Finding Palindromes!

Since, this is a 16 digit palindrome every digit should be at least 2 times present in the number so that it can be in both side of the number.

Thus, we can make a multiple of 16 only by 2 fours or 4 twos. The summation can be made by taking ones, as they won't change the multiplication.

$\color{teal}{\text{Case 1}}$ : Taking 1 four, 4 ones, 3 zeroes in each side of the palindrome.

We can get the number of arrangements by subtracting the number of arrangements when 0 is the first digit from the total number of arrangements : $\boldsymbol {\frac{8!}{4!\times 3!}-\frac{7!}{4!\times2!}=175}$

$\color{teal} {\text {Case 2}}$ : Taking 2 twos, 4 ones & 2 zeroes in each side of the palindrome.

Similarly, the total number of arrangements we get this time is, $\boldsymbol {\frac{8!}{2!\times 4!\times2!}-\frac{7!}{2!\times4!}=315}$

So, the number of total possible arrangements will be, $\boldsymbol { 175+315=\color{#69047E} {\boxed {490}}}$

A 16-digit palindrome must have the form $d_1d_2...d_7d_8d_8d_7...d_2d_1$ , with $d_i \in \{0,1,2,...,9\}$ with $d_1 \neq 0$ . For the sum of the digits to be 16, the sum of first 8 digits $d_1$ to $d_8$ must be 8. Simultaneously, the product of the first 8 digits should be 4.

It can be easily seen that the first 8 digits of the palindrome must be the arrangements of either of the two sets of digits:

• $\{2,2,1,1,1,1,0,0\}$ and $\{4,1,1,1,1,0,0,0\}$

where the first digit isn't 0.

Now, if the first 8 digits of the palindrome contains two 2(s), four 1(s), two 0(s), then the number of arrangements excluding $0$ as the first digit is $\dfrac{6(7!)}{2!2!4!} = 315$ .

If the first 8 digits of the palindrome contains one 4, four 1(s), three 0(s), then the number of arrangements excluding $0$ as the first digit is $\dfrac{5(7!)}{3!4!} = 175$ .

Total possible arrangements = $\boxed{490}$ .