Find the largest six-digit palindrome that is divisible by 45.

Step-by-step solution

1. 1

   Write the general six-digit palindrome. It has the digit pattern $abccba$, so

   $N=\overline{abccba},\qquad a\neq 0$

   Only $a$, $b$, $c$ are free — three digits, not six.

2. 2

   Split 45 into coprime factors. Since $45=9\times 5$ and $\gcd(9,5)=1$, the number must be divisible by $9$ and by $5$. These give two independent digit conditions.

3. 3

   Apply divisibility by 5 to pin down $a$. The last digit must be $0$ or $5$. In a palindrome the last digit equals the first, and the first digit of a six-digit number cannot be $0$. Therefore

   $a=5,\qquad N=\overline{5bcc b5}$

4. 4

   Apply divisibility by 9 to constrain $b+c$. The digit sum is

   $5+b+c+c+b+5=10+2b+2c=2(b+c+5)$

   This must be a multiple of $9$. Since $\gcd(2,9)=1$, the factor of $2$ can be dropped:

   $b+c+5\equiv 0\pmod 9\ \Longrightarrow\ b+c\equiv 4\pmod 9$

   so $b+c=4$ or $b+c=13$ (the only values reachable with two digits).

5. 5

   Maximise place value from the left. With $a$ fixed at $5$, the next most significant free digit is $b$, so take $b=9$. Then $c\\equiv 4-9\\equiv 4\\pmod 9$ requires $b+c=13$, giving $c=4$:

   $N=594495$

6. 6

   Verify both divisibility conditions. Digit sum: $5+9+4+4+9+5=36$, and $36=4\times 9$ $\\checkmark$. Last digit $5$ $\\checkmark$. Directly:

   $594495\div 45=13211\quad\checkmark$

   A scan downward through all six-digit palindromes confirms nothing larger works, so $594495$ is the answer.