What shall be the number of zeroes at the right end if 1000! is expand

Question

What shall be the number of zeroes at the right end if 1000! is expanded ?

A. 200
B. 201
C. 240
D. 248
E. 249

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ankushsambare · Answer

To find the number of zeroes at the end of 1000!, count how many times 5 is a factor:

Number of trailing zeros =
1000/5 + 1000/25 + 1000/125 + 1000/625
= 200 + 40 + 8 + 1
= 249

Answer: E

ravi1522 · Answer

Hello,
What shall be the number of zeroes at the right end if 1000! is expanded ?

basically question is how many 2and 5 are there which will form 10
but there are more two than 5 so lets calculate no of 5

so 1000/5+1000/25+1000/125+1000/625
 200+40+8+1
Hence there will 249 Zeroes at right

Hence E is correct answer

007Phoenix007 · Answer

To find trailing zeroes in n!, count the number of times 5 appears in the prime factors of all numbers from 1 to n, 
Since 2s are more than enough to pair with 5s to form 10s.
For 1000! :
 =200
 =40
 =8
 =1

Adding them, we get
200+40+8+1= 249

Therefore option E

Dereno · Answer

1000! Seems to be a big number.

To start with let’s take 10! to understand the pattern.

10! = 1*2*3*4*5*6*7*8*9*10

To make 10, we need a (2*5). In 10! , the number of 2’s is GREATER than the number of 5’s.

So, to calculate the exact number of 10s, we need to find the number of 5s.

The number of 2’s in 10!

= 2*4*6*8*10

= 2*2^2* (3*2)*2^3* (2*5)

= 2^8

10! On successive division will give = 5+2+1 = 8

The number of 5s in 10! = 2

The number of 10s in 10! = the minimum powers of (2,5) = min(8,2) = 2 = number of 5s in 10! .

So, the number of 10s in 1000!, we divide successively by 5 = 200+40+8+1 =  249

Option E

rohitz98 · Answer

In order to find the number of zeros in 1000!
Dividing by powers of 5 we get 249 zeros  Option E

ExpertsGlobal5 · Answer

Video Explanation:

https://www.youtube.com/watch?v=eXblrsJPq1o

rohitz98 · Answer

Thanks, your explanation seems detailed

NuttyIan · Answer

We're looking for the amount of prime factors that make up 10, or 2 and 5. Now there are over 500 factors of 2 in 1000!, because every other number is a factor of 2 and there are numbers which are factors of 2^x like 4 (2*2). So the limiting factor here is 5. The number of factors of 5^1 or greater is 200, since 1000/5= 200. But there are also factors of 5^2. So we also divide 1000/25, which is 40. There is also a factor of 5^3 (125), which I thought we would have to double count as in 2 extra but note that we already counted all the 5^2 numbers. So the first 5^2 of 5^3 was already counted. So we just count all the 5^3 numbers as 1 extra. hence 1000/125, or 8. The last factor we have is 5^4, which is 625. 5^5 is not included since if we even attempt to double 5^4 we get a number larger than 1000. Hence there is only 1 5^4 number. So

5^1= 1000/5= 200
5^2= 1000/25= 40 extra
5^3= 1000/125= 8 extra
5^4= 1000/625= 1 extra
249 Grand total.

5^1= 1000/5= 200
5^2= 1000/25= 40 extra
5^3= 1000/125= 8 extra
5^4= 1000/625= 1 extra
249 Grand total.

What shall be the number of zeroes at the right end if 1000! is expand

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What shall be the number of zeroes at the right end if 1000! is expand

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