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(B) Let the arithmetic progression be $a, b, c, d, e$ with common difference $D$. Since $a, b, c, d, e$ are in arithmetic progression,we can write the

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$3$

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(B) Let the arithmetic progression be $a, b, c, d, e$ with common difference $D$. Since $a, b, c, d, e$ are in arithmetic progression,we can write them as $c-2D, c-D, c, c+D, c+2D$. Given $a+b+c+d+e = 5c = \lambda^3$ for some integer $\lambda$. Given $b+c+d = 3c = u^2$ for some integer $u$. From $3c = u^2$,we have $c = \frac{u^2}{3}$. Substituting this into $5c = \lambda^3$,we get $5(\frac{u^2}{3}) = \lambda^3$,which implies $5u^2 = 3\lambda^3$. For this to hold,$u$ must be a multiple of $3$ and $\lambda$ must be a multiple of $5$. Let $u = 3k$ and $\lambda = 5m$. Then $5(9k^2) = 3(125m^3)$,which simplifies to $45k^2 = 375m^3$,or $3k^2 = 25m^3$. For the smallest natural number solution,we set $m=3$ and $k=5$. Then $u = 3(5) = 15$ and $\lambda = 5(3) = 15$. However,checking the condition $5c = \lambda^3$,if $\lambda=15$,$5c = 3375$,so $c = 675$. Checking $3c = u^2$,$3(675) = 2025 = 45^2$. Thus,$c = 675$,which has $3$ digits.

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

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