How to find out if a number is prime - SAT Math

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 value of , or , is the product of and , 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 _p_2 has exactly three factors.

(Alternatively, we can plug in any prime number for p and see how many factors _p_2 has. For example, if p is 3, then the factors of _p_2, or 9, are 1, 3, and 9.)

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.

Let x represent the smallest of the four numbers.

Then we can set up the following equation:

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

Plug in a prime number such as and evaluate all the possible solutions. Note that the question asks which value COULD be prime, not which MUST BE prime. As soon as your number-picking yields a prime number, you have satisfied the "could be prime" standard and know that you have a correct answer.

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:

-> series length 2: 2,3

-> series length 3: 3,5,7

-> series length 5: 5,7,11,13,17

-> 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 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 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 will thus end on a number larger than 31, meaning we will never finish on a 31.

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

Let's try to find the factors.

It may not be easy to see as a composite number, but if you know the divisibility rule for which is double the last digit and subtract from the rest , you will see is not prime.

The smallest prime number is actually . 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.

It may not be easy to see as a composite number, but if you know the divisibility rule for which is double the last digit and subtract from the rest, you will see is not prime. The divisibility rule for is add the outside digits and if the sum matches the sum then it is divisible . The divisibility rule for is if the digits have a sum divisible by , then it is . All even numbers are composite numbers with the exception of . So with these analyses, answer is .

Since all the numbers are odd and don't end with a , let's check the basic divisbility rule. The divisibility rule for is if the digits have a sum divisible by , then it is.

Based on this analysis, only is divisible by and therefore not prime and is our answer.

This will require us to know the divisibility rule of . The reason for this choice is that some of the numbers are palindromes like so we eliminate . For the three digit numbers, the divisibility rule for is add the outside digits and if the sum matches the sum then it is divisible. Let's see.

Based on this test, is not divisible by and is our answer.

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

Only is divisible by and is not prime and therefore our answer.

The smallest prime number is actually . is not a prime nor a composite number. It is a unit. This will eliminate the choices with a in them. The next prime numbers are . Our answer is then . is a perfect square and has more than two factors .

Prime numbers are integers. So doing division and square roots will not generate integers. By doing multiplication and exponents, we involve more factors. The only possibility is subtraction. If was and we subtracted we get which is also prime.

Other than the number itself and one, we also need to have prime factors in that number. Since it's not a perfect square, we need to find the smallest possible prime numbers. That will be or which is our answer.

Since all the numbers are odd and don't end with a , let's check the basic divisbility rule. The divisibility rule for is if the digits have a sum divisible by , then it is.

Based on this analysis, only is divisible by and therefore not prime and is our answer.

The order of the prime numbers start from . is not prime as it's a unit. is a composite number. So our fourth smallest prime number is .

Let's work backwards. All even numbers are not prime so we skip . is clearly divisible by . is definitely a composite number as it's divisible by . is divisible by because of the divisibility rule . is divisbile by . The divisibility rule for is double the last digit and subtract from the rest . From the remaining answers, is prime and is the largest prime number under and is our answer.

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 value of , or , is the product of and , 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 _p_2 has exactly three factors.

(Alternatively, we can plug in any prime number for p and see how many factors _p_2 has. For example, if p is 3, then the factors of _p_2, or 9, are 1, 3, and 9.)