feruz77 wrote:
Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468
b) 10^469
c) 10^470
d) 10^467
e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT?
# of trailing zeros in 20!, 21!, 22!, 23!, and 24! will be 4 (20/5=4. For 21!, 22!, 23! and 24!, instead of 20 you'll have 21, 22, ... but the result will be the same) --> total of 4*5=20 trailing zeros for these 5 terms. (Note here that this won't always be correct: for example 20 and 50 have one trailing zero each but 20*50=1,000 has three trailing zeros not two. That's because extra 2 in 20 and extra 5 in 50 "produced" one more trailing zero. In our case though, we won't have any extra 5-s in any factorial, as all are already used for existing trailing zeros);
# of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) --> total of 6*5=30 trailing zeros for these 5 terms;
# of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) --> total of 7*4=28 trailing zeros for these 5 terms;
So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28}*something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\).
Total of 468 trailing zeros.
Answer: A.
Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this.
Theory on this topic:
everything-about-factorials-on-the-gmat-85592.htmlAs you mentioned that any extra 2 or 5 left out from any of the factorial will result in One more Trailing Zero.
Also if possible please direct me to such a question where we have to take care of extra 2 or 5 being left out .