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The number of consecutive zeros at the end of 77!x42! is

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New post 16 Sep 2018, 03:18
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The number of consecutive zeros at the end of 77!x42! is

1) 9
2) 18
3) 24
4) 27
5) 29

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Re: The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 16 Sep 2018, 03:35
Trialing zeros depends on 2 and 5 couples

As 77!, as well as 42!, has more twos than fives, we have to get the number of fives. There always will be enough twos.

First of all find fives in 77!
The best approach is to divide 77 by \(5^x\). where x is an integer and \(5^x\) doesn't exceed 77. Second, we need only quotient of the division and ignore remainder. Third, we add the quotients. So:
\(\frac{77}{5^1}\)=15
\(\frac{77}{5^2}\)=3

In 77! we have got 15+3=18 Fives

The same with 42!
\(\frac{42}{5}\)=8
\(\frac{42}{25}\)=1

8+1=9

Expression 77!*42! will have 18+9=27 fives

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Re: The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 16 Sep 2018, 03:36
Afc0892 wrote:
The number of consecutive zeros at the end of 77!x42! is

1) 9
2) 18
3) 24
4) 27
5) 29


number of consecutive zeroes = number of 5*2 pairs
since number of 5 < number of 2
finding number of 5 will suffice

77/5 + 77/25 = 15+3 = 18
+
42/5+42/25 = 8+1 = 9

total = 27
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Re: The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 24 May 2019, 20:27
Chethan92 wrote:
The number of consecutive zeros at the end of 77!x42! is

1) 9
2) 18
3) 24
4) 27
5) 29


How to use BOX function: https://gmatclub.com/forum/how-many-tra ... l#p2190991

Let's find the number of zeroes in 77!-> 18
Let's find the number of zeroes in 42!-> 9

Total 4) 27
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Re: The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 26 May 2019, 07:54
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why are you dividing by 25?
whats the logic behind 5^x
isnt 25 already being counted in 5's
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The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 26 May 2019, 08:01
ayusharora96 wrote:
why are you dividing by 25?
whats the logic behind 5^x
isnt 25 already being counted in 5's


Let's solve one sum together:- Number of trailing zeroes in 40!

Now as per box function you have to calculate:-

40/5 + 40/25= to get the number of trailing zeroes.

If you see carefully-
5 contributes one 5,
10 contributes one 5,...

...But 25 contributes two fives in 40! so we divide it by 25. Check the hyperlink in my previous post.
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The number of consecutive zeros at the end of 77!x42! is   [#permalink] 26 May 2019, 08:01
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