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vishal chugh
- Junior | Next Rank: 30 Posts
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- Location: Ludhiana ,Punjab, India
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in 6! we have 6*5*4*3*2*1 OR 1 ten, we may also deduce ten is only possible if 2*5=10, since we cannot prime factorize 6,5,4,3,2,1 to more than one 2 and one 5 we have only one pair of 2 and 5, hence 1 ten
let's look further, 7!, we have here additional 7 but again only 1 ten
8! here again only 1 ten and 9! again only 1 ten
Thus, 10^6 + 10^7 +10^8 +10^9 = 10^6(1+10+100+1000) OR 1,111*10^6 only 6 zeros
vishal chugh wrote:what is the no. of zeros at the end of the expression
(6!)^6! + (7!)^7!+(8!)^8!+(9!)^9!












