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Is there a shortcut for this problem?

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by indi » Thu Sep 22, 2011 6:27 pm
Hi
What is the OA for this question?
prachi.sakhare wrote:How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?
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by Anurag@Gurome » Thu Sep 22, 2011 6:49 pm
prachi.sakhare wrote:How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

Now, 43! Is 1* 2* 3* 5 * ....* 10...... * 15...... * 20.......* 25.......* 30.........* 35....*40*41*42*43.
This product has 5^9 as factor.
All these 9 5's will combine with 9 2's to give 9 zeros at the end of 43!.

The correct answer is 9.
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by rohit_gmat » Thu Sep 22, 2011 10:10 pm
prachi.sakhare wrote:How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?
the Q is askin how many 10s are in there... 10 = 5 x 2 (primes) ...
so we need to find out how many 5s & 2s are there... obviously the number of 5s will be lower, so lets focus on that.

numbers w/ 5 in it : 5, 10 ,15 , 20 ,25, 30, 35, 40
number of 5s_______: 1 1 1 1 2 1 1 1 = 9
so there are 9 5s and (obviously) 9 2s... so there are 9 10s & so 9 zeros
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by shankar.ashwin » Thu Sep 22, 2011 10:21 pm
A shortcut to these sums would be to factorize these number by 5

For Eg;

5|43

5|8 - 3 (Neglect Reminder)
5|1 - 3

Answer should be = 8+1 =9
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by amsm25 » Fri Sep 23, 2011 1:40 pm
43/5 + 43/5^2 = (approx)8+1 = 9
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by gabrieldoria » Wed Sep 04, 2013 5:49 pm
There is, and its very simple!

Check the explanation on this link:

https://puzzles.nigelcoldwell.co.uk/nineteen.htm

Every 5, 10, 15... gives an additional zero, and 25, 50, 75, 100.. gives two additional zeros.

Hope this helps.
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