Question

Let a, b, c, d, e be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and b + c + d is square of an integer. The least possible value of the number of digits of c is?(a) 2(b) 3(c) 4(d) 5

Hint: We know that in an arithmetic series, the numbers are in such sequence that the difference between two consecutive terms is constant throughout the series.

Let a, b, c, d, e be natural numbers in an arithmetic progression. We know that arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between two consecutive terms is constant.
Let us assume that the common difference between them is D and the first term of series is A.
So according to our assumption, the numbers can be written as,
a = A – 2D
b = A – D
c = A
d = A + D
e = A + 2D
We know that it is given in the question that $a+b+c+d+e$ is the cube of an integer. So, let us suppose that the integer is p.
So, we can write that,
$a+b+c+d=e={{p}^{3}}....\left( i \right)$
Also, it is given in the question that b + c + d is the square of an integer. So, let us suppose that the integer is q.
So, we can write that,
$b+c+d={{q}^{2}}....\left( ii \right)$
Now we can substitute the values of A, B, C and D in equation (i) as shown below,
$A-2D+A-D+A+A+D+A+2D={{p}^{3}}$
On simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with the variable A in the left-hand side of the equation.
$5A={{p}^{3}}....\left( iii \right)$
Now substituting the value of B, C and D in equation (ii), we will get
$A-D+A+A+D={{q}^{2}}$
Again, on simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with variable A in the left-hand side of the equation.
$3A={{p}^{2}}....\left( iv \right)$
Now we have to divide equation (iv) by equation (iii). We will get,
$\dfrac{5A}{3A}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
We can observe that variable A is cancelled out and we will be left with
$\dfrac{5}{3}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
$\dfrac{{{p}^{3}}}{5}=\dfrac{{{q}^{2}}}{3}$
It is said in the question that we have to find the least possible value of the number of digits of c.
So by hit and trial method, the least possibility is
\begin{align} & p=5\times 3 \\ & q=5\times 3\times 3 \\ \end{align}
As it will satisfy the equation, we have $p=15\text{ and }q=45$.
Now we know that
$5A={{p}^{3}}$
So putting the value of p = 15 in it, we will get
\begin{align} & 5A={{15}^{3}} \\ & 5A=3375 \\ & A=675 \\ \end{align}
Since we have the value of c = A, we get that c = 675.
It consists of 3 numbers of digits, so the answer to the question is 3. Hence, we get option b as the correct answer.

Note: We can approach this question with a different method. We can take elements of the A.P series as $a,a+d,a+2d...$ instead of $a=A-2D,\text{ }b=A-D,c=A,d=A+D,e=A+2D$. But, this method will increase the number of unknown variables in the equation and make it difficult to solve the question.