Zeros at the End of 600!

In order for a number to have a zero at the end of it, it must be a multiple of $10$ , or a multiple of both $2$ and $5$ , and it will have one zero at the end for every pair of powers of $2$ and $5$ that divide it. Because there are many more powers of $2$ than $5$ in $600!$ , it will be sufficient to count the powers of $5$ . There is one power of $5$ for every multiple of $5$ from $5$ to $600$ , giving $120$ powers. There is an additional power of $5$ for every multiple of $5^2$ from $25$ to $600$ , giving $24$ powers. Finally, there is another additional power of $5$ for every multiple of $5^3$ from $125$ to $600$ , giving $4$ powers. Because $5^4$ is greater than $600$ , there are no more powers of $5$ unaccounted for. $120+24+4=148$ .

the highest power of a prime number p contained in n! is given by k(n!)= [n/p]+[n/p^2]+[n/p^3]+------ until the series the terminates.

(here k(n!) denotes the highest power of p, and [x]=greatest integer function less than or equal to x.)

So to find out the number of zeros at the end of 600! we need to find the highest power of 5 in it, as it is obvious that the highest power of 2, will be greater than that of 5,hence the highest power of 5 multiplied with the similar power of 2 will give the number of zeros.

highest power of 5 in 600! is [600/5]+[600/25]+[600/125]+[600/625]+---= 120+24+4+0+----= 148

Therefore, the number of zeros at the right end of 600! is 148. (QED)

We know that a number gets a zero at the end of it if the number has 10 as a factor. For instance, 10 is a factor of 50, 120, and 1234567890. So we need to find out how many times 10 is a factor in the expansion of 600! .
But since 5×2 = 10, I need to account for all the products of 5 and 2. Looking at the factors in the expansions, there are many more numbers that are multiples of 2 (2, 4, 6, 8, 10, 12, 14,...) than are multiples of 5 (5, 10, 15,...). That is, if we take all the numbers with 5 as a factor, we'll have way more than enough even numbers to pair with them to get factors of 10 (and another trailing zero on the factorial). So to find the number of times 10 is a factor, all we really need to worry about is how many times 5 is a factor in all of the numbers between 1 and 600.

Okay, how many multiples of 5 are there in the numbers from 1 to 600? d 600 ÷ 5 = 120, so there are 120 multiples of 5 between 1 and 600.

But wait: 25 is 5×5, so each multiple of 25 has an extra factor of 5 that we need to account for. How many multiples of 25 are between 1 and 600? Since 600 ÷ 25 = 24, there are 24 multiples of 25 between 1 and 600. Similarly we need to account for an extra five from every multiple of 125 between 1 and 600 i.e one extra 5 from 125,250,375 and 500 which amounts to 4 more 5 as factors. Therefore total number of factors of 5 = 120+24+4 =148 and that is the number of zeroes at the end of 600!

We all know that 600!=600x599x598x597x...x3x2x1. To count the number of zeros on the right end of 600!, we must know how many times the number 10 was used as a factor. Since 10=5x2, we must count how many times the number 5 and 2 were used as a factor. In 600!, 2 was used more than 5 as a factor that's why the number of times number 5 was used a factor will tell how many factors of 10 are there in 600! because to form the number 10, we must have pair of 5 and 2 as factors. There are 120 multiples of 5 in 600! since 600/5=120. In 120 multiples of 5, there are 24 multiples multiple of 25 since 600/25=24 and a multiple of 25 is also a multiple of 5. In 24 multiples of 25, there are 4 multiples of 125 since 600/125=4 and a multiple of 125 is also a multiple of 25. In all multiples of 25, the number 5 was used as a factor twice and in all multiples of 125, the number 5 was used as a factor thrice. It implies that the number 5 was used {[(120-24)x1]+[(24-4)x2]+(4x3)}=148 times. Therefore, there are 148 zeros on the right end of 600!.

Using prime factorization, assume that $600!=2^x5^yA$ , where $A$ contains no factor of 2 and 5. Let $k=min(x,y)$ , we have $600!=2^x5^yA=(2*5)^kB=10^kB$ , where $B$ is not divisible by 10. So the number of $0$ on the right end of $600!$ is $k$ .

To find $y$ , we need to find the number of multiples of power of $5$ . Notice that $5^4=625>600$ , so we have 3 cases: Multiples of $5^3$ ( $\left \lfloor \frac{600}{5^3} \right \rfloor=4$ numbers); multiples of $5^2$ ( $\frac{600}{5^2}-4=20$ numbers) and multiples of $5$ ( $\frac{600}{5}-4-20=96$ numbers). So $y=4*3+20*2+96=148$

Now we need to find $x$ . There are $300$ even number between $1$ and $600$ , this implies $x>300>y$ . Thus, $k=148$

In each multiple of 5, there will be always a preceding factor of 2 that makes 10 and will add a 0 in our final number. Therefore; we can count the number of zeros with the multiples of 5, thus 600/5 is a solution. However, there are those numbers that 5 does not only appear once like 5^2 and 5^3 which are 25 and 125 respectively, therefore we add also the multiples of 25 and 125 in 600. : (600/5) + (600/25) + (600/125) = 120 + 24 + 4.8 *however there are only 4 125's in 600 there 4 is accepted only =120 + 24 + 4 =148

for getting a zero at the end we need a pair of 5 and 2... we concentrate on how many 5 s 600! has..(becoz no. of 2 s will always exceed no. of 5`s) 600/5=120 600/25=24 600/125=4.8 (4)

120+24+4=148

The highest power of a prime, p that divides n! is given by [\frac {n}{p}] + [\frac {n}{p^2}] + \ldots + [\frac {n}{p^k}] , where p^k is the largest power of p which is less than or equal to n.

