Terminate Half the Numbers
5 0 ! 1 0 0 !
How many trailing zeros are there in the value of the number above?
Details and Assumptions
Trailing zeros are a sequence of 0 s in the decimal representation of a number, after which no other digits follow. For example, the trailing zeros of the number 1 3 , 9 5 0 , 2 0 0 , 0 0 0 is 5
The answer is 12.
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4 solutions
Discussions for this problem are now closed
60 * 70 * 80 * 90 * 100 = 6 * 7 * 8 * 9 * 10 ^6,Thus only six zeros. But you have mentioned 7 zeros, Please explain.
Man you forget that 2x5=10, for example 52x55 make one more trailing zero :D
Actully, I was getting wrong answer because I was taking 75 * 72 as only one zero, but then I realize it is having 2 trailing zero. :-)
Why 12? I got 11.
5 2 ∗ 5 5
6 0
6 2 ∗ 6 5
7 0
7 2 ∗ 7 5
8 0
8 2 ∗ 8 5
9 0
9 2 ∗ 9 5
1 0 0 (counts 2 times)
That's 11 zero's right? Please explain.
Moderator note:
As Sri Vishnu Yandrapati already pointed out below, this solution is wrong.
75 factors are 5 5 3. it contains two 5's and another even number makes it two 0's
That makes sense, Thanks!
Now it's clear, Thanks @SRI VISHNU YANDRAPATI
answer is 12...... simple stuff.... the 50! terms will cancel off.... So, there are 5 series...... from 51 to 60 you will have 2 zeroes at the end for the entire multiplication of the 10 terms.... Similarly, for 61 to 70, there are 2 zeroes, from 71 to 80 there are 3 zeroes at the end, for 81 to 90 there are 2 zeroes and for last series 91 to 100 there will 3 zeroes at the End.... so, overall multiplication of the 5 series u will have 12 zeroes at the End....
Trailing zeros come from 5 and its multiples >>
no. of zeros in 100! = 5 1 0 0 + 2 5 1 0 0 = 24
no. of zeros in 50! = 5 5 0 + 2 5 5 0 = 12
the trailing zeros in 5 0 1 1 0 0 ! = 24 -12 = 12
Moderator note:
This solution is wrong and incomplete. You did not explain why you stop at 2 5 1 0 0 and 2 5 5 0 . It does not show a clear pattern to what you're doing. The correct way is to apply floor functions accompanied with De Polignac's formula . You would have gotten the wrong answer if the fraction is replaced with numbers that aren't a multiple of 5 , say: 1 1 1 ! 2 2 2 ! .
i didnt get how you found out the number of trailing zeros
The main source of zeros is digit FIVE. Each FIVE will produce zero when multiplied by any even number. Also Each 5 2 , 5 3 . . . e t c . will add more zeros.
100!/50! =51 * 52 * 53 * 54 * 55 * 56 * 57 * 58 * 59 * 60 * ..... * 100
so., we have 60,70,80,90,100 ---> 7 zeros
and., the products with 5 in units place (55,65,75,85,95) with even number in Units place(like.,52,62,72,82,92)
i.e., 55*52 = results 0 in units place. so total 5 pairs will give -----> 5 zeros
7+5 = 12 zeros in the result