Five consecutive positive integers have the property that the sum of the second, third and fourth is a perfect square, while the sum of all five is a perfect cube. If $m$ is the third of these five integers, then the minimum possible value of $m$ satisfies

(A) $m \leq 200$

(B) $200800$
Since $m$ is the third of the five integers, then the five integers are $m-2, m-1, m, m+1$, and $m+2$.

The sum of all five is thus $(m-2)+(m-1)+m+(m+1)+(m+2)=5 m$ and the sum of the middle three is $(m-1)+m+(m+1)=3 m$.

Therefore, we want to find the smallest integer $m$ for which $3 m$ is a perfect square and $5 m$ is a perfect cube.

Consider writing $m, 3 m$ and $5 m$ each as a product of prime numbers.

For $3 m$ to be a perfect square, each prime must occur an even number of times in the product. Thus, the prime 3 must occur an odd number of times in the product that represents $m$.

For $5 \mathrm{~m}$ to be a perfect cube, the number of times that each prime occurs in the product must
be a multiple of 3 . Thus, the prime 5 must occur a number of times which is one less than a multiple of 3 in the product that represents $m$.

Since both of the primes 3 and 5 are factors of $m$, then to minimize $m$, no other prime should occur.

For $3 m$ to be a perfect square, 5 must occur an even number of times in $3 m$.

For $5 m$ to be a perfect cube, the number of times that 3 occurs in $5 m$ is a multiple of 3 .

Therefore, $m$ is a number which contains an odd number of $3 \mathrm{~s}$ and an even number of $5 \mathrm{~s}$ (since $3 m$ does), and at the same time contains 3 a number of times which is a multiple of 3 (since $5 m$ does) and 5 a number of times that is 1 less than a multiple of 3 .

To minimize $m, m$ should contain as few $3 \mathrm{~s}$ and $5 \mathrm{~s}$ as possible, so should contain three $3 \mathrm{~s}$ and two 5 s, so $m=3^{3} 5^{2}=675$.

ANSWER: (D)