Question

If $\alpha ,\beta ,\gamma$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also, $\alpha <\beta <\gamma$. Then, $\gamma$ is equals to:

• A. $3$
• B. $14$
• C. $5$
• D. $11$

Answer 1

Answer: (C)

$5$

Let the numbers be $n-1,n,n+1$

As per the given information, we have

$(n-1)+n^2+{(n+1)}^3={(n-1+n+n+1)}^2$

$\Rightarrow n^3-5n^2+4n=0$

$\Rightarrow n=0,1,4$

For $n=0,$ the numbers are: $-1,0,1$ this is out as all the numbers should be positive

For $n=1,$ the numbers are: $0,1,2$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)

For $n=4,$ the numbers are: $3,4,5$

We have got largest number which is $5$

Hence, the value of $\gamma$ is $5$.