Palindrome numbers with conditions

Question

If $n$ is a palindrome of three digits and $n+32$ is palindrome of four digits then find $n$.

Answer 1

$n=969$

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:

$969,979,989,999$

Answer 2

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.

Answer 3

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 $n = 1001 - 32 = 969$.

Answer 4

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.