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How many trailing zeroes would be found in 63! upon expansion?

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How many trailing zeroes would be found in 63! upon expansion?  [#permalink]

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Updated on: 12 Dec 2018, 04:03
00:00

Difficulty:

5% (low)

Question Stats:

80% (00:53) correct 20% (00:56) wrong based on 113 sessions

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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.
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Re: How many trailing zeroes would be found in 63! upon expansion?  [#permalink]

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12 Dec 2018, 03:37
1
EgmatQuantExpert wrote:
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

63/5+63/5^2 = 12 + 2 = 14 IMO C
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Re: How many trailing zeroes would be found in 63! upon expansion?  [#permalink]

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14 Dec 2018, 21:50
1
2

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.

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Re: How many trailing zeroes would be found in 63! upon expansion?  [#permalink]

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15 Dec 2018, 05:19
No of trailing zeroes
63/5 --> 12/5 --> 2
So, trailing zeroes = 12 + 2 = 14
Re: How many trailing zeroes would be found in 63! upon expansion?   [#permalink] 15 Dec 2018, 05:19
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