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how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40)

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how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 11 Jan 2017, 05:30
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how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

1) 30
2) 60
3)40
4)45
5)53
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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 11 Jan 2017, 06:09
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Here we have to find the number of trailing zeroes.
The concept - a term with zero is made when there's a presence of 2 and 5, because 2*5 is 10. If there's 1 2 and 1 5, there will be one trailing zero. Hence, we will find the number of 2s and 5s in the term mentioned in the question to arrive at the solution.

Number of 2s - 5^5*(2*5)^5*(3*5)^5* (4*5)^5* (5*5)^5* (2*3*5)^5* (5*7)^5* (8*5)^5* (9*5)^5* (2*5*5)^5. This gives us 5+10+5+15+5=40 2s. Because 4 is 2^2 and (2^2)^5 has 10 zeroes. Similarly, 8 gives 15 zeroes.

Number of 5s - we can find it in a similar manner. 5+5+5+5+10+5+5+5+5+10= 60 5s.

Since we need a pair of 2 and 5 to get a trailing zero, we will exhaust all 2s after 40 zeroes because we can't pair any more 2s with 5s to form a trailing zero.

So, is the answer 40? Option C?


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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 11 Jan 2017, 06:10
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yezz wrote:
how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

1) 30
2) 60
3)40
4)45
5)53


We have more than enought 5, hence we need to add powers of 10 and powers of 2.

\(10^5*20^5*30^5*40^5 = 10^{25}*2^{15}\)

we have 25 "pure" zeros and 15 twos, which will find their counterpart 5s.

Total number of trailing zeros = \(25 + 15 = 40\)

3) (C)
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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 11 Jan 2017, 06:50
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yezz wrote:
how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

1) 30
2) 60
3)40
4)45
5)53


Rewrite \(N = 5^5\times 10^5 \times 15^5 \times 20^5 \times 25^5 \times 30^5 \times 35^5 \times 40^5 \times 45^5 \times 50^5\)
\(N=(5 \times 10 \times 15 \times ... \times 50 )^5\)
\(N=M^5\)

\(M=5 \times 10 \times 15 \times ... \times 50\)
\(M= (5 \times 1) \times (5 \times 2) \times ... \times (5 \times 10)\)
\(M=5^{10} \times (1 \times 2 \times 3 \times ... \times 10)\)
\(M=5^{10} \times 10!\)

We need to find what is the maximum value of integer number \(n\) that \(10!\) is divisible by \(2^n\)

\(n=\bigg [\frac{10}{2}\bigg ] + \bigg [\frac{10}{2^2}\bigg ] + \bigg [\frac{10}{2^3}\bigg ]\)
\(n=5 + 2 + 1 = 8\)

Hence \(10! = 2^8 \times k\) where \(k\) is not divisible by 2

Hence \(M=5^{10} \times 2^8 \times k = 10^8 \times 5^2 \times k = 10^8 \times k'\)

\(\implies N= M^5 = (10^8 \times k')^5 = 10^{40} \times k'^5\)

\(N\) has 40 trailing zeros
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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 14 Jan 2017, 06:38
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my take
N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) = (5^5)^10 * (10!)^5

powers of 5 in 5^50 = 50 , powers of 5 in 10!^5 = 2*5 , thus power of 5's in N = 50+10 = 60

powers of 2 in 10!^5 = (10/2 + 10/4+10/8)* 5 ( ignoring quotient ) = 8*5 =40

thus number of trailing zeros limited by the number of 2's = 40
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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 16 Jan 2017, 17:46
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yezz wrote:
how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

A) 30
B) 60
C)40
D)45
E)53


Let’s first rewrite the given expression:

N = (5^5)(10^5)(15^5)(20^5)(25^5)(30^5)(35^5)(40^5)(45^5)(50^5)

To determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number. Each 5-and-2 pair yields 10, and each 10 contributes one trailing zero to the product.

5^5 = 5^5

10^5 = 2^5 x 5^5

15^5 = 3^5 x 5^5

20^5 = (4 x 5)^5 = (2^2 x 5^1)^5 = 2^10 x 5^5

25^5 = (5^2)^5 = 5^10

30^5 = (5 x 3 x 2)^5 = 5^5 x 3^5 x 2^5

35^5 = 5^5 x 7^5

40^5 = (8 x 5)^5 = (2^3 x 5^1)^5 = 2^15 x 5^5

45^5 = (9 x 5)^5 = (3^2 x 5^1)^5 = 3^10 x 5^5

50^5 = (25 x 2)^5 = (5^2 x 2^1)^5 = 5^10 x 2^5

Since we see there are fewer 2s than 5s in the factorization above, we simply count the number of 2s and thus we can determine the number of 5-and-2 pairs.

We have:

2^5, 2^10, 2^5, 2^15, 2^5

Since we have 40 2s, we have 40 5-and-2 pairs and thus there are 40 trailing zeros in the product.

Answer: C
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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 21 Jan 2017, 14:42
The statement in direct form generates power of base 10 with exponent 5, each one adds 5 zeros to the left., 20 ^ 5 = 2 ^ 5 * 10 ^ 5, taking the same analysis for the whole sample in direct form we have a Total of 25 zeros to the left, we must see if with the other powers we can form base powers 10.
Effectively 40 ^ 5 * 50 ^ 5 = 4 ^ 5 * 10 ^ 5 * 5 ^ 5 * 10 ^ 5, grouping 4 ^ 5 * 5 ^ 5 = 20 ^ 5 = 2 ^ 5 * 10 ^ 5, which yields 5 zeros more to the left.
Iterating the previous process, we get a total of 40 zeros on the left.

Correct answer C

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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40) [#permalink]

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New post 18 Apr 2018, 04:27
broall wrote:
yezz wrote:
how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40^5*45^5*50^5?

1) 30
2) 60
3)40
4)45
5)53


Rewrite \(N = 5^5\times 10^5 \times 15^5 \times 20^5 \times 25^5 \times 30^5 \times 35^5 \times 40^5 \times 45^5 \times 50^5\)
\(N=(5 \times 10 \times 15 \times ... \times 50 )^5\)
\(N=M^5\)

\(M=5 \times 10 \times 15 \times ... \times 50\)
\(M= (5 \times 1) \times (5 \times 2) \times ... \times (5 \times 10)\)
\(M=5^{10} \times (1 \times 2 \times 3 \times ... \times 10)\)
\(M=5^{10} \times 10!\)

We need to find what is the maximum value of integer number \(n\) that \(10!\) is divisible by \(2^n\)

\(n=\bigg [\frac{10}{2}\bigg ] + \bigg [\frac{10}{2^2}\bigg ] + \bigg [\frac{10}{2^3}\bigg ]\)
\(n=5 + 2 + 1 = 8\)

Hence \(10! = 2^8 \times k\) where \(k\) is not divisible by 2

Hence \(M=5^{10} \times 2^8 \times k = 10^8 \times 5^2 \times k = 10^8 \times k'\)

\(\implies N= M^5 = (10^8 \times k')^5 = 10^{40} \times k'^5\)

\(N\) has 40 trailing zeros


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Re: how many trailing zero's in N = (5^5*10^5*15^5*20^5*25^5*30^5*35^5*40)   [#permalink] 18 Apr 2018, 04:27
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