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. $5$

Answer 1

Answer: (B)

$3$

Let the numbers be

$a=C-2Db=C-Dc=Cd=C+De=C+2D$

Then,

$a+b+c+d+e=5C={\alpha }^3....\left(i\right)b+c+d=3C={\beta }^2......\left(ii\right)$

From (i) and (ii)

$\frac{{\alpha }^3}{5}=\frac{{\beta }^2}{3}$

For this least possibility is

$\alpha =5\times 3,\beta =5\times 3\times 3\alpha =15,\beta =45C=\frac{{\alpha }^3}{5}=\frac{{15}^3}{5}=675c=C=675$

So option $B$ is correct.