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by sudhir3127 » Tue Aug 05, 2008 8:16 am
My answer is C. 34 do let me know if its correct ..i dont wanna mislead others..
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by parallel_chase » Tue Aug 05, 2008 8:32 am
Even I think the answer is 34.

(3!*4!)!= 144!

we have to find the number of 5's and 2's to determine the consecutive 0's

multiples of 2 = 72
multiples of 5 = 28

We just have to concentrate on 5's since we have enough number of 2's

Therefore number of 5's will be the number of consecutive 0's

25=5*5 (one five is already counted, we just have to add 1 more 5)
50=5*5*2 (one five is already counted, we just have to add 1 more 5)
75=5*5*3 (one five is already counted, we just have to add 1 more 5)
100=5*5*4(one five is already counted, we just have to add 1 more 5)
125=5*5*5 (one five is already counted we just have to add 2 more 5's)

Therefore total number would be =28+1+1+1+1+2=34

whats the OA?
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by sudhir3127 » Tue Aug 05, 2008 8:36 am
good work chase...

just an improvement.. i dont think we need to look for 2's as well

the method can be

144/5 + 144/25 +144/125 = 34..

thats it !!!
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by vaivish » Sat Aug 09, 2008 6:53 am
Oa is 34.
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