If n = 5 * 10 * 15 * 20 * 25 ........... * 500, how many trailing

Thanks Chetan . Updated the answer choices.

Hi,
you have gone wrong in highlighted portion

n = 5 * 10 * 15 * 20 * 25 ........... * 500
is NOT equal to n =5 (1*2*3*4*5.......100) BUT 5^100*(1*2*...100).. since you are taking out 5 from each term..

Hi,
bunuel has already given you a link to check on.
Otherwiseto check zeroes in th eend of a number, we require to know how many 10s are there in it..
400 has two 10s, 10000 has four 10s and so on..
Now what is 100!-- It is 1*2*3*4*...*99*100 -- product of all numbers till 100..
How many 10s will it have.. will depend on number of 2s and 5s it will have..
we have seen that it has excess 5s in 5^100, so we check for number of 2s..
\([\frac{100}{2}]+[\frac{100}{4}]+[\frac{100}{8}]+[\frac{100}{16}]+[\frac{100}{32}]+[\frac{100}{64}] = 50+25+12+6+3+1 = 97..\)

Now let me explain you withslightly smaller number..
say we are to find 2s in 15!..
\([\frac{15}{2}]+[\frac{15}{4}]+[\frac{15}{8}]+[\frac{15}{16}]\)
Now 15/2 gives us how many multiple of 2s are there till 15 so 15/2 = 7.5 , we take only integer value , so 7 what are they 2,4,6,8,10,12,14--7 values
now 15/4 gives us those number of 2s that add on due to multiples of 4; 15/4 = 3.75 what are they - 4,8,12-- 3 values
similarly number of multile of 8s, since they have three 2s- 8 = 2*2*2- we have counted two 2s through above method, so 15/8 = 1.8 , so 1 value- 8
beyond that 15/16 = 0.9, since it does not hane any 16s in it..
so TOTAL 2s = SUM of all the above

Hi Chetan,

How did you go from 5^100*100! to 97?

Thanks,
Diego.

Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope it helps.

Hey Vyshak and Everyone ...!!!
I have some serious concerns about this question.
I have seen many questions on the gmatclub to calculate the trailing zeros which can be done via easy manipulation of the number of two's and five's ad the numbers involved were small.
Don't you think The Formula you guys use for calculating the number of trailing zeros is a bit out of concept here ; Since the numbers involved are always small.
Do you really think GMAT will ask us this ?
I am seriously asking question as if the answer is YES then i might well learn the formula stated above .

Regards
StoneCold

Hi,

This concept tests the 2's and 5's required to form zeros in a given expression and I don't see it as a formula. Its always better to understand the question and logic behind it rather than to learn the logic as a trick. This question can be solved in under 2 mins and can very well be tested in exam.

No offence I still disagree
Anyways Please Provide an alternative solution for people who don't want to use the formula
I tried counting two's and zero's
Now without using the formula - How can someone count the number of two's between 1 and 100 in two minutes?

When its a small umber then counting the number of two;s is easy
Any other way out?


Regards
StoneCold

Hi StoneCold,

Whenever I type a reply to you, I am always afraid of that STUNNER :D

About the question, yes you cannot calculate the number of 2's in between 1 and 100 in under 2 minutes,
That is the precise reason why you need to remember the formula.

And, yes this can very well be tested on the GMAT.

Hi,

I would say even if I do not know the formula, I will get to that by just moving on because formula comes out of the approach itself..
lets count 2s in 1 to 100-- it could be in the form of 2s, 4s,8s etc..
how may multiples of 2 in 100 = 100/2 = 50..
how many multiples of 4 in 100 = 100/4 = 25....Actually its 25 of 4s which means 50 of 2s, but 25 of these have been counted in multiples of 2..
how many multiples of 8 in 100 = 100/8 = 12.5....so 12
how many multiples of 16 in 100 = 100/8 = 6....so 6
how many multiples of 32 in 100 = 100/32 = 3....so 3
how many multiples of 64 in 100 = 100/64 = 1....so 1.

Add all of them to get your answer..
the formula is actually the addition of the step wise approach you take to find the number of 2s..

=5^100×100!
In 100! There are
50 multiples of 2
25 multiples of 4
12 multiples of 8
6 multiples of 16
3 multiples of 32
1 multiple of 64
So total 97, 2's are there .
So number of trailing zero's are 97 as number of 5's are more then 97.
Option D is the answer

Posted from my mobile device

Hint : to find out the no of zeroes , count nos of 2s and 5s

Take either of the nos of 2s or 5s whichever is minimum .

Make the problem statement in a factorial

then divide either with 2 or 5 (whichever is minimum )

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