, 2014

Solution 1

Since $900 = 30^2$ and $30 = 2\times 3 \times 5$, then $900 = 2^2 3^2 5^2$.

The positive divisors of $900$ are those integers of the form $d=2^a 3^b 5^c$, where each of $a,b,c$ is 0, 1 or 2.

For $d$ to be a perfect square, the exponent on each prime factor in the prime factorization of $d$ must be even.

Thus, for $d$ to be a perfect square, each of $a,b,c$ must be 0 or 2.

There are two possibilities for each of $a,b,c$ so $2 \times 2 \times 2 = 8$ possibilities for $d$.

These are $2^0 3^0 5^0 = 1$, $2^2 3^0 5^0 = 4$, $2^0 3^2 5^0 = 9$, $2^0 3^0 5^2 = 25$, $2^2 3^2 5^0 = 36$, $2^2 3^0 5^2 = 100$, $2^0 3^2 5^2 = 225$, and $2^2 3^2 5^2 = 900$.

Thus, 8 of the positive divisors of 900 are perfect squares.

Solution 2

The positive divisors of 900 are $$1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, 900$$ Of these, 1, 4, 9, 25, 36, 100, 225, and 900 are perfect squares ($1^2,2^2,3^2,5^2,6^2,10^2,15^2,30^2$, respectively).

Thus, 8 of the positive divisors of 900 are perfect squares.
In isosceles triangle $ABC$, $\angle ABC = \angle ACB$, so the sides opposite these angles ($AC$ and $AB$, respectively) are equal in length.

Since the vertices of the triangle are $A(k,3)$, $B(3,1)$ and $C(6,k)$, then we obtain $$\begin{aligned} AC & = AB \\ \sqrt{(k-6)^2 + (3-k)^2} & = \sqrt{(k-3)^2 + (3-1)^2}\\ (k-6)^2 + (3-k)^2 & = (k-3)^2 + (3-1)^2 \\ (k-6)^2 + (k-3)^2 & = (k-3)^2 + 2^2 \\ (k-6)^2 & = 4\end{aligned}$$ Thus, $k-6 = 2$ or $k-6 = -2$, and so $k=8$ or $k=4$.

We can check by substitution that each satisfies the original equation.