How many zeros does 100! end with?
20
24
25
30
32
Ans is 24. Can anyone explain in simple terms? Thanks!
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- Chak De
100! = 1 x 2 x 3 x ... x 9 x 10 x ... x 15 x 20 x .... 90 x ... x 95 ... x 100
Now each of 10, 20, 30 ie.e every multiple of ten will add one zero, and hundred will add one extra. Hence total zeros = 11 (10 multiples and one extra for 100)
Now each pair of 2,5 ... 12,15... and so on will give one zero. there are 10 such pairs. So Ten more zeroes. But 75 x 72 = 5400 gives two zeroes, hence total zeroes from ths step is also 11.
now we have to account for those products groups which still leave a number 5 in tens place, this is because we have several numbers (10 total) ending in 4 which multiplied with these can give more zeroes. Remember we have already counted the 2's and the 5's in units place.
25 x 22 = 550, which multiplied by any number with 4 in units place will give one more zero
50 , which multiplied by any number with 4 in units place will give one more zero
this means two more zeroes.
Hence total number of zeroes = 11 + 11 + 2 = 24
its all trial and error !
Now each of 10, 20, 30 ie.e every multiple of ten will add one zero, and hundred will add one extra. Hence total zeros = 11 (10 multiples and one extra for 100)
Now each pair of 2,5 ... 12,15... and so on will give one zero. there are 10 such pairs. So Ten more zeroes. But 75 x 72 = 5400 gives two zeroes, hence total zeroes from ths step is also 11.
now we have to account for those products groups which still leave a number 5 in tens place, this is because we have several numbers (10 total) ending in 4 which multiplied with these can give more zeroes. Remember we have already counted the 2's and the 5's in units place.
25 x 22 = 550, which multiplied by any number with 4 in units place will give one more zero
50 , which multiplied by any number with 4 in units place will give one more zero
this means two more zeroes.
Hence total number of zeroes = 11 + 11 + 2 = 24
its all trial and error !
- JonathanSchneider
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- kylo
i think there is a much simpler approach to this problem.
now we get 11 zeroes from numbers such as 10,20,30,...........,90,100)
now a zero can be added if we have a 2 and a 5 i.e, 2 x 5 = 10.
how many such 2s & 5s we get -
between 1 to 10 we get one set of 2 & 5.
similarly we get one set each - from 11 to 20, from 31 to 40, from 51 to 60, from 61 to 70, from 81 to 90 & from 91 to 10.
total 7 such sets or 7 zeroes can be added.
now 21 to 30, 41 to 50 & 71 to 80 are different.
from 21 to 30 we get two 5s from 25. hence we will get two sets of 2 & 5.
from 41 to 50 we get one 5 from 45 & one more 5 from 50. hence we get two sets of 2 & 5.
from 71 to 80 we get two 5s from 75 or 5 x 5 x 3. hence we get two sets of 2 & 5.
total 6 sets or 6 zeroes can be added.
hence total no of zeroes = 11 + 7 + 6 = 24.
Thanks!
now we get 11 zeroes from numbers such as 10,20,30,...........,90,100)
now a zero can be added if we have a 2 and a 5 i.e, 2 x 5 = 10.
how many such 2s & 5s we get -
between 1 to 10 we get one set of 2 & 5.
similarly we get one set each - from 11 to 20, from 31 to 40, from 51 to 60, from 61 to 70, from 81 to 90 & from 91 to 10.
total 7 such sets or 7 zeroes can be added.
now 21 to 30, 41 to 50 & 71 to 80 are different.
from 21 to 30 we get two 5s from 25. hence we will get two sets of 2 & 5.
from 41 to 50 we get one 5 from 45 & one more 5 from 50. hence we get two sets of 2 & 5.
from 71 to 80 we get two 5s from 75 or 5 x 5 x 3. hence we get two sets of 2 & 5.
total 6 sets or 6 zeroes can be added.
hence total no of zeroes = 11 + 7 + 6 = 24.
Thanks!
- RAJESHTL
100!
100! - How many Zeroes is the question. Right?
A zero will be formed when 5 and 2 get multiplied
In 1*2*3*4*5*6*7..........98*99*100
we have
1) Number of single powere of fives = 100/5 = 20
2) And Number of powers of 5(5 square) less than 100 = 100/25 = 4
The next power of 5 is 125. But 125 > 100
So there are 24 distinct 5s in 100!
Now count for single, double, thrible powers of 2
Obviously number of 2s in 100! will be more than the number of 5s in 100!
However we are interested only in the minimum
For example
5*5*5*2*2 = 500 (only two zeroes) as number of 5s = 3, number of 2s = 2
So the number of Zeroes is 24
A zero will be formed when 5 and 2 get multiplied
In 1*2*3*4*5*6*7..........98*99*100
we have
1) Number of single powere of fives = 100/5 = 20
2) And Number of powers of 5(5 square) less than 100 = 100/25 = 4
The next power of 5 is 125. But 125 > 100
So there are 24 distinct 5s in 100!
Now count for single, double, thrible powers of 2
Obviously number of 2s in 100! will be more than the number of 5s in 100!
However we are interested only in the minimum
For example
5*5*5*2*2 = 500 (only two zeroes) as number of 5s = 3, number of 2s = 2
So the number of Zeroes is 24
- ahistegt4
Excellent explanations!
Each pair of 5 and 2 creates one zero. We have more 2's than 5's between 1 and 100, so lets focus how many 5's we have.
We have 100/5=20
We also have 25, 50, 75 and 100 which have two fives in them one of which have already been counted above. So we have 4 more.
Total=20+4=24
Each pair of 5 and 2 creates one zero. We have more 2's than 5's between 1 and 100, so lets focus how many 5's we have.
We have 100/5=20
We also have 25, 50, 75 and 100 which have two fives in them one of which have already been counted above. So we have 4 more.
Total=20+4=24
- esledge
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Please post question source
Somebody please post the source of this question (soon!) or we'll have to delete it to avoid potential copyright issues. Thanks!
Emily Sledge
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- JonathanSchneider
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