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Chapter 5: Problem 391

The number of zeros at the end of \(100 !\) is (a) 20 (b) 22 (c) 24 (d) 26

Short Answer

Expert verified
The number of zeros at the end of \(100!\) is determined by the number of 5s in its prime factors. Using the formula \[f_p(n) = \frac{n - s_p(n)}{p - 1}\], we can find the number of 5s in \(100!\) as 24. Therefore, there are 24 zeros at the end of \(100!\). Answer: \((c) 24\).

Step by step solution

01

Write the formula to find the number of occurrences of a prime factor in the factorial

To find the number of occurrences of a prime factor 'p' in the factorial, we can use the formula \[f_p(n) = \frac{n - s_p(n)}{p - 1}\] where \(s_p(n)\) is the sum of the digits of \(n\) when expressed in base p and \(f_p(n)\) is the number of times a prime factor p occurs in the factorial n.
02

Find the number of 5s in 100!

We have to find the number of 5s in 100! using the formula from Step 1. Here, \(n = 100\) and \(p = 5\). First, let's find \(s_p(n) = s_5(100)\), which is the sum of the digits of 100 when expressed in base 5. \[100_{10} = 400_{5}\] The sum of the digits of 100 in base 5 is 4. Now we can find the number of occurrences of 5 in 100! as: \[f_5(100) = \frac{100 - 4}{5 - 1} = \frac{96}{4} = 24\]
03

Determine the number of zeros at the end of 100!

The number of zeros at the end of 100! is equal to the number of 5s in the 100!, which we found in Step 2. Therefore, there are 24 zeros at the end of 100!. The correct option is: (c) 24

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Most popular questions from this chapter

Ten different letters of an alphabet are given Words with 5 letters are formed from the given letters. The no, of words which have at least one letter repeated is (a) 69760 (b) 30240 (c) 9948 (d) 10680

In an examination a question paper consists of 10 questions which is divided into two parts. i.e. part \(i\) and part ii containing 5 and 7 questions respectively. A student is required to attempt 8 question in all selecting at least 3 from each part. In how many ways can a student select the questions (a) \({ }^{5} \mathrm{C}_{2} \cdot{ }^{7} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{1} \cdot{ }^{7} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{0} \cdot{ }^{7} \mathrm{C}_{4}\) (b) \({ }^{12} \mathrm{C}_{5} \cdot{ }^{12} \mathrm{C}_{7}\) (c) \({ }^{5} \mathrm{C}_{3} \cdot{ }^{7} \mathrm{C}_{5}\) (d) \({ }^{12} \mathrm{C}_{8}\)

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