## Consider the natural numbers having at least $5$ prime factors, which don’t have to be distinct. Sorting these numbers by size gives a list which starts with:

$32=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

$48=2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$

$64=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

$72=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$

$80=2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$

$96=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$

$\cdots$

So, for example, the fifth number with at least $5$ prime factors is $80$.

Find the millionth number with at least one million prime factors. Give your answer modulo $123454321$.

### Due to the size of the numbers involved in this problem, a direct approach would be complex and not feasible. Instead, we’ll use combinatorics and the Principle of Inclusion-Exclusion (PIE).

Let’s call a number with at least one million prime factors a “bad number”. Our task is to find the smallest good number, or a number which has fewer than a million prime factors.

There’s a total of $N = 2^{1000000}$ numbers to consider since for any of the 1000000 prime factors, we have the option of choosing 2 or not (hence 2 options per factor, raised to the power of a million).

We need to subtract from $N$ the count of “bad numbers”. Calculating this involves calculating the count of numbers that have at least $i$ prime factors where $i$ ranges from 1 to 1000000. This count will be $C(1000000, i) \cdot 2^{1000000-i}$.

However, simply subtracting this from $N$ will undercount the number of good numbers due to double-counting bad numbers with more than one level of badness (for example, numbers with both 500,000 and 700,000 prime factors). We correct this with the Principle of Inclusion-Exclusion, which gives us an alternating sum for the count of bad numbers:

$B = \sum_{i=1}^{1000000} (-1)^{i+1} \cdot C(1000000, i) \cdot 2^{1000000-i}$.

We then find the count of good numbers as $G = N – B$.

An algorithm to compute these sums efficiently doesn’t need to calculate each term from scratch but can update each successive term based on the previous term, using the identity $C(n, k) = C(n, k-1) \cdot \frac{n-k+1}{k}$. A similar identity for the power of 2 can be used to update that factor. We start with $C(1000000, 0) = 2^{1000000} = 1$ and can calculate each successive term from there.

In the end, the millionth good number is the millionth number not excluded by the count of bad numbers: $N – (G – 1000000) \mod 123454321$.

(This problem is an example of a math problem at the intersection of combinatorics and number theory. Its solution requires both clever manipulation of the combinatorial identities used to count possibilities and use of the Principle of Inclusion-Exclusion to correct for overcounting caused by the overlap of the conditions that define “bad” numbers. Computationally, it’s a big number problem: while the problem definition and the calculations each step along the way are simple in principle, the numbers involved—1000000 prime factors, the binomial and power terms—would be far too large to handle directly in practice. The solution requires both mathematical insights to simplify the calculations and careful attention to the properties of the calculations to keep them within the range of practical computation.)

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