Prime Numbers

Prime Numbers exercise

• Determine which numbers are prime.

  Hints

  11 is a prime number, as its factors are 1 and 11. 10 is not a prime number, as its factors are 1, 2, 5, and 10.

  There is only one number which is neither prime or composite.

  Each number divisible by 2 is at least divisible by 1, 2, and itself. So it can't be a prime number.

  But what about 2 itself?

  Solution

  A prime number is defined as a number which is divisible only by 1 and itself.

  Each number which has more factors than 1 and itself is called a composite number.

  Remember, though, that 1 is neither a prime or a composite number.

  The smallest prime number is 2 and it's the only even prime number. All other even numbers are divisible by 2 as well, and thus are composite numbers.

  To determine all prime numbers from 1 up to 120, you can proceed as follows:

  • First, sift out 1.
  • Next, look at all multiples of 2: you can sift out all those multiples except 2 itself.
  • Next, take all the multiples of 3 and sift all of them out except 3.
  • Now sift out the multiples of 5, except 5 itself.
  • Lastly, sift out the multiples of 7, except 7 itself.

  All numbers left are prime numbers.

  Roscoe checks one of the rocks left over. He takes a hammer and brakes it to find a diamond inside!

• Use the sieve to find all the prime numbers.

  Hints

  The numbers which are not highlighted are prime numbers.

  All numbers with a green cross are prime numbers. There are five more prime numbers in the sieve.

  A prime number is a number which is only divisible by $1$ and itself.

  Solution

  All numbers with a green cross are prime numbers.

  To cancel out composite numbers step by step:

  • First, cancel out the multiples of $2$, i.e. $4$, $6$, $8$, $10$, $12$, $14$, $16$, $18$, $20$, $22$, $24$
  • Next, cancel out the remaining multiples of $3$, i.e. $9$, $15$, $21$
  • Lastly, cancel out the only remaining multiple of $5$, i.e. $25$

• Decide if each number is prime or composite.

  Hints

  The only even prime number is $2$.

  Any number ending in $5$ or $0$ is divisible by $5$.

  If the sum of the digits of a number is divisible by $3$, then so is the number itself.

  Let's have a look at the following example: $1236$.

  The sum of the digits is $1+2+3+6=12$, and $12$ is divisible by $3$. Thus, $1236$ is divisible by $3$ too.

  Solution

  How can you tell if a number is composite or prime?

  Check the divisibility rules:

  • Every even number is divisible by $2$.
  • Every number whose digits sum to a number divisible by $3$ is also divisible by $3$.
  • Each number ending in $5$ or $0$ is divisible by $5$.

  Next you can check the multiples of other numbers:

  • The multiples of $7$ are $7$, $14$, $21$, $35$, $42$, $49$, ...
  • The multiples of $11$ are $11$, $22$, $33$, $44$, $55$, $66$, ...
  • The multiples of $13$ are $13$, $26$, $39$, $52$, ...

  For $\mathbf{123}$, the digit sum $1+2+3=6$ is divisible by $3$ so $123$ is divisible by $3$. $123$ is a composite number.

  For $\mathbf{124}$, it is an even number and thus divisible by $2$.

  For $\mathbf{55}$, it ends in $5$ and so is divisible by $5$.

  For $\mathbf{17}$, it is neither divisible by $2$ nor by $3$ nor by $5$, so it's only divisible by $1$ and itself. $17$ is a prime number.

  For $\mathbf{37}$, it is neither divisible by $2$ nor by $3$ nor by $5$ nor by $7$ nor by $13$ nor by $17$, so it's only divisible by $1$ and itself. $37$ is a prime number.

  For $\mathbf{127}$, it is not divisible by $2$, nor $3$, nor $5$, ... $127$ is a prime number.

  For $\mathbf{53}$, we can check it in the same way and see that it is prime as well.

• Decide the possible side lengths.

  Hints

  Keep in mind the area of a rectangle is given by $A_{\text{rectangle}}=(\text{length})(\text{width})$.

  A prime number is a number which is only divisible by $1$ and itself.

  If the area is given by a prime number then there is only one choice of side lengths.

  If the given area is a composite number then there exists more than one choice of side lengths.

  Solution

  The possible side lengths depend on if the area is a prime or composite number:

  • If the number is prime there only exists one combination.
  • If the number is composite there exist more than one combination.

  The numbers $\mathbf{11}$, $\mathbf{17}$ and $\mathbf{19}$ are prime numbers. Thus all the given areas are only given by the lengths $1$ and the regarding prime number.

  $\mathbf{15}$ has the factors $1$, $3$, $5$ and $15$. The possible combinations are $1$ and $15$ as well as $3$ and $5$.

  $\mathbf{21}$ has the factors $1$, $3$, $7$ and $21$. The possible combinations are $1$ and $21$ as well as $3$ and $7$.

• Explain what "being divisible by" means.

  Hints

  The multiples of $3$ are: $3$, $6$, $9$, $12$, and so on...

  Keep in mind that $12\div 3=4$.

  Solution

  If a number $a$ is a multiple of another number $b$, then we can say that $a$ is divisible by $b$.

  In other words, if $a\div b=c$, then

  • $a$ is the dividend.
  • $b$ is the divisor.
  • $c$ is the quotient.

  We say that a number $a$ is divisible by another $b$ if there is no remainder when $a$ is divided by $b$.

  For instance, $14$ is divisible by $7$ if $14\div 7=2$. But $10$ is not divisible by $4$ because $10\div 4=2$ remainder $2$.

• Check the following statements.

  Hints

  Look for a contradiction if you think that a statement isn't true.

  A prime number has no other factors besides $1$ and itself.

  Solution

  Let's check the statements about prime numbers.

  • The product of two prime numbers can't be a prime number because the resulting product has $1$ and itself as factors, as well as the two prime numbers in which it is a product of.

  • The sum of two prime numbers could be a prime number, for example $2+5=7$. But in general it isn't. With the exception of $2$, all prime numbers are odd. The sum of two odd numbers is an even number and thus not prime.

  • The factors of a composite number are prime numbers. For example, $24=2\times 2\times 2\times 3$.

  • There exist pairs of prime numbers whose difference is $2$. For instance, $3$ and $5$, $5$ and $7$, $11$ and $13$, and $17$ and $19$. Such pairs are called twin primes.

  • For the last statement, we look for a contradiction. Lets take $13$. It is a prime number and $2\times 13+1=26+1=27$. $27$ is divisible by $3$ and thus not prime. So, this statement isn't true.