Question

Show that “the range of the first seven multiples of the smallest odd prime number is also a multiple of that prime number.”

Answer 1

Hint:The smallest odd prime number is 3. The range of the first seven multiples of 3 will be the difference between its seventh multiple and its first multiple. Calculate the first and seventh multiple of 3 and get the difference between the two and check if it is divisible by 3.

Complete step-by-step answer:
According to the question, we have to show that the range of the first seven multiples of the smallest odd prime number is also a multiple of that prime number.
We know that the smallest odd prime number is 3.
So,
To show – “The range of the first seven multiples of 3 will also be a multiple of 3.”
Let us first find first seven multiples of 3,
We know that nth multiple of a number $=n\times (that\ number)$
First seven multiples of 3 will be, $\left( 3\times 1 \right),\left( 3\times 2 \right),\left( 3\times 3 \right),\left( 3\times 4 \right),\left( 3\times 5 \right),\left( 3\times 6 \right)\ and\ \left( 3\times 7 \right)$
$\Rightarrow$ First seven multiples of 3=3,6,9,12,15,18 and 21.
We know that range of a group of number = (largest number in the group) – (smallest number in the group)
In the first seven multiples of 3,
Largest number = 21 and
Smallest number = 3.
Putting largest number = 21 and smallest number = 3 in the above formula for range, we will get,
Range = 21 – 3 =18.
We know, $18=3\times 6$. So, 18 is the 6th multiple of 3. Hence, we have shown that “the range of the first seven multiples of the smallest odd prime number (i.e. 3) is also a multiple of that prime number (i.e. 3)”.

Note: Another method,
Let us assume that the smallest prime number is ‘P’.
Range of its first seven multiples = (its 7th multiple) – (its 1st multiple)
$\begin{align}
  & =\left( P\times 7 \right)-\left( P\times 1 \right) \\
 & =7P-P \\
 & =6P \\
\end{align}$
We know that nth multiple of a number $=n\times (that\ number)$
So, 6P will be also a multiple of P.