Finding the Palindromes

Let the numbers be $\overline {aba}$ and $\overline {cdc}$ . Then

$(101a+10b)(101c+10d)=436995\implies 10201ac+1010(ad+bc)+100bd=436995$ .

Therefore either $a$ is $5$ , $c$ is odd, or $c$ is $5$ , $a$ is odd. Let us choose the first. Then, since $51005c<436995, c\leq 8$ . Also, since $0\leq b, c, d\leq 9$ , therefore $c>6$ . So the only possible value of $c$ is $7$ .

Therefore $357035+1010(7b+5d)+100bd=436995\implies 101(7b+5d)+10bd=7996$ . Hence $b$ must be $8$ , as the only possibility to get a unit's place digit $6$ is from $7,7\times 8=56$ . Then $d=4$ .

Hence $\overline {aba}=585, \overline {cdc}=747$ , and their sum is $585+747=\boxed {1332}$ .

This problem has a few steps.

Step 1: Find the prime factorization of 436995.

3x3x3x3x5x13x83

Step 2: Use the numbers in the prime factorization to find the 2 palindromes that the problem is asking about.

3x3x83=747

3x3x5x13=585

Step 3: Add the palindromes you found in step 2.

747+585=1332

So the sum of the 2 palindromes that multiply to 436995 is 1332.