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# How many consecutive zeros will appear at the end of 43! if that numbe

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Manager
Joined: 23 Sep 2016
Posts: 93
How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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02 Dec 2016, 14:09
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Difficulty:

15% (low)

Question Stats:

68% (00:41) correct 32% (01:25) wrong based on 138 sessions

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How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8
Senior Manager
Joined: 13 Oct 2016
Posts: 364
GPA: 3.98
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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02 Dec 2016, 23:14
$$\frac{43}{5} + \frac{43}{5^2} = 8 + 1 = 9$$

Intern
Joined: 18 Sep 2016
Posts: 2
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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04 Dec 2016, 09:18
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8

Thanks to user Bunuel, unforunately first post so cannot link, but formula copied below:
$$\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n

For number of zeroes denominator must be less than 43, therefore use:
-> 5 as 5 < 43
-> 5^2 as 25 < 43
-> 5^3 cannot as 125 >43

=$$\frac{43}{5} + \frac{43}{5^2}$$
= 8 + 1
=9

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Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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04 Dec 2016, 10:39
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8

43/5 = 8
8/5 = 1

Hence, no of zeroes will be 9

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Posts: 55670
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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05 Dec 2016, 01:40
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8

Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope it helps.
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Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

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14 Sep 2018, 17:42
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8

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 creates a 10, and each 10 creates an additional zero.

Since we know there are fewer 5s in 43! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 43!, we can use the following shortcut in which we divide 43 by 5, then divide the quotient of 43/5 by 5 and continue this process until we no longer get a nonzero quotient.

43/5 = 8 (we can ignore the remainder)

8/5 = 1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 43!.

Thus, there are 8 + 1 = 9 factors of 5 within 43! This means we have nine 5-and-2 pairs, so there are 9 consecutive zeros at the end of 43! when it is expanded to its final result.

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Re: How many consecutive zeros will appear at the end of 43! if that numbe   [#permalink] 14 Sep 2018, 17:42
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