Question

The product of 4 consecutive natural numbers is 5040. Find those numbers.

Answer 1

Hint: We know that consecutive numbers are the number which follow each other in order, without any gap like 22,24,26,28 are the example of consecutive even numbers. Similarly we will find 4 consecutive natural numbers as $n$, $n+1$, $n+2$ and $n+3$. According to the question $n\times (n+1)\times (n+2)\times (n+3)=5040$.

Complete step-by-step answer:
It is given in the question that the product of four consecutive natural numbers is 5040. We know that any consecutive number following order like 1,3,5,7,9 are the consecutive odd natural numbers. Also the natural numbers start from 1. Now, let us assume that the first number out of 4 numbers is $n$, therefore, the second number will be $n+1$, third number will be $n+2$, fourth number will be $n+3$.
Given that the product of these consecutive numbers is 5040. Therefore, in mathematical form, we can write as following equation -
$n\times (n+1)\times (n+2)\times (n+3)=5040$. On solving we get following steps
= $({{n}^{2}}+3n)\times ({{n}^{2}}+3n+2)=5040$
Now, we assume that $a=({{n}^{2}}+3n)$, then, we get $a\times (a+2)=5040$
= ${{a}^{2}}+2a=5040$, adding 1 to both sides, we get
= ${{a}^{2}}+2a+1=5041$
= ${{(a+1)}^{2}}=5041$.
Therefore, taking under root on both sides
$(a+1)=\pm 71$ .
According to our assumption, $a=({{n}^{2}}+3n)$, so using this equation, we get
${{n}^{2}}+3n+1=\pm 71$, now we have two equations, ${{n}^{2}}+3n-70=0$ and ${{n}^{2}}+3n+72=0$. For equation ${{n}^{2}}+3n+72=0$, we get Discriminant, D, to be negative,
 $D={{b}^{2}}-4ac$
$\begin{align}
  & D=9-4\times 1\times 72 \\
 & D=9-288 \\
 & D=-279 \\
\end{align}$
therefore no real roots.
Hence only the equation ${{n}^{2}}+3n-70=0$ is left. Solving this equation, we get
$\begin{align}
  & {{n}^{2}}+10n-7n-70=0 \\
 & n\times (n+10)-7\times (n+10)=0 \\
 & (n+10)\times (n-7)=0 \\
\end{align}$
Therefore, $n=7$ and $n=-10$. We know that -10 is not a natural number, therefore $n=7$ is the only solution for the given question. Hence, the consecutive numbers that satisfy the given equation are $n=7$, $n+1=8$, $n+2=9$, $n+3=10$ , that is, $7\times 8\times 9\times 10=5040$ .

Note: If we do not assume $a=({{n}^{2}}+3n)$ in the solution part then our calculation becomes more complex and the chances of error are increased. So, in order to reduce our effort we must proceed with the made assumption.Students should know the concept of discriminant to check whether roots are real or imaginary.To find discriminant we use ${D}^2={b}^2-4ac$, If $D>0$ and $D=0$ then roots are real and if $D < 0$ then roots are imaginary.