Squares, Cubes And A Toddler

$a,b,c,d$ and $e$ are consecutive positive integers such that $b+c+d = 3c$ and $a+b+c+d+c = 5c$ .

Now $3 | 3c \Rightarrow 3^2 | 3c \because 3c$ is a perfect square as given that $b+c+d$ is a perfect square ).

$\Rightarrow 3 | c \Rightarrow 3 | 5c \Rightarrow 3^3 | 5c \because 5c$ is a perfect square.

Also $5 | 5c \Rightarrow 5^3 | 5c \Rightarrow 5^2 | c$

$\therefore 3^3 5^2 | c$ or equivalently $675 | c$

$\therefore 675$ is the smallest possible value of $c$ .

QED.

Let $b+c+d = 3a + 6 = p^2$ and $a+b+c+d+e = 5a + 10 = q^3$ , which yield:

$a+2 = c = \frac{p^2}{3}$ or $\frac{q^3}{5}$ (i).

If (i) is true, then we require $p = 3^{w}5^{x}, q = 3^{y}5^{z}$ for the smallest possible $w,x,y,z \in \mathbb{N}.$ Substituting these expressions into (i) now produces:

$\frac{p^2}{3} = \frac{q^3}{5} \Rightarrow \frac{3^{2w}5^{2x}}{3} = \frac{3^{3y}5^{3z}}{5} \Rightarrow 3^{2w-1}5^{2x} = 3^{3y}5^{3z-1}$ (ii).

Setting the corresponding exponents equal to one another next gives:

$2w-1 = 3y \Rightarrow w =2, y =1$

$2x = 3z-1 \Rightarrow x = z =1$

Hence, $p = 3^25^1 = 45$ and $q = 3^15^1 = 15 \Rightarrow c = \frac{45^2}{3} = \frac{15^3}{5} = \boxed{675}.$

b+c+d=3c & a+b+c+d+e=5c
since we want to get the lowest possible value for c,let c=(3^x)X(5^y)(since 3&5 must divide c). So,3c=3^(x+1)X(5^y) is a square.
Therefore 2I(x+1) & 2Iy
5c is cube.So 3Ix & 3I(y+1).
So least possible value of x=3 & y=2.
So.c=(3^3)X(5^2)=675(ANSWER).