Palindrome numbers with conditions If $n$ is a palindrome of three digits and $n+32$ is palindrome of four digits then find $n$.
 A: For some digits $A,B,C,D$, we have:
$$
A ~ B ~ A \quad \xrightarrow{~+~32} \quad C~ D ~ D ~ C
$$
Since $n$ changes from three digits to four digits when adding $32$, we know that $968 \leq n \leq 999$ so that $1000 \leq n+32 \leq 1031$. Hence, just by looking at the range of $n+32$, we know for sure that $C = 1$ and $D = 0$. Thus, since $n + 32 = 1001$, we have that $n = 969$, which is indeed also a palindrome.
A: $n=969$
Since you know the number goes from $3$ digits to $4$ digits by the addition of $32$, then the number $n$ must be between $968$ and $999$. From here, just add $32$ to any palindrome numbers that are between the $968$ and $999$.
The options being:
$969,979,989,999$
A: Since $32$ is smaller than $100$, we know that $n+32\lt 1100$ and that $n+32$ has to be $1001$. So $n=1001-32=969,$ which is a palindrome three digits.
A: The smallest four-digit palindromic number is $1001$ and the second smallest is $1111$.  Since $1001$ is the only four-digit palindromic number you can reach by adding $32$ to a three-digit number (palindromic or not!) your answer must be $n = 1001 - 32 = 969$.
