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Find the greatest possible number of consecutive zeros at the end of the product of three positive integers if the sum of these three integers equals 939.
(The number 202100 has exactly 2 consecutive zeros at the end.)


Sagot :

Answer:

A. Of first 100 multiples of 10

B. 5*10*15*20*25*30*35*40*40*45

C. 100!

I would also like to know whether this kind of question is possible in GMAT and what would be the complexity level.

1. 124

2. 10

3. 24