Largest 2 digit prime

99 is not a prime number. Same goes to 98. But 97 can't be divide by any number.

Just back-calculate from $99$ . We keep on calculating until we find a prime. $99 \ is \ not \ a \ prime$ $98 \ is \ not \ a \ prime$ $97 \ is \ a \ prime$ Thus the answer is $97$ .

$99$ is not prime since digits add up to $18$ which is a multiple of $3$ hence $99$ is a multiple of $3$ ....also $98$ is not a prime because it is an even number and all even numbers are multiples of $2$ ..now we know that if a number $n$ is not divisible but any prime less then the square root of $n$ it must be a prime. Now $\sqrt{97}=9.8488$ and primes less then $9.8488$ are $2,3,5,7$ 97 is odd so it is not a multiple of $2$ , also $9+7=16$ which is not a multiple of 3 hence 97 is not a multiple of 3.for a number to be a multiple of 5 it digit at one's place must be 0 or 5 but in this case it is 7 so 97 is not a multiple of 5....similerly it is easy to see that 97 is not a multiple of 7...hence 97 is not divisible by any prime less then square root of 97 hence 97 must be a prime...:-)

An easy approach:- Just observe that 97 is a prime

just divide and multiply. problem officer?

OEIS A000040: The prime numbers

u can see a sequence in the prime digits for eg 13 the difference is 2 there largest 2digit prime number is 97

As we can see that every prime p>3 can can be written as 6n+1 or 6n-1. Case 1 6n-1<100 n<16.83333 This implies n=16 p=6×16-1 p=95,which is not a prime since it is a multiple of 5. Case 2 6n+1<100 n<16.5 This implies n=16 And p=6×16+1 p=97

If you know the prime numbers it is easy.Otherwise start testing the numbers from 99 .The answer will be 97. 98 is not a prime number. 99 is not a prime number. 97 is a prime number.