Let $N$ be the number of triples $(x, y, z)$ of positive integers such that $x<y<z$ and $x y z=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2} \cdot 11^{2} \cdot 13^{2} \cdot 17^{2} \cdot 19^{2}$. When $N$ is divided by 100 , the remainder is
(A) 28
(B) 88
(C) 8
(D) 68
(E) 48
Suppose that $a, b$ and $c$ are positive integers with $a b c=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2} \cdot 11^{2} \cdot 13^{2} \cdot 17^{2} \cdot 19^{2}$.

We determine the number of triples $(a, b, c)$ with this property. (We are temporarily ignoring the size ordering condition in the original question.)

Since the product $a b c$ has two factors of 2 , then $a, b$ and $c$ have a total of two factors of 2.

There are 6 ways in which this can happen: both factors in $a$, both factors in $b$, both in $c$, one each in $a$ and $b$, one each in $a$ and $c$, and one each in $b$ and $c$.

Similarly, there are 6 ways of distributing each of the other squares of prime factors.

Since $a b c$ includes exactly 8 squares of prime factors and each can be distributed in 6 ways, there are $6^{8}$ ways of building triples $(a, b, c)$ using the prime factors, and so there are $6^{8}$ triples $(a, b, c)$ with the required product.

Next, we include the condition no pair of $a, b$ and $c$ should be equal. (We note that $a, b$ and $c$ cannot all be equal, since their product is not a perfect cube.)

We count the number of triples with one pair equal, and subtract this number from $6^{8}$.

We do this by counting the number of these triples with $a=b$. By symmetry, the number of triples with $a=c$ and with $b=c$ will be equal to this total.

In order to have $a=b$ and $a \neq c$ and $b \neq c$, for each of the squared prime factors $p^{2}$ of $a b c$, either $p^{2}$ is distributed as $p$ and $p$ in each of $a$ and $b$, or $p^{2}$ is distributed to $c$.

Thus, for each of the 8 squared prime factors $p^{2}$, there are 2 ways to distribute, and so $2^{8}$ triples $(a, b, c)$ with $a=b$ and $a \neq c$ and $b \neq c$.

Similarly, there will be $2^{8}$ triples with $a=c$ and $2^{8}$ triples with $b=c$.

This means that there are $6^{8}-3 \cdot 2^{8}$ triples $(a, b, c)$ with the required product and with no two of $a, b, c$ equal.

The original problem asked us to the find the number of triples $(x, y, z)$ with the given product and with $x<y<z$.

To convert triples $(a, b, c)$ with no size ordering to triples $(x, y, z)$ with $x<y<z$, we divide
by 6 . (Each triple $(x, y, z)$ corresponds to 6 triples $(a, b, c)$ of distinct positive integers with no size ordering.)

Therefore, the total number of triples $(x, y, z)$ with the required properties is

$$
N=\frac{1}{6}\left(6^{8}-3 \cdot 2^{8}\right)=6^{7}-2^{7}=279808
$$

When $N$ is divided by 100 , the remainder is 8 .

ANSWER: (C)