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Difficulty:
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40%
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Find the number of trailing zeros in the expansion of (20!*21!*22! ......... *33!)^3!.
A. 468
B. 469
C. 470
D. 467
E. 471
A. 468
B. 469
C. 470
D. 467
E. 471
feruz77
# 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: https://gmatclub.com/forum/everything-ab ... 85592.html
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gurpreetsingh
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feruz77
Excellent explanation by Bunuel.
Let me give you one quick way to solve such questions.
3! = 6 => the answer must be of the form \(10^{6n}\)
Only A and C have even powers. Hence B C E are out
Out of A and C only A's power is divisible by 3, hence C is out.
Hence A is the answer.
General Discussion
Thank You Bunuel.
I have one more remark:
This series is not a series of numbers, it is a factorial and in this respect your approach makes sense because in a factorial, as a ganeral rule, a number of trailing zeros will depend on the highest power of 5.
Thank You one more time.
I have one more remark:
This series is not a series of numbers, it is a factorial and in this respect your approach makes sense because in a factorial, as a ganeral rule, a number of trailing zeros will depend on the highest power of 5.
Thank You one more time.
Bunuel - i understand the rule of trailing zeroes, but how did u deduct from that
that if its ^6 - it have be a multiply of 6?
thanks.
that if its ^6 - it have be a multiply of 6?
thanks.
144144
For example:
100 has 2 trailing zeros, 100^6=(10^2)^6=10^12 will have 2*6 trailing zeros.
Now, we have (something)^6: if # of trailing zeros of that something is x then # of trailing zeros of (something)^6 will be 6x, so multiple of 6.
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oh... thanks. +1
do u ever sleep btw?
do u ever sleep btw?
gurpreetsingh
Hi gurpreet
can u pls tell me , out of A,C ,why did u check whether the ans is div by 3
I know their is smthng silly i am missing ,....but pls tell me what is that.....
gurpreetsingh
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read my solution again.
he given impression is raised to the power 3! which is 6.
=> the answer should be of the form - something raised to the power 6n.
out of all the given powers,which are raised on 10, only A and C are eligible to be divisible by 6 as all others are odd numbers.
Now out of A and C, 470 is not divisible by 6n, hence it can not be the answer.
let me know if you are still not able to make it.
he given impression is raised to the power 3! which is 6.
=> the answer should be of the form - something raised to the power 6n.
out of all the given powers,which are raised on 10, only A and C are eligible to be divisible by 6 as all others are odd numbers.
Now out of A and C, 470 is not divisible by 6n, hence it can not be the answer.
let me know if you are still not able to make it.
mbaiseasy
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Last visit: 29 Dec 2013
Posts: 316
Concentration: Marketing, Finance
Schools: Full Time MBA (A$)
GPA: 3.23
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I love trailing zeroes.
20! has 4 factors of 5 = 5^4
21! to 24! also have 4 factors of 5 each = 5^16
25! has 6 factors of 5 = 5^6
26! to 29! also have 6 factors of 5 each = 5^24
30! has 7 factors of 5 = 5^7
31! to 33! has 7 factors of 5 = 5^21
There are 5^78 then raised to 3!=6 so we have 5^468. Obviously we have more than 468 factors of 2 so the count of 5 is our limiting factor.
Answer: A
20! has 4 factors of 5 = 5^4
21! to 24! also have 4 factors of 5 each = 5^16
25! has 6 factors of 5 = 5^6
26! to 29! also have 6 factors of 5 each = 5^24
30! has 7 factors of 5 = 5^7
31! to 33! has 7 factors of 5 = 5^21
There are 5^78 then raised to 3!=6 so we have 5^468. Obviously we have more than 468 factors of 2 so the count of 5 is our limiting factor.
Answer: A
I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471 (without the 10^), Isn't it? Maybe I am missing something.
mbaiseasy
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joe123
20! = 20*19*18*17*...*4*3*2*1
We know that from the above has 5^4.
We know that from the above it has 10 even numbers and some of them like 8 = 2^3. Thus, there are at least 10 factors of 2 or 2^17 to be exact.
To get the trailing zero, you have to capture a pair of 5 and 2. Choose the limiting factor.
Thus, we have 5^4*2^17=(5^4)(2^4)(2^13) giving 10^4...
Continue to do this in the other factorials.
21!,22!,23!,24! will have a total of 10^16
25! will have 10^6 since 25 has two factors of 5.
Do it until 33! and we will have 78 factors of 10.
But we have to raise by 3! = 6. 78*6= 468
do we get such questions on gmat?
joe123
Lets say
20! = 10^x
21! = 10^y
22! = 10^z
Then
(20!*21!*22! ……… *33!)^3! = (10^x * 10^y * 10^z....10^n)^3!
= (10)^(3!(x*y*z*...n)
= (10)^6(xyz...n)
In given options, 6 is the multiple of 468 only.
So Ans
(A)
Sachin9
Trailing zeros concept is tested on the GMAT (check here: everything-about-factorials-on-the-gmat-85592.html), though this particular question is a bit out of scope.
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joe123
You are correct. The answer is 468, not \(10^{468}\). The problem statement is wrong.
hitman5532
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joe123
That threw me for a loop as well. I was going crazy trying to figure out how there would be so many damn zeros.
The answers should be 468 - 471 or the problem should written differently.
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Superb question. Captured the essence of GMAT in a single shot.
As soon as I saw the ^ 3! I made a note (some number) x6 on my scrap paper.
Started going through the choices for anything dividing by 6 and voila!
468 was the first choice and was divisible. Saw that the rest are consecutive so marked it directly.
+1 Cheers for the poster
As soon as I saw the ^ 3! I made a note (some number) x6 on my scrap paper.
Started going through the choices for anything dividing by 6 and voila!
468 was the first choice and was divisible. Saw that the rest are consecutive so marked it directly.
+1 Cheers for the poster
Buneul, here's my doubt:
# 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;
for calculating trailing zeros up til 24! you did just 20/5=4.
but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7)
Suppose I want # of trailing zeros in 310!
using your concept 310/5+310/5^2=62+12=74 trailing zeroes
BUT using the factorial calculator below I am getting 76 trailing zeroes
https://www.nitrxgen.net/factorialcalc.php
please suggest a fool proof method for calculating trailing zeroes of any +ve integer.
# 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;
for calculating trailing zeros up til 24! you did just 20/5=4.
but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7)
Suppose I want # of trailing zeros in 310!
using your concept 310/5+310/5^2=62+12=74 trailing zeroes
BUT using the factorial calculator below I am getting 76 trailing zeroes
https://www.nitrxgen.net/factorialcalc.php
please suggest a fool proof method for calculating trailing zeroes of any +ve integer.
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