There are k zeros at the end of 34! = 34 x 33 x 32 x .... 4 x 3 x 2 x 1: What
is the value of k?
I know the answer is 7 zeros but I can't show it with a complete solution. Can anybody help me with this,please...........
Consider the prime factorization of 34!. Since 2's are in abundance, we only need to calculate the number of 5's. 5, 10, 15, 20 and 30 yield one 5 each while 25 yields two 5's. So there are seven 5's in the prime factorization of 34!. Hence there are seven zeros at the end of 34!
no. of zeros means no. of tens (10)
to get 10, two factors are required 2 and 5.(10=2X5)
so it is enough to find how many 2's and 5's are there.
1. find no. of 2's
34/2=17
17/2=8
8/2=4
4/2=2
2/2=1
do the successive division and add all quotient, we get no. of 2's =17+8+4+2+1=32.
2. find no. of 5's
34/5=6
6/5=1
no. of 5's =6+1=7
so 34!=(some numbers) X 2^32 X 5^7.
therefore no of zeros or tens is maximum no of combination of 2 &
maximum 7 combination is possible.
$\displaystyle \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{34} {5^i}\right\rfloor=\left\lfloor\frac{34}{5}\right\ rfloor+\left\lfloor\frac{34}{5^2}\right\rfloor=7$
At a certain point, the values will all be zero.
That notation is the floor and the formula finds the number of trailing zeros.
For instance, let's look at numbers we can easily see 5! = 120 and 10! = 3628800. We have 1 and 2 trailing zeros, respectively.
$\displaystyle \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{5}{ 5^i}\right\rfloor=\left\lfloor\frac{5}{5}\right\rf loor=1$
$\displaystyle \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{10} {5^i}\right\rfloor=\left\lfloor\frac{10}{5}\right\ rfloor=2$