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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 $\mathrm{b}+\mathrm{c}+\mathrm{d}$ is square of an integer. The least possible value of the number of digits of $\mathrm{c}$ is
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The correct answer is:
$3$
$a=C-2 D$
$b=C-D$
$c=C$
$d=C+D$
$e=C+2 D$
$a+b+c+d+e=5 c=\lambda^{3}$
$b+c+d=3 c=\mu^{2}$
$\Rightarrow 3 \lambda^{3}=5 \mu^{2}$
$\frac{\lambda^{3}}{5}=\frac{\mu^{2}}{3}$ least possibility
$\lambda=5 \times 3, \mu=5 \times 3 \times 3$
$\lambda=15$ $\mu=45$
$C=\frac{(45)^{2}}{3}=15 \times 45=675$
$b=C-D$
$c=C$
$d=C+D$
$e=C+2 D$
$a+b+c+d+e=5 c=\lambda^{3}$
$b+c+d=3 c=\mu^{2}$
$\Rightarrow 3 \lambda^{3}=5 \mu^{2}$
$\frac{\lambda^{3}}{5}=\frac{\mu^{2}}{3}$ least possibility
$\lambda=5 \times 3, \mu=5 \times 3 \times 3$
$\lambda=15$ $\mu=45$
$C=\frac{(45)^{2}}{3}=15 \times 45=675$
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