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 possible value of c-

Question

Toppr Admin · Accepted Answer

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series,

we know $b + c + d = 3 c$ and $a + b + c + d + e = 5 c$ .

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 $c$ .

$3^{3} 5^{2} = 675$

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 possible 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 possible value of c.