Question

How many natural numbers ‘n’ are there, such that ‘n!’ ends with exactly 30 zeroes?

Answer 1

Step-by-step explanation:

Natural number do not include negative numbers or zero

Example natural numbers

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 7 18 19 20 21 22 23 24 25 26 27 28 29 30

Answer 2

Answer:

    The number of factors increases by $3$.  so it becomes $31 > 30$. Thus, we find a number  $n$ that $n!$ has $30$ factors of $5$.

Step by step explanation:

•  If $n=100$ taking randomly keeping in view $30$ the number of zeroes.
• When $100!$ has ${\frac{100}{5}+\frac{100}{5^{2} } =24$.So $n$ should be greater than $100$.
• The next multiple  $5$ is $105$. But $105=5$ × $21$ has only one extra $5$.
• When the number of zeroes will increase by $1$ only.
• Similarly $110,115$ and $120$ also have one extra $5$. Number of zeroes (from $120!$ to $124!$) $=28$.
• Now, the next multiple of $5$ is $125$ and $125$ contains three $5's$.
• So, the number of zeroes will increase by $3$.
• The number of zeroes is $125!==28+3=31$. So, there is no factorial of a number that ends with $30$ zeroes.