Question

What is the largest power of $ 20 $ contained in $ 100 $ factorial?

Answer 1

Hint: In the given problem, we have to find the largest power of $ 20 $ contained in $ 100 $ factorial. We first break down $ 20 $ into its constituent prime factors and then proceed further. Prime factors are factors of a number that are, themselves, prime numbers. Then, we find the power of prime factors in factorial using the formula $ n = \left[ {\dfrac{N}{p}} \right] + \left[ {\dfrac{N}{{{p^2}}}} \right] + \left[ {\dfrac{N}{{{p^3}}}} \right] + ... $ where $ \left[ . \right] $ represents the greatest integer function.

Complete step-by-step answer:
The first step is to find out the prime factors of $ 20 $ .
 $ 20 = 2 \times 2 \times 5 $
So, we will find the power of prime factors $ 2 $ and $ 5 $ in $ 100 $ factorial.
So, the power of $ 2 $ in $ 100! $ $ = \left[ {\dfrac{{100}}{2}} \right] + \left[ {\dfrac{{100}}{{{2^2}}}} \right] + \left[ {\dfrac{{100}}{{{2^3}}}} \right] + \left[ {\dfrac{{100}}{{{2^4}}}} \right] + \left[ {\dfrac{{100}}{{{2^5}}}} \right] + \left[ {\dfrac{{100}}{{{2^6}}}} \right] + ... $
Computing the powers of two, we get,
 $ = \left[ {\dfrac{{100}}{2}} \right] + \left[ {\dfrac{{100}}{4}} \right] + \left[ {\dfrac{{100}}{8}} \right] + \left[ {\dfrac{{100}}{{16}}} \right] + \left[ {\dfrac{{100}}{{32}}} \right] + \left[ {\dfrac{{100}}{{64}}} \right] + ... $
Dividing the numbers, we get,
 $ = \left[ {50} \right] + \left[ {25} \right] + \left[ {12.5} \right] + \left[ {6.25} \right] + \left[ {3.125} \right] + \left[ {1.5625} \right] + ... $
Taking out the integral parts of all the numbers, we get,
 $ = 50 + 25 + 12 + 6 + 3 + 1 $
Simplifying the calculations,
 $ = 97 $
So, the highest power of $ 2 $ in $ 100! $ is $ 97 $ .
Similarly, finding power of five in $ 100! $ .
So, the power of $ 5 $ in $ 100! $ $ = \left[ {\dfrac{{100}}{5}} \right] + \left[ {\dfrac{{100}}{{{5^2}}}} \right] + ... $
Computing the powers of $ 5 $ ,
 $ = \left[ {\dfrac{{100}}{5}} \right] + \left[ {\dfrac{{100}}{{25}}} \right] + ... $
 $ = \left[ {20} \right] + \left[ 4 \right] + ... $
Taking the integral parts of all numbers, we get,
 $ = 24 $
So, the highest power of $ 5 $ in $ 100! $ is $ 24 $ .
No, since the power of $ 2 $ in $ 100! $ is $ 97 $ . So, the highest power of $ {2^2} $ or $ 4 $ in $ 100! $ is $ 48 $ .
Now, we know that $ 20 = 2 \times 2 \times 5 $ . We can see here the highest prime factor is $ 5 $ . So, the least power in $ 100! $ is $ 5 $ . So, this will determine the power of $ 20 $ contained in $ 100 $ factorial.
Hence, the largest power of $ 20 $ contained in $ 100 $ factorial is $ 24 $ .
So, the correct answer is “24”.

Note: We must have a clear understanding of concepts related to factorials and permutations and combinations so as to solve the problem. We must take care of the calculations while doing the question as it can alter our final answer. One must know how to compute the greatest integer function of numbers to get to the final answer.