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Updated on: 12 Dec 2018, 04:03
Originally posted by EgmatQuantExpert on 12 Dec 2018, 03:34.
Last edited by EgmatQuantExpert on 12 Dec 2018, 04:03, edited 3 times in total.
Last edited by EgmatQuantExpert on 12 Dec 2018, 04:03, edited 3 times in total.
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C
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Variations in Factorial Manipulation - Exercise Question #3
How many trailing zeroes would be found in 63! upon expansion?
To read the article: Variations in Factorial Manipulation
How many trailing zeroes would be found in 63! upon expansion?
- A. 6
B. 12
C. 14
D. 53
E. 57
To read the article: Variations in Factorial Manipulation
Archit3110
12 Dec 2018, 03:37
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EgmatQuantExpert
63/5+63/5^2 = 12 + 2 = 14 IMO C
14 Dec 2018, 21:50
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Solution
Given:
- • We are given a factorial value: 63!
To find:
- • The number of trailing zeroes that is present in 63! upon expansion.
Approach and Working:
In a product of two or more numbers, trailing zeroes are generated from the instances of 10's only.
- • Hence, we need to find the highest power of 10, that can divide 63!
As 10 is a composite number, first we need to prime factorize it in terms of the prime factors and their corresponding powers.
- • \(10 = 2^1 * 5^1\)
Next, we need to find the individual instances of 2 and 5.
- • Instances of 2: \(\frac{63}{2^1} + \frac{63}{2^2} + \frac{63}{2^3} + \frac{63}{2^4} + \frac{63}{2^5} = \frac{63}{2} + \frac{63}{4} + \frac{63}{8} + \frac{63}{16} + \frac{63}{32} = 31 + 15 + 7 + 3 + 1 = 57\)
• Similarly, instances of 5: \(\frac{63}{5^1} + \frac{63}{5^2} = \frac{63}{5} + \frac{63}{25} = 12 + 2 = 14\)
Now, the number 10 is formed by one instance of 2 and one instance of 5.
- • Hence, number of pairs possible of \(2^1\) and \(5^1\) = minimum (57, 14) = 14
Hence, the correct answer is option C.
Answer: C
