【Question 18】
If $a, b, c, d, e$ are consecutive positive integers, and $a<b<c<d<e, b+c+d$ is a perfect square, $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$?
【Analysis and Solution】
Number Theory, Perfect Square, Perfect Cube.
$a, b, c, d, e$ are consecutive positive integers;
By the middle term theorem, $b+c+d=3c$ is a perfect square, $a+b+c+d+e=5c$ is a perfect cube; hence $c$ must contain the prime factors 3 and 5;
To make $c$ as small as possible, let $c=3^{m} \times 5^{n}$; $b+c+d=3c=3 \times 3^{m} \times 5^{n}=3^{m+1} \times 5^{n}$ is a perfect square; $m+1, n$ are both multiples of 2; i.e., $m \div 2$ leaves a remainder of 1, $n \div 2$ leaves a remainder of 0; $a+b+c+d+e=5c=5 \times 3^{m} \times 5^{n}=3^{m} \times 5^{n+1}$ is a perfect cube; $m, n+1$ are both multiples of 3;
i.e., $m \div 3$ leaves a remainder of 0, $n \div 3$ leaves a remainder of 2;
From $m \div 2$ leaves a remainder of 1, $m \div 3$ leaves a remainder of 0, we get $m \div 6$ leaves a remainder of 3, the smallest $m$ is 3; from $n \div 2$ leaves a remainder of 0, $n \div 3$ leaves a remainder of 2, we get $n \div 6$ leaves a remainder of 2, the smallest $n$ is 2; the minimum value of $c$ is $3^{3} \times 5^{2}=675$.