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Let a,b,c,d,e be consecutive positive integers ,such that, (b+c+d) is a perfect square and (a+b+c+d+e) is a perfect cube. Find the smallest value of c.

Let a,b,c,d,e be consecutive positive integers ,such that, (b+c+d) is a perfect square and (a+b+c+d+e) is a perfect cube. Find the smallest value of c.

Grade:12th pass

1 Answers

Venkat
273 Points
5 years ago
Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers. $c$ is minimized if it’s prime factorization contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = 675$, which works as the solution.

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