Find the highest power of prime which divides a factorial.

Question

Find the highest power of $10$ that divides $50!$.

I know by De-Polignac's formula ,the highest power of $5$ that divides $50!$ is $12$ & the highest power of $2$ that divides $50!$ is $47$.

I am able to find this because $5$ & $2$ are primes but here $10$ is not prime but $10=5 \times 2$ so what can we say about the power of 10

My observation say that it may be $47-12$ Please correct me

Answer 1

Well to create a 10, we need a five and two. Since there are only 12 fives in the prime factorization (via De-Polignac), we can pair each of these fives with two. Thus the greatest $n$ such that $10^n|50!$ is $n=12$.

In general, if we want to find the greatest power of $k$, a composite number, that divides into $n!$, all we have to do is apply De-Polignac to the greatest greatest prime $p_i$ that divides $k$, as all of the other primes must appear more often.

Answer 2

Twelve $10$s requires twelve $5$s and twelve $2$s. Since you only have twelve $5$s, you cannot make a larger power of $10$ than $10^{12}$.

In general, let $N = p_1^{j_1} p_2^{j_2} \cdots p_n^{j_n}$ be a decomposition into distinct primes $p_i$ (having positive integer exponents, $j_i$). Then $N^k$ requires $k j_1$ copies of $p_1$, $k j_2$ copies of $p_2$, and so on, up to $k j_n$ copies of $p_n$.

Answer 3

For every $n\in\mathbb{N}$ there exists a unique $m\in\mathbb{N}$, which is neither a multiple of $2$ nor of $5$, such that $$n!=2^{e_2(n!)}\cdot 5^{e_5(n!)}\cdot m=(2^{e_2(n!)-e_5(n!)}\cdot m)\cdot 10^{e_5(n!)}$$

Since by the Polignac's formula $$50!=2^{47}\cdot 5^{12}\cdot m=(2^{35}\cdot m)\cdot 10^{12},$$ the number of trailing zeroes in $50!$ is $12$.

Answer 4

As you found:

The highest power of $2$ that divides $50!$ is $\left\lfloor \frac{50}{2} \right \rfloor + \left\lfloor \frac{50}{2^2} \right \rfloor + \left\lfloor \frac{50}{2^3} \right \rfloor + \cdots$ which evaluates as $ 25+12+6+3+1$ $= 47$, each term being half-rounded-down of the previous one.

The highest power of $5$ that divides $50!$ is $\left\lfloor \frac{50}{5} \right \rfloor + \left\lfloor \frac{50}{5^2} \right \rfloor = 10+2= 12$, similarly dividing through by $5$ repeatedly.

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So $50! = k\cdot 2^{47}\cdot 5^{12}$, where $k$ is not divisible by either $2$ or $5$. Clearly we can also write this as $50! = k\cdot 2^{35}\cdot 10^{12}$ - and again $k$ is coprime to both $2$ and $10$.

So the result is that $50!$ is divisible by $10^{12}$ but not by $10^{13}$, since here the powers of $5$ determine how many powers of $10$ can be extracted from $50!$.