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by vishal.pathak » Sat Nov 19, 2011 1:15 am
vishal chugh wrote:what is the no. of zeros at the end of the expression
(6!)^6! * (7!)^7!*(8!)^8!*(9!)^9!
No of 0's in an addition = no of 0's in the smallest number of the addition

EG: 500+60 = 560 (No of 0's = no of 0's in the smallest number)

Smallest no here is 6!^6!
No of 0's in 6! = 1.
6! = 720. So no of 0's in 6!^6! = 720*1 = 720

PS: DUPLICATE POST
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by vishal chugh » Sat Nov 19, 2011 2:24 am
total no. of zeros in 6! is 1;
so total no. of zeros in 6!^6! is 6!;
similarly 7!^7! has 7! zeros and so on...
hence total no. of zeros in the given expression is 6! +7! +8! +9!
this can be explained in this way
10 * 10 * 10 =1000 or 3 zeros.. mean zeros get added while we multiply the numbers
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by vishal.pathak » Sat Nov 19, 2011 2:29 am
vishal chugh wrote:total no. of zeros in 6! is 1;
so total no. of zeros in 6!^6! is 6!;
similarly 7!^7! has 7! zeros and so on...
hence total no. of zeros in the given expression is 6! +7! +8! +9!
this can be explained in this way
10 * 10 * 10 =1000 or 3 zeros.. mean zeros get added while we multiply the numbers
Oh yes. This is a product. Somehow, I read it as addition of the terms. My mistake
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