3 digit Prime Palindrome Numbers. Question. How many three digit palindrome number are prime?
Ans. Any 3 digit palindrome number is of type "aba" where b can be chosen from the numbers 0 to 9 and a can be chosen from 1 to 9. So the totality of these type of numbers are 10×9=90. 
But, how can I find the prime numbers out of these 90s……!!!
Thankyou.
 A: $90$ is not so many, so you can just check them.  We know that primes end in $1,3,7,$ or $9$, so $a$ must be one of those and you are down to $40$.  You should be able to find a condition on $a,b$ that will guarantee that $aba$ is divisible by $3$, which will cut the number down a few more.
A: You need to check only numbers of the form $xyx$, where $x\in[1,3,7,9]$ and $y\in[0,\dots,9]$.
This comes down to $4\cdot10=40$ numbers.
You can reduce this even further, by checking only numbers of the following forms:


*

*$1y1$, where $y\in[0  ,2,3  ,5,6  ,8,9]$ (the rest are divisible by $3$)

*$3y3$, where $y\in[  1,2  ,4,5  ,7,8  ]$ (the rest are divisible by $3$)

*$7y7$, where $y\in[0  ,2,3  ,5,6  ,8,9]$ (the rest are divisible by $3$)

*$9y9$, where $y\in[  1,2  ,4,5  ,7,8  ]$ (the rest are divisible by $3$)


This comes down to $7+6+7+6=26$ numbers.

The remaining numbers can be tested only against $7,11,13,17,19,23,29,31$:


*

*$121$ is divisible by $11$

*$161$ is divisible by $7$

*$323$ is divisible by $17$

*$343$ is divisible by $7$

*$707$ is divisible by $7$

*$737$ is divisible by $11$

*$767$ is divisible by $13$

*$949$ is divisible by $13$

*$959$ is divisible by $7$

*$979$ is divisible by $11$

*$989$ is divisible by $23$


So $11$ palindromes are not prime, which leads to $26-11=15$ prime palindromes.
