Chapter 9.1, Problem 51E

Solution

i.

To state: The number of 0’s in the string if $50!$ ends in a string of consecutive 0’s.

The resultant answer is 12.

Given information:

It is given that $50!$ ends in a string of consecutive 0’s.

Explanation:

As it is known that factorial notation represents: $n!=n\left(n-1\right)\left(n-2\right)\cdots 3\cdot 2\cdot 1$

Then $50!$ can be written as:

  $50!=50\cdot 49\cdot 48\cdot 47\cdot \dots \cdot 3\cdot 2\cdot 1$

The string of $50!$ contains 12 zeros.

Therefore, the string contains 12 zeros.

ii.

To state: The reason how do you know that the string contains 12 zeros.

The answer is $50!$ contains 12 zero.

Given information:

The string contains 12 zeros.

Explanation:

The factors of 50 are:

  $5,10,15,20,25,30,35,40,45,50$

Every one of the factors $5,10,15,20,25,30,35,40,45,50$ are divisible by 5 and only 25 and 50 are divisible by 25.

As it is known that the multiplication of 5 and 2 makes a zero. The factors $5,10,15,20,30,35,40,45$ contains eight 5’s and 25 and 50 contains four 5’s.

Then total numbers of 5’s are 12. Then the total number of 0’s it will form is 12.