Largest Power of 10 dividing the number 70!

Question

Find out what is the largest Power of 10 dividing the number 70!

Answer 1

It is sufficient to find the power of $5$ in the prime factoring of $70!$ (the power of 2 is obviously larger than the power of 5)

This is $\lfloor70/5\rfloor+\lfloor70/25\rfloor=14+2=16$. The answer is $10^{16}$

Answer 2

How many power of 5 are there? The naive answer would be $\frac{70}{5}=14.$

But consider $25!.$ That number has 6 powers of 5 since $5^2=25.$

So with 70, 25 and 50 are multiples of 25 below it so the number of factors of 5 is $14+2=16.$ Obviously there are many more factors of 2 (70/2=35 is already more than 16), so there's 16 factors of 10.

Answer 3

The answer is same as the highest power of $5$ in $70! $. Highest power of prime number in factorial can be found out by $$\lfloor \frac {70}{5} \rfloor + \lfloor \frac {70}{5^2} \rfloor +\lfloor \frac {70}{5^3} \rfloor +\cdots $$ $$=14+2+0+0\cdots =16$$ Hope it helps.

Answer 4

The prime factorisation of $10$ is $2^1\cdotp 5^1$.

We must then count factors are divisible by 5 (and by how many times) in

$$\small 1{\times}2{\times}3{\times}2^2{\times}5{\times}2\cdotp 3{\times}7{\times}2^3 {\times}3^2{\times}2\cdotp 5{\times}\cdots{\times}3^3{\times}19{\times}2^2\cdotp 5{\times}3\cdotp 7{\times}2\cdotp 11{\times}23{\times}2^3\cdotp 3{\times}5^2{\times}\cdots{\times}2\cdotp 5\cdotp 7$$

There are fourteen factors containing $5$, but of these two contain $5^2$ and so must be counted twice.

There are also thirty five factors divisible by $2$ (at least once each), of which only seven are divisible by $2$ and $5$. So we are covered.

$10^{16}$ is the largest exponential of $10$ dividing the factorial of $70$.

And indeed, living in the age of electronics we can now easily find that: $\begin{align}70! =&~{10^{16}{\times}\small 1197857166996989179607278372168909873645893814254642585755536286462800958278984531968} \\=&~ {2^{67}\cdotp 3^{32}\cdotp 5^{16}\cdotp 7^{11}\cdotp 11^6\cdotp 13^5\cdotp 17^4\cdotp 19^3\cdotp 23^3\cdotp 29^2\cdotp 31^2\cdotp 37\cdotp 41\cdotp 43\cdotp 47\cdotp 53\cdotp 59\cdotp 61\cdotp 67}\end{align}$

But it's good to understand the basics.

Answer 5

$10=\color\red{2}\cdot\color\green{5}$

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The multiplicity of $\color\red{2}$ in $70!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}70}\Big\lfloor\frac{70}{\color\red{2}^n}\Big\rfloor=35+17+8+4+2+1=\color\red{67}$

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The multiplicity of $\color\green{5}$ in $70!$ is $\sum\limits_{n=1}^{\log_{\color\green{5}}70}\Big\lfloor\frac{70}{\color\green{5}^n}\Big\rfloor=14+2=\color\green{16}$

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Therefore:

• The maximum value of $n$ such that $\color\red{2}^n$ divides $70!$ is $\color\red{67}$
• The maximum value of $n$ such that $\color\green{5}^n$ divides $70!$ is $\color\green{16}$

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Therefore, the maximum value of $n$ such that $(\color\red{2}\cdot\color\green{5})^n$ divides $70!$ is $\min(\color\red{67},\color\green{16})=16$.