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Expert replies
by vishal chugh » Fri Nov 18, 2011 11:49 pm
what is the no. of zeros at the end of the expression
(6!)^6! + (7!)^7!+(8!)^8!+(9!)^9!
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Source: — Problem Solving |

by pemdas » Fri Nov 18, 2011 11:58 pm
the number 0 pops up at the end when we have full tens
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!
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by pemdas » Sat Nov 19, 2011 12:06 am
yet the question is powers of factorials, meaning each has 1 ten there and several numbers ...
i addressed only (6!)^6 + (7!)^7+(8!)^8+(9!)^9

to come up with factorials in the powers we need to apply full factorial calc which is time consuming and is unlikely to appear on GMAT (my opinion)

10^6! + 10^7! +10^8! +10^9! = 10^6!(1+10^7+10*56+10^504) -> 720 zeros (1+727 zeros +776 zeros +1224 zeros) ...

pemdas wrote:the number 0 pops up at the end when we have full tens
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!
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by shankar.ashwin » Sat Nov 19, 2011 12:14 am
YEah! I agree, I don't think this is a GMAT problem, its way beyond scope. Could you post the source and answer choices? Pls stick to GMAT problems here as thats the entire point of this forum. You can find help in many other forums for general maths problems, this is not the place IMO.

Since you add numbers here, the term with least number of 0's would be the number of 0's in the expression.

6! has 1 zero
6!^6! will have 1*6! (or) 720 zeros.
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by vishal chugh » Sat Nov 19, 2011 12:51 am
6! has 1 zero
so total zeros in 6!^6! is 6!;
7! has 1 zero
so total zeros in 7!^7! is 7!
8! has 1 zero;
so total zeros in 8!^8! is 8!
9! has 1 zero
so total zeros in 9!^9! is 9!
hence there are total of 6!+ 7!+ 8!+ 9! has total of 6! zeros as after which it will give some no. which will be non zeros.....
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