Q:

The License Plate Problem

The 7-digit license plate of a car happens to be a palindrome. Its first digit on the far left is twice the digit next to it. The 3rd digit from the right is odd and 5 more than 2nd digit from the left. The middle digit is the average of the first three digits on the far left. What is this license plate number?
HINT: A palindrome is a number that reads the same forward and backward. For example, 123321 and 6776 are palindromes.

A:

The license plate is of the form $abcdcba$, where each $a,b,c,d$ is a digit.
From the information in the question, we know
$$
a=2b \quad\text{and}\quad c=5+b \quad\text{and}\quad c\ \text{is odd}
\quad\text{and}\quad d=(a+b+c)/3.
$$
Since $c$ is odd and $c=5+b$, we know $b$ is even. Thus $b$ is one of $0,2,4,6,8$. It can't be $6$ or $8$, because then $a$ would then not be a digit. This leaves that $b$ is $0$, $2$, or $4$.
If $b=0$, then $a=0$ and $c=5$ so that $d=(0+0+5)/3$ is not a digit.
If $b=2$, then $a=4$ and $c=7$ so that $d=(4+2+7)/3$ is not a digit.
If $b=4$, then $a=8$ and $c=9$ so that $d=(8+4+9)/3=7$.
Therefore the number is $8497948$.

A:

Ok lets do this systematically. Note that as we have a palindrome the third digit from the right is the same as the third digit from the left. If we let $a$ be the second digit then using the relations we are given we can see that the number plate must be $(2a,a,5+a,\frac{5+4a}{3},5+a,a,2a)$. Note that we immediately get that $a < 5$ as otherwise the third digit is more than 9 so not a single digit. Also note that if $a \neq 1,4$ then the fraction is not an integer so we get that either $a = 1$ or $a = 4$. Also if $a = 1$ we have that our third digit from the right is $5+1 = 6$ which is not odd. Hence we have that $a = 4$ and the number plate is $8497948$ . 

