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

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Senior Manager
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Joined: 18 Jul 2018
Posts: 377
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
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The number of consecutive zeros at the end of 77!x42! is  [#permalink]

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New post 16 Sep 2018, 02:18
2
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A
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Difficulty:

  45% (medium)

Question Stats:

56% (01:31) correct 44% (01:46) wrong based on 25 sessions

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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, 02:35
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

Imo
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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, 02:36
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
D
GMAT Club Bot
Re: The number of consecutive zeros at the end of 77!x42! is &nbs [#permalink] 16 Sep 2018, 02:36
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The number of consecutive zeros at the end of 77!x42! is

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