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by selango » Fri Nov 12, 2010 10:41 pm
We need to find how many 10's are in 203!.

The prime factors of 10 are 2 and 5.since there are many 2's let us find how many 5's are in 203!

203/5+203/25+203/125 [Take only the quotient]

40+8+1=49 5's are available are in 203!

-->The highest power of 10 in 203! is 49

Pick C
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by Rezinka » Fri Nov 12, 2010 10:56 pm
Can you please explain how have you calculated the no. of 5s?
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by selango » Fri Nov 12, 2010 11:07 pm
The formula is like this:

To calculate the number of a's in n!,divide n by increasing power of 'a' till the highest power b where a^b<n


In this example we need to find number of 5's in 203!

So divide 203 by 5,5^2,5^3 [5^3<203]

203/5+203/25+203/125 [Note that here we need to take only the quotient]

40+8+1=49

Another example:

We need to find how many 5's are in 50!

50/5+50/25=10+2=12 5's are in 50!

Hope this clarify!!
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by Rezinka » Fri Nov 12, 2010 11:18 pm
Thats great..!!
I didn't know that..!!
Thanks :D
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by Deepthi Subbu » Sat Nov 13, 2010 5:42 am
Hello Selango ,

Thats a great formula . Are there some links or notes that provide such shortcuts??
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by Deepthi Subbu » Sat Nov 13, 2010 5:50 am
Also , by the formula - >

203 / 10 + 203 / 100 - > 20 + 2 - > 22 (Why is this not the answer )???
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