IIT JEE , AIEEE Forum - Mathematics , Algebra

pantpranav

How many 0's are there in the number 1000! ?

ankur_30

hi,

generally there are three zeros in '1000'. but, if you are asking this question on the basis of some theorem or axiom the its answer may be different.

byeeeeeeeeeeeee.......

pollp2

24
iff right do rate me

pantpranav

Sorry boy.
Its a question of class 11.
Not for you.
Please check the answer of your question and rate my reply.

pantpranav

Sorry its also incorrect.

bsgdabest

is it 249?

bsgdabest

highest power of a prime "p" in factorial n:( [t]=greatest integer function)
is [n/p] + [n/p^2] +.... ( continue till u get 0)
generally in 1000!, there is lower power of 5 than power of 2 ( u want highest 10^ans) if p<q and u get 5^p*2^q from my abv relation, your answer will obviously be p, u can leave out the extra 2,s they dont contribute to zero, if u calculate for 5 u will get 249., try for power of 2, u will more than that...this is a method only to find zeros at the end, but u cannot find no. of zeros in between using this method.
for proof, refer TMH or "Higher algebra" by hall and knight...Hope this helped

anchitsaini

edit
i have got it now( don't rate me for this cos iberis and bsgdabest have given the answer before me)

here is the method--

 find the total number of factors of 5 and the total number of factors of 
2 in all the numbers from 1 to 
1000; the smaller of those two numbers will be the number of 0's at 
the end of 1000!
we know that factors of 5 will be less , hence lets find that

1) 1 out of every 5 numbers 1 to 1000 inclusive (200 of them) 
   contains at least one factor of 5.

2) Of the 200 numbers 1 to 1000 inclusive that contain at least one 
   factor of 5, 1 out of every 5 (40 of them) contains a second 
   factor of 5.

3) Of the 40 numbers 1 to 1000 inclusive that contain a second 
   factor of 5, 1 out of every 5 (8 of them) contains a third factor 
   of 5.

4) Of the 8 numbers 1 to 1000 inclusive that contain a third factor 
   of 5, 1 out of every 5 (1 of them) contains a fourth factor of 5.

The total number of factors of 5 contained in the numbers 1 to 1000 
inclusive is then

   200 + 40 + 8 + 1 = 249

iberis22

To find the number of 0's we have to find what is the highest power of 2 and 5 in 1000!.

But the power of 2 will be very high so the limiting factor will be power of 5.

To find the power of 5, we divide the number by 5, find the greatest integer.

n then again, divide the quotient(greatest integer) by 5 n so on until we get 0.

[1000/5] = 200

[200/5] = 40

[40/5] = 8

[8/5] = 1

[1/5] = 0

total power of 5 = 200 + 40 + 8 + 1 = 249

Decoder

take it as a chemical reaction where one is in excess and other is a limiting reagent..here 5 act as limiting reagent..

for a zero..u have to get a 2 and a 5...

u will have 500 even no.s ..then 250 no.s having a extra power of 2...then 125 with 2^3..and so on..

similarly u have to calculate for 5...

actually it is [1000/5] + [1000/25] .+ [1000/125] +.[1000/625]..
from here u extract all the 5's or exponent of 5 in 1000!

solving u get 249..

sboosy

The general formula ..if ur keen on that one line formula is ...
the number of zeros in n! is
[n/5] + [n/5²] + [n/5³] + ....till as long as the denominator is less than the numerator ..obviously ..even if u proceed beyond that ..u ll geta 0 everytime

so 1000! ..no of zeroes ..

[1000/5] + [1000/25] +[1000/125] + [1000/625] = 200 + 40 + 8+ 1= 249

..and this method as mentioned by iberis will give only the ending zeroes in a

number's factorial

for example

8! = 40320

using the above method we get 1 and not 2

this shows that it accounts for only ending zeroes

Decoder

this is actually the formula for exponent of 5..
exponent of 5 end up as exponent of 10..from there this zero question arises..
similarly exponent of any no . can be found..

ardra

I am just a starter.So please pardon me if Igot it wrong.

1000/5=200

200/5=40

40/5=8

8/5=1

200+40+8+1=249

ananth_patri

to find the no. of zero's we have to find the power of 2 and power of 5 then the no. of 0's is min of power of 2 or 5.....to find the power of a prime no...we have the formula (n/a) + (n/a²) + (n/a³) + ...... so on ....till a^m such tht a^m < n < a^m+1 ...

Where n=the term

a= the power which we want to find....

rate me if usefulll

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