Palindrome paranoia

A number is divisible by 45 = 9  5 if and only if it is divisible by both 5 and 9. If our palindromic number is divisible by 5, it must end in a 5, and hence also starts with a 5. Next, for it to be divisible by 9, the sum of the digits of the number must be divisible by 9. The largest palindromic 8-digit number that starts with a 5 and has digit sum divisible by 9 equals 59944995.

at first sight we clearly find out that since the number is divisible by 45 it must be divisible by 9 and 5 both. since it is divisible by 5 it will have unit digit 5 or 0 , not 0 because it is palindromic and if it would be 0 then the palindrome of it would start with first digit as 0 , so the unit digit is 5 and because it is a palindrome first digit would also be 5 to be read same from both sides. now, $let\quad us\quad take\quad the\quad number\quad to\quad be\quad k\\ k\quad =\quad 5abccba5\\ 9\quad |\quad k\quad so,\quad \\ 9\quad |\quad (5+a+b+c+a+b+c+5)\\ 9\quad |\quad 10+2(a+b+c)\quad let\quad ,\quad 2(a+b+c)\quad =\quad m\\ 10\quad \equiv \quad 1\quad (mod\quad 9)\\ m+10\quad \equiv \quad 9\quad (mod\quad 9)\quad \\ subtracting\quad we\quad get;\\ 2(a+b+c)\quad \equiv \quad 8\quad (mod\quad 9)\\ m\quad =\quad 9k+8\quad \\ now\quad we\quad have\quad to\quad determine\quad largest\quad possible\\ of\quad "m".\quad \\ m\quad may\quad be\quad equal\quad to\quad N\quad =\quad (8,17,26,35,44,53,62......)\\ but\quad since\quad m\quad is\quad even\quad 17,35,53\quad are\quad not\quad possible\\ if\quad m\quad =\quad 62\quad then\quad a+b+c\quad =\quad 31\quad which\quad is\quad \\ not\quad possible\quad as\quad a,b,c\quad are\quad digits\quad from\quad 1\quad to\quad 9\\ so\quad 62\quad and\quad onwards\quad are\quad not\quad the\quad values\quad of\quad m\\ so,\quad n\quad gets\quad modified\quad to:\\ N\quad =\quad (8,26,44)\\ let\quad us\quad take\quad the\quad largest\quad of\quad set\quad N\\ then\quad m\quad =\quad 44\\ a+b+c\quad =\quad 22\\ now\quad because\quad a\quad is\quad before\quad b\quad for\quad having\quad greatest\quad value\\ of\quad this\quad 8\quad digit\quad number\quad a>=b>=c\\ taking\quad largest\quad value\quad of\quad a=b=9\quad we\quad get\quad c=4\\ and\quad the\quad number\quad k\quad =\quad 59944995\\$