
The prime number that comes just after 43 is
a. 49
b. 45
c. 47
d. None of these
Answer
485.4k+ views
Hint: In this type of problem, you must know the definition of prime number as well as the difference between a prime number and composite number. A prime number has only two factors (i.e 1 and itself) while a composite number has more than two factors. We write the factors for the given numbers between 44 and 50 and see which satisfies the prime numbers property.
Complete step-by-step answer:
We know that prime numbers are the positive integers having only two factors (i.e 1 and the number itself.)
Or, in other words, a prime number is the positive integers greater than 1 that is not a product of two smaller positive integers.
Now let’s write the numbers after 43 and try to find their factors.
Let’s take 44 and try to find their factors.
To find the factor of any number we have to divide that number by smallest possible number that divides that number by giving remainder =0
So, firstly, divide 44 by 2, we get, Quotient =22 and Remainder =0
Now take quotient and divide it by 2
So, after dividing 22 by 2, we get, Quotient =11 and Remainder =0
Now again take quotient and divide it by 11, we get, Quotient =1 and Remainder =0
We know that,
\[\text{Number = Divisor }\!\!\times\!\!\text{ Quotient + Remainder}\]
So, we can say that,
$\Rightarrow 44=2\times 22+0$.
\[\Rightarrow 44=2\times 2\times 11\].
\[\Rightarrow 44={{2}^{2}}\times 11\].
Similarly, we can find the factors of all numbers.
Number-Factors
\[\Rightarrow 44={{2}^{2}}\times 11\].
\[\Rightarrow 45={{3}^{2}}\times 5\].
\[\Rightarrow 46=2\times 23\].
\[\Rightarrow 47=1\times 47\].
\[\Rightarrow 48={{2}^{4}}\times 3\].
\[\Rightarrow 49={{7}^{2}}\].
So, from the above calculation, we can easily say that 47 is the first prime number after 43 because it has only two factors that are 1 and 47.
So, the correct answer is “Option C”.
Note: The prime numbers always satisfy the relation $6n-1$ and $6n+1$ giving the value of n as a non-negative integer. The solution of this question can be found using this relation by equating it to 47 each of the options and finding the solution.
For example:
Let us check for option (a)
Equate 49 to $6n-1$ and solve for n.
\[\begin{align}
& 6n-1=49 \\
& \Rightarrow 6n=50 \\
& \Rightarrow n=\dfrac{50}{6} \\
\end{align}\]
The term \[n=\dfrac{50}{6}\] is not an integer this option (a) is invalid.
Let us check for option (b)
Equate 45 to 6n-1 and solve for n.
\[\begin{align}
& 6n-1=45 \\
& \Rightarrow 6n=46 \\
& \Rightarrow n=\dfrac{46}{6} \\
\end{align}\]
The term \[n=\dfrac{46}{6}\] is not an integer this option (b) is invalid.
Equate 47 to 6n-1 and solve for n.
\[\begin{align}
& 6n-1=47 \\
& \Rightarrow 6n=48 \\
& \Rightarrow n=\dfrac{48}{6} \\
& \Rightarrow n=8 \\
\end{align}\]
The term n=8 is an integer this option (c) is valid.
Let us check for option (a)
Equate 49 to $6n+1$ and solve for n.
\[\begin{align}
& 6n+1=49 \\
& \Rightarrow 6n=48 \\
& \Rightarrow n=8 \\
\end{align}\]
The term \[n=8\] is an integer this option (a) is valid but 47 comes before 49. so, 49 is not immediate prime after 43.
Let us check for option (b)
Equate 45 to $6n+1$ and solve for n.
\[\begin{align}
& 6n+1=45 \\
& \Rightarrow 6n=44 \\
& \Rightarrow n=\dfrac{44}{6} \\
\end{align}\]
The term \[n=\dfrac{44}{6}\] is not an integer; this option (b) is invalid.
Complete step-by-step answer:
We know that prime numbers are the positive integers having only two factors (i.e 1 and the number itself.)
Or, in other words, a prime number is the positive integers greater than 1 that is not a product of two smaller positive integers.
Now let’s write the numbers after 43 and try to find their factors.
Let’s take 44 and try to find their factors.
To find the factor of any number we have to divide that number by smallest possible number that divides that number by giving remainder =0
So, firstly, divide 44 by 2, we get, Quotient =22 and Remainder =0
Now take quotient and divide it by 2
So, after dividing 22 by 2, we get, Quotient =11 and Remainder =0
Now again take quotient and divide it by 11, we get, Quotient =1 and Remainder =0
We know that,
\[\text{Number = Divisor }\!\!\times\!\!\text{ Quotient + Remainder}\]
So, we can say that,
$\Rightarrow 44=2\times 22+0$.
\[\Rightarrow 44=2\times 2\times 11\].
\[\Rightarrow 44={{2}^{2}}\times 11\].
Similarly, we can find the factors of all numbers.
Number-Factors
\[\Rightarrow 44={{2}^{2}}\times 11\].
\[\Rightarrow 45={{3}^{2}}\times 5\].
\[\Rightarrow 46=2\times 23\].
\[\Rightarrow 47=1\times 47\].
\[\Rightarrow 48={{2}^{4}}\times 3\].
\[\Rightarrow 49={{7}^{2}}\].
So, from the above calculation, we can easily say that 47 is the first prime number after 43 because it has only two factors that are 1 and 47.
So, the correct answer is “Option C”.
Note: The prime numbers always satisfy the relation $6n-1$ and $6n+1$ giving the value of n as a non-negative integer. The solution of this question can be found using this relation by equating it to 47 each of the options and finding the solution.
For example:
Let us check for option (a)
Equate 49 to $6n-1$ and solve for n.
\[\begin{align}
& 6n-1=49 \\
& \Rightarrow 6n=50 \\
& \Rightarrow n=\dfrac{50}{6} \\
\end{align}\]
The term \[n=\dfrac{50}{6}\] is not an integer this option (a) is invalid.
Let us check for option (b)
Equate 45 to 6n-1 and solve for n.
\[\begin{align}
& 6n-1=45 \\
& \Rightarrow 6n=46 \\
& \Rightarrow n=\dfrac{46}{6} \\
\end{align}\]
The term \[n=\dfrac{46}{6}\] is not an integer this option (b) is invalid.
Equate 47 to 6n-1 and solve for n.
\[\begin{align}
& 6n-1=47 \\
& \Rightarrow 6n=48 \\
& \Rightarrow n=\dfrac{48}{6} \\
& \Rightarrow n=8 \\
\end{align}\]
The term n=8 is an integer this option (c) is valid.
Let us check for option (a)
Equate 49 to $6n+1$ and solve for n.
\[\begin{align}
& 6n+1=49 \\
& \Rightarrow 6n=48 \\
& \Rightarrow n=8 \\
\end{align}\]
The term \[n=8\] is an integer this option (a) is valid but 47 comes before 49. so, 49 is not immediate prime after 43.
Let us check for option (b)
Equate 45 to $6n+1$ and solve for n.
\[\begin{align}
& 6n+1=45 \\
& \Rightarrow 6n=44 \\
& \Rightarrow n=\dfrac{44}{6} \\
\end{align}\]
The term \[n=\dfrac{44}{6}\] is not an integer; this option (b) is invalid.
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