Binomial Coefficients Divisible by 10

Let $T(m, n)$ be the number of the binomial coefficients $^iC_n$ that are divisible by $10$ for $n \le i \lt m$ ($i$, $m$ and $n$ are positive integers).
You are given that $T(10^9, 10^7-10) = 989697000$.

Find $T(10^{18}, 10^{12}-10)$.

The number of binomial coefficients that are divisible by ten depends on the combination of prime numbers (specifically, multiples of 2 and 5) that make up the numerator and the denominator of the coefficient.

According to the binomial theorem, each binomial coefficient $^iC_n$ in the expansion of $(x+y)^n$ is computed as $^nC_i = \frac{n!}{i!(n-i)!}$, where $n!$ is the factorial of $n$. If $^nC_i$ is divisible by $10$, then for some $k$, $^nC_i = 10^k$.

Note that $10^k$ = $2^k * 5^k$. So if $n!$ in the numerator of the binomial coefficient contains an equal or larger amount of both 2 and 5 factors than those in $i!(n-i)!$ in the denominator, the coefficient will be divisible by 10.

However, since factorials of numbers always contain more factors of 2 than 5 (as there are more multiples of 2), the key factor here is the powers of 5.

We can therefore simplify the problem to finding how many binomial coefficients have a sufficient number of factors of 5.

Unfortunately, the problem you provided as given, $T(10^9, 10^7-10) = 989697000$, can’t directly help us compute $T(10^{18}, 10^{12}-10)$, without a clear pattern or relationship between $m$, $n$, and $T(m,n)$ that regulates the change in the number of tens-divisible coefficients.

The value of $T(10^{18}, 10^{12}-10)$ can be theoretically computed using a computer algorithm or program that adheres to the principles outlined above, but it would be incredibly resource-intensive.

Without additional context or constraints, it’s difficult to provide a direct mathematical solution for $T(10^{18}, 10^{12}-10)$. The numbers involved are astronomically large, thus it’s practically impossible to calculate this by hand.

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