An interesting problem on arithmetic progression and maxima-minima

Question

Came across this problem in national based examination.

$a,b,c,d,e$ are natural numbers in AP.

Now it is given that $a+b+c+d+e$=cube of an integer

And $b+c+d$ = square of an integer

The question asks us to find the least number of digits of c

Now here his my approach:

Let $a,b,c,d,e$ be denoted as $x-2,x-1,x,x+1,x+2$ respectively.

The $a+b+c+d+e=5x$

$b+c+d=3x$;

Then we can write $5x=a^3$ and $3x=b^2$(As the question suggests that the integers may not be same)

After that i could not proceed..

Answer 1

We need to solve $5Y^2=3X^3$. Assume $Y=3^{p}5^{q}$ and $X=3^{r}5^{s}$, so $2p=3r+1$ & $2q+1=3s$ ... The smallest values give $p=2$, $r=1$, $q=1,s=1$ which gives $Y=45$ and $X=15$.

This gives the central term in the AP is $\color{red}{675}$. So the least number of digits is $\color{blue}{3}$.