Symmetrical Primes

Number Theory Level 3

Let $p$ be the primes which are having its first digit equal to the last digit. Find the largest prime $p$ for $p< 1000$ .

1 solution

Tijmen Veltman
Jun 13, 2015

We can simply work our way down from $1000$ to see what the largest possible value of $p$ is:

$999$ : divisible by $3$ ;

$989$ : divisible by $23$ ;

$979$ : divisible by $11$ ;

$969$ : divisible by $3$ ;

$959$ : divisible by $7$ ;

$949$ : divisible by $13$ ;

$939$ : divisible by $3$ ;

$929$ : prime!

Therefore the answer is $\boxed{929}$ .

Simple standard approach.

Aah, did the same way. But earlier thought that it needed some special formula.

Swapnil Das - 5 years, 9 months ago