3 digit Prime Palindrome Numbers.

Question

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.

Answer 1

$90$ 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.

Answer 2

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.

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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.