# How do you list all the prime numbers between 80 and 90?

Dec 27, 2016

#### Answer:

You evaluate the 5 odd numbers from 81 to 89.

#### Explanation:

In division testing you may skip $2$ now, we can also skip $5$ if the number doesn't end in $5$ or $0$

$81 / 3 = 27$ comes out evenly

$83 / 3 = 27$ and 2 left
$83 / 7 = 11$ and 6 left
We don't have to test any further: 83 is prime

$85 / 5 = 19$ comes out evenly

$87 / 3 = 29$ comes out evenly

$89 / 3 = 29$ and 2 left
$89 / 7 = 12$ and 5 left
We don't have to test any further: 89 is prime

Note : We didn't have to test for the next prime 11, because if the number has a factor of 11 or higher, it also has a factor smaller than 11, and we would have found that. The highest prime you test for is smaller/equal to the square root of the number.

Aug 7, 2017

#### Answer:

$83 \mathmr{and} 89$

#### Explanation:

To list the prime numbers between $80 \mathmr{and} 90$ we can use the rules of divisibility as a start;

$80 , 81 , \text{ "82," "83," "84," "85," "86," "87," "88," "89," } 90$

All even numbers are divisible 2, they are not prime. apart from $2$
This leaves all the odd numbers.

$81 , \text{ "83," "85," "87," } 89 ,$

A number ending in $5$ is divisible by $5$

$81 , \text{ "83," "87," } 89 ,$

If the sum of the digits is divisible by $3$, $3$ is a factor

$81 \text{ "rarr 8+1=9." }$ 81 is not prime

$83 \text{ "rarr 8+3 = 11" }$ 83 might be prime

$87 \text{ "rarr 8+7=15" }$ 87 is not prime

$89 \text{ "rarr 8+9 = 17" }$89 might be prime

As factors are always in pairs we only need to consider factors less than the square roots.

$\sqrt{83} = 9. \ldots .$

$2 , 3 , 4 , 5 , 6 , 7 , 8 , 9$ are not factors: $83$ is prime

$\sqrt{89} = 9. \ldots . .$

$2 , 3 , 4 , 5 , 6 , 7 , 8 , 9$ are not factors: $89$ is prime