.write( "integers are 1*44, 2*22, and 4*11. But the larger factor,\r\n" );
document.write( "t+u cannot be more than 9+8 or 17, thus that would leave\r\n" );
document.write( "only 4*11.  4 is even and 11 is odd. However t-u and t+u \r\n" );
document.write( "are either both even or both odd.  So we have ruled out \r\n" );
document.write( "m = 2.\r\n" );
document.write( "\r\n" );
document.write( "That leaves only the possibility \r\n" );
document.write( "\r\n" );
document.write( "m = 1 and\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Since 11 is prime, the only two positive integer factors it has\r\n" );
document.write( "are 1 and 11.  Therefore the smaller factor  must be 1\r\n" );
document.write( "and the larger factor 11.  So we have the system of equations:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Solving that gives us t=6 and u=5, which both can be digits.  \r\n" );
document.write( "So the only solution is the two digit number 65.\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );