To find least possible value of a no. in A.P. 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 the square of an integer. The least possible value of the number of digits of $c$ is
 A: If you let the numbers be $p-2q, ~p-q, ~p, ~p+q,~ p+2q$, their treatment will be considerably easier. 
You want $5p$ to be a perfect cube and $3p$ to be a perfect square. It is also desirable to get the minimum possible value of $5p$ under these conditions. It is important to notice the primes 3 and 5. We should look at those boundary cases where their presence is critical.
This problem is an application of prime factorization of naturals. We will try to predict the least value of $p$ by looking at its prime factorization. 
$3p$ is a perfect square, so we stick to an odd power of $3$. Other primes must also be in even multiples for this to happen. We keep this in mind for the next condition, i.e., $5p$ is a perfect cube. For this to happen, we need a power of 5 that is 2 (mod 3) and also even (because of the previous condition). A perfect candidate for the power of 5 is $2$ itself. And also, the power of $3$ must be a multiple of $3$, since $5p$ is a perfect cube. So a perfect candidate for the power of $3$ is $3$.
Overall, these conditions are satisfied by $p = 3^3 \cdot 5^2 = 675$, or $c = 675$ as originally designated. 
A: Let the common difference be $\alpha$.
$$a+b+c+d+e = 5c$$
$$b+c+d=3c$$
Hence $5c$ is a cube and $3c$ is a square.
Hence $c=5^2k^3$ and $c=3l^2$.
$k$ must be divisible by $3$ and $l$ must be divisible by $5$.
$c=5^2.3^3.k_1^3$ and $c=5^2.3.l_1^2$
$$c=5^2.3.3^2k_1^3$$
Picking $k_1=1$ satisfies the condition.
