Let $$n=2^{19}3^{12}$$. Let $$M$$ denote the number of positive divisors of $$n^{2}$$ which are less than $$n$$ but would not divide $$n$$. What is the number formed by taking the last two digits of $$M$$ (in the same order)?

Write $$n$$ in its prime-factor form:
$$n = 2^{19}\,3^{12}$$.

1. Total divisors of $$n^{2}$$

The exponent of 2 in $$n^{2}$$ is $$2\times19 = 38$$ and that of 3 is $$2\times12 = 24$$. Using the divisor-count formula $$\prod (e_i+1)$$,

$$\tau(n^{2}) = (38+1)(24+1)=39\times25 = 975.$$

2. Divisors of $$n^{2}$$ that are < $$n$$

For every divisor $$d$$ of $$n^{2}$$ the number $$\dfrac{n^{2}}{d}$$ is also a divisor. Exactly one of the pair $$\{d,\dfrac{n^{2}}{d}\}$$ is smaller than $$n$$ (except when $$d=n$$). Hence the divisors of $$n^{2}$$ fall into $$\dfrac{975-1}{2}$$ such pairs plus the single divisor $$d=n$$ itself.

Number of divisors of $$n^{2}$$ that are strictly less than $$n$$:

$$\frac{975-1}{2}= \frac{974}{2}=487.$$

3. Among these, how many already divide $$n$$?

Divisor count of $$n$$ itself:

$$\tau(n) = (19+1)(12+1)=20\times13 = 260.$$

Exactly one of those 260 divisors equals $$n$$; the remaining 259 are less than $$n$$. Therefore 259 of the 487 numbers obtained in step 2 are common to both $$n^{2}$$ and $$n$$.

4. Required count $$M$$

$$M = 487-259 = 228.$$

5. Last two digits of $$M$$

The number formed by the last two digits of 228 is 28.

Answer: 28