Now, for our problem, if 10^k is the highest power of 10 that divides 600!, then we will have k zeroes at the end of the decimal representation of 600! Now, 10^k = 2^k . 5^k the highest power of 2 which divides 600! = [\frac {600}{2}] + [\frac {600}{2^2}] + [\frac {600}{2^3}] + \ldots + [\frac {600}{2^k}] = [\frac {600}{2}] + [\frac {600}{4}] + [\frac {600}{8}] + \ldots + [\frac {600}{512}] = 200 + 125 + 75 + 37 + 18 + 9 + 4 + 2 + 1 = 471. The highest power of 5 that divides 600! = [\frac {600}{5}] + [\frac {600}{5^2}] + [\frac {600}{5^3}] = 120 + 24 + 4 = 148 Now, the highest power of 2 that divides 600! (2^471) is far more than the highest power of 5 that divides 600! (5^148). So the number of 2's is enough to match each 5 to get a 10. Thus, the highest power of 10 that divides 600! = the highest power of 5 that divides 600! = 148 Thus we conclude there are 148 zeroes at the end of 600!

We know that if $p$ is a prime number and $n$ is a non negative integer, the highest power of $p$ that divides $n!$ is $[n/p] + [n/p^2] + [n/p^3] + ...$ . Now $10= 2*5$ . Note that highest power of $2$ that divides $600!$ is $[600/2] + [600/4] + [600/8] + [600/16] + [600/32] + [600/64] + [600/128] + [600/256] + [600/512]$ , and the highest power of $5$ that divides $600!$ is $[600/5] + [600/25] + [600/125]$ . Since $2<5$ , we can easily conclude that the highest power of $2$ that divides $600!$ is less than the highest power of $5$ that divides $600!$ . Note that the highest power of $10$ that divides $600!$ is equal to the highest power of $5$ that divides $600!$ . This is because if $\alpha$ is the highest power of $5$ that divides $600!$ , i.e $5^\alpha | 600!$ , and $\beta$ is the highest power of $2$ that divides $600!$ , $\beta > \alpha$ and $2^\alpha | 2^ \beta$ , and hence $2^ \alpha | 600!$ . Since $g.c.d (2,5)= 1$ , $g.c.d (2^\alpha , 5^\alpha) = 1$ . Thus $2^ \alpha * 5^\alpha | 600! \implies 10^\alpha | 600!$ . Now if $\gamma$ is some integer greater than $\alpha$ such that $10^ \gamma | 600!$ , since $5^ \gamma | 10^ \gamma$ , we obtain $5^ \gamma | 600!$ , a contradiction of the fact that $\alpha$ is the highest power of $5$ that divides $600 !$ . Hence the answer= $[600/5] + [600/25] + [600/125] = 148$ .

To count the 5s we first count the number 5s then 25s then 125s under 600 and add these numbers together. This is because all the numbers that are multiples of only 5 are counted once. The numbers that are multiples of 25 and not 125 are counted twice and the numbers that are multiples of 125 are counted 3 times. Since 625>600 we need not consider any powers of 5 greater than 125. (600/5)+(600/25)=120+24=144. And (600/124)=4.8 meaning there are really only 4 multiples of 125 less than 600. So the inal answer is 120+24+4=148

The number of zeros at the right end of an integer is equal to the number of times that 10 divides the number, which is equal to the minimum of the number of times that 2 and 5 divide the number.

For any integer $d>0$ , let $600 = q \cdot d + r$ , where $q$ and $r$ are integers and $0 \leq r \leq d-1$ is the remainder of 600 divided by $d$ . Then, there are $q$ numbers less than or equal to 600, that are multiples of $d$ .

We will calculate the number of times that 5 divides $600!$ . There are 120 multiples of 5, 24 multiples of 25, 4 multiples of 125 and 0 multiples of 625. The multiples of 5 will contribute 1 factor of 5 each. The multiples of 25 will contribute 2 factors of 5, but we have already counted 1 factor from the multiple of 5, so they will contribute 1 additional factor of 5. The multiples of 125 will contribute 3 factors of 5, but we have already counted 1 factor from the multiple of 5, and another factor of 5 from the multiple of 25, so they will contribute 1 additional factor of 5. Hence, there are $120 \times 1 + 24 \times 1 + 4 \times 1 = 148$ factors of 5 in $600!$ .

We will calculate the number of times that 2 divides $600!$ . There are 300 multiples of 2, 150 multiples of 4, 75 multiples of 8, 37 multiples of 16, 18 multiples of 32, 9 multiples of 64, 4 multiples of 128, 2 multiples of 256, 1 multiple of 512. Using a similar method as before, there are $300 \times 1 + 150 \times 1 + 75 \times 1 + 37 \times 1 + 18 \times 1 + 9 \times 1 + 4 \times 1 + 2 \times 1 + 1 \times 1 = 596$ factors of 2 in $600!$ .

Thus, the number of times that 10 divides $600!$ is 148. This is the number of zeros at the right end of $600!$ .

Note: It should be obvious that $600!$ has many more factors of 2 than of 5. The calculations were done as a further example for students who have not seen this before. The solution may be made clearer using floor notation $( \lfloor \cdot \rfloor )$ , which we are avoiding for the sake of those who are unfamiliar with it.