Determine analytically a positive four digit whole number with no leading zeroes which is such that the value obtained by reversing the number is 81 less than twice the original number.

Since 2(1000k+100h+10t+u)-81 = 1000u+100t+10h+k, the value obtained by reversing must be odd so k is odd. By combining variables in the equation we get

1999k+190h = 80t+998u+81. Since t<=9 and u<=9, 1999k+190h < 9783 so k < 5. Either k=1 or k=3. If k=1 then 1918+190h = 80t+998u  ==>  3=3u mod5 so u=1 or u=6. Now u=1  ==>  92+19h = 8t which is impossible since t<10. Also  u=6  ==>  19h = 8t+407 again impossible since h<10. So k must equal 3 and our original equation becomes  5916+190h = 80t+998u forcing 3u=1 mod5 so u=2 or u=7. Now u=2 implies that 392+19h = 8t which is impossible since t<10 so u must equal 7. Now we have 5916+190h = 80t+6986 which reduces to 19h = 8t+107. This last equation forces h=9 and t=8. Therefore original number is 3987.                          

Posted by Dennis on 2006-12-08 08:55:44