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Consecutive zeroes

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by krishnasty » Sat Jun 18, 2011 12:11 am
How many consecutive zeroes will be present at the end of the following number?

153 × 170 × 187 × 204 × 221 × ...... × 850

1) 9
2) 40
3) 21
4) 11
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by Frankenstein » Sat Jun 18, 2011 2:36 am
Hi,
153 X 170 X 187 X 204 X 221 X ...... X 850 = (17x9)x(17x10)x...(17x50)=(17^42)(50!)/8!
17^42 will not have any impact on the number of zeroes at the end
Lets consider the number of zeros at the end for (50!)/8!
The number of zeroes at the end of n! is [n/5]+[n/5^2]+[n/5^3]+..... where [p] represents the integer part of p.
For 50!, number of zeros is [50/5] + [50/5^2] = 10+2 =12
For 8!, number of zeros is [8/5] = 1
So, number of zeros in (50!)/8! is 12-1 =11

Hence, D
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by krishnasty » Sat Jun 18, 2011 3:32 am
that is correct...
OA: D
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by amit2k9 » Sun Jun 19, 2011 4:42 am
good question indeed.

17^42 ( 50 ! / 8 !)

50!/5 = 12
8!/5 = 1

thus 12-1 = 11 it is.
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