The key here is to find the prime factorizations of both numbers and multiplying the common prime factors together:

$\displaystyle 81: 3^4$

$\displaystyle 216: 2^3*3^3$

Both prime factorizations have $\displaystyle 3^3$ in common, so $\displaystyle 3^3 = 27$ is our answer.

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don't forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

There are 8 possible numbers; 4,6,8,10,12,14,16,18.

One is not a prime number, so only 8, 10, 12, 14, 16, and 18 can be the sum of two different prime numbers.

The primes, in order, are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ...

We create a few series:

$\displaystyle x=2$ -> series length 2: 2,3

$\displaystyle x=3$ -> series length 3: 3,5,7

$\displaystyle x=5$ -> series length 5: 5,7,11,13,17

$\displaystyle x=7$ -> series length 7: 7,11,13,17,19,23,29

etc.

We can then see that, of the answers, only 47 and 31 remain possibly correct answers.  Now we need to decide which of those two are impossible. 

We could do another series, but the $\displaystyle x=11$ series has 11 terms requiring us to go further and further up.  If we do this, we'll find that it terminates at 47, meaning that 31 must be the correct answer.

Another way, however, is to notice that 29 is the end of the $\displaystyle x=7$ series.  Since 31 is the very next prime number, if we start on 11, the series that terminates in 31 would have to have length 7 as well.  Every series after $\displaystyle x=7$ will thus end on a number larger than 31, meaning we will never finish on a 31.

The value of $\displaystyle p^{2}$, or $\displaystyle p\cdot p$, is the product of $\displaystyle p$ and $\displaystyle p$, so it will be divisible by 1, p, p * p, and nothing else (we know that the p’s are not divisible because they are prime). Therefore p² has exactly three factors.

(Alternatively, we can plug in any prime number for p and see how many factors p² has. For example, if p is 3, then the factors of p², or 9, are 1, 3, and 9.)

Let x represent the smallest of the four numbers. 

Then we can set up the following equation:

$\displaystyle \dpi{100} x + (x+1) + (x+2)+ (x+3) = 210$

$\displaystyle \dpi{100} 4x + 6 = 210$

$\displaystyle \dpi{100} 4x = 204$

$\displaystyle \dpi{100} x = 51$

Therefore the four numbers are 51, 52, 53, 54.  The only prime in this list is 53.

A prime number is a number with factors of one and itself.

Let's try to find the factors.

$\displaystyle 19=1, 19$

$\displaystyle 21=1, 3, 7, 21$

$\displaystyle 25=1, 5, 25$

$\displaystyle 33=1, 3, 11, 33$

$\displaystyle 91=1, 7, 13, 91$

It may not be easy to see $\displaystyle 91$ as a composite number, but if you know the divisibility rule for $\displaystyle 7$ which is double the last digit and subtract from the rest $\displaystyle (91=9-1*2=9-2=7)$, you will see $\displaystyle 91$ is not prime. 

The smallest prime number is actually $\displaystyle 2$. $\displaystyle 1$ is not a prime nor a composite number. It is a unit. 

A prime number is a number with factors of one and itself.

Let's try to find the factors.

$\displaystyle 89=1, 89$

$\displaystyle 343=1, 7, 49, 343$

$\displaystyle 121=1, 11, 121$

$\displaystyle 998=1, 2, 499, 998$

$\displaystyle 867=1, 3, 17, 51, 289, 867$

It may not be easy to see $\displaystyle 343$ as a composite number, but if you know the divisibility rule for $\displaystyle 7$ which is double the last digit and subtract from the rest$\displaystyle (343=34-3*2=34-6=28)$, you will see $\displaystyle 343$ is not prime. The divisibility rule for $\displaystyle 11$ is add the outside digits and if the sum matches the sum then it is divisible $\displaystyle (121=1+1=2)$. The divisibility rule for $\displaystyle 3$ is if the digits have a sum divisible by $\displaystyle 3$, then it is $\displaystyle (867=8+6+7=21)$. All even numbers are composite numbers with the exception of $\displaystyle 2$. So with these analyses, answer is $\displaystyle 89$. 

Since all of the digits don't add to a sum of $\displaystyle 3$, and we dont see any numbers een or end in $\displaystyle 5$, let's try the divisibility rule for $\displaystyle 7$ which is double the last digit and subtract from the rest.

$\displaystyle 91=9-1*2=7$

$\displaystyle 103=10-2*3=4$

$\displaystyle 89=8-2*9=-10$

$\displaystyle 83=8-2*3=2$

$\displaystyle 101=10-1*2=8$

Only $\displaystyle 91$ is divisible by $\displaystyle 7$ and is not prime and therefore our answer.