Factorial's product: Too huge

Let $N=125,411,328,000$ . For $N$ to have 3 trailing zeros, $5^3$ must be a factor. The power 3 of 5 comes from $5!$ , $6!$ and $7!$ . Therefore, $n=\boxed{7}$ . If $n > 7$ , trialing zeros is more than 3 and if $n < 7$ , trialing zeros is less than 3. By computing, we find that $\displaystyle \prod_{k=1}^7 k! = 125,411,328,000$ .

Prime factorization of $125,411,328,000 = 2^{16} \times 3^7 \times 5^3 \times 7$ . To get exactly one $7$ , $n$ must be $7$

Lemma If a decimal numeral ends in $n$ zeroes not directly preceded by $5$ , it may be divided by $5^n$ but not by $5^{n+1}$ .

Our number therefore contains three factors 5, but not more.

For $5 \leq n < 9$ , $n!$ contains precisely one factor 5.

Therefore the desired answer is $1!\times 2!\times 3!\times 4!\times 5!\times 6!\times 7!$ .

The number of trailing zeroes indicate the number of $10$ s in the product. There are $3$ zeroes at the end, so $10^3$ must be a factor .

Prime factorisation of $10=5 \times 2$ . Then $10^3=(2 \times 5)^3$ i.e. to get $10^3$ , you need at least three $2$ s and $5$ s .

Let $n$ be the highest number to be factorised. To find out how many times $x$ (any number strictly less than $n$ ) appears in this product, use this formula: $x[n-(x-1)]$ .

(This may not seem very intuitive to you the first time, but if you continue to play with this you'll get that eventually. To get you started, consider this — how many 2s will be there in 4!? Think about this. $2$ appears in the factorial of every number, but $1$ . Oh, but why? Ask yourself again. Bring home the actual point by asking why repeatedly. Why does this happen? Because $1$ is smaller than $2$ and its factorial cannot hold the number. But the $3$ other numbers greater than $2$ do. How do you find that 3? Because you subtract that numbers less than 2 from the total number. Any integer $z$ in the number line has only one predecessor — $z-1$ . Hope you get the idea.)

$\displaystyle {1 \times (2 \times 3) \times (2 \times 3 \times 4) \times (2 \times 3 \times 4 \times 5)...}$

In all of the options, the number of $2$ s in the product is more than necessary. (Warning! Too much of both the numbers will yield a number greater than $10^3$ . We don't want that.) The number of $5$ s must be exactly three, since there's way too much $2$ s. In all the options, $5$ is present at least once in the product. We don't even need to use the formula at this point; common sense is enough to guide us.

Trying out the options

If $n=5$ , we'll have one $5$ .

If $n=6$ , we'll have two $5$ s.

If $\boxed{n=7}$ , we'll have three $5$ s.

$125,411,328,000$ is clearly product of $1000$ or $10^{3}$ .

Since $10=2\times 5$ , first $n!$ which is a product of $10$ is clearly $5!=5\times ...\times 2\times 1$ . Then, $10$ is also a factor of $6!$ and $7!$ because $6!=6\times 5!$ and $7!=7\times 6\times 5!$ .

Hence, $1000\mid 7!\times 6!\times 5!$ , so we can conclude that $n=7$ .

Since $125,411,328,000$ ends with 3 zeros, it is a multiple of 1000. $1000=2^3\times~5^3$ So we need to extend the string of factorials until we get at least three 2s and three 5s show up.

The 2s occur at 2!, 3!, and 4!. The 5s finally show up at 5!, 6!, and 7!.

Look at the number of zeroes. Each zero must be contributed by (i! for i>=0)

as we have three zeroes, so,

n-5+1=3

or,n-4=3

or,n=7

LaTeX: $\text {First of all, let's have a look at} \; \Large \color{#3D99F6}{000}$

LaTeX: $\text {We have} \; \Large {\color{#3D99F6}{1000 = (5\times 2)^3 = 2^3\times 5^3}}$

LaTeX: $\text {Let's decompose into prime numbers} \; \Large \color{#3D99F6}{1!.2!.3!.4!.5!.6!.7!}$

LaTeX: $\large 1(2^{\color{#D61F06}{1}}.1).(3.2^{\color{#D61F06}{1}}.1).(3.2^{\color{#D61F06}{3}}.1).(5^{\color{#3D99F6}{1}}.3.2^{\color{#D61F06}{3}}.1).(5^{\color{#3D99F6}{1}}.3^2.2^{\color{#D61F06}{4}}.1).(7.5^{\color{#3D99F6}{1}}.3^2.2^{\color{#D61F06}{4}}.1)$

LaTeX: $\text {We have to find 3 times the number} \color{#3D99F6}{\;5\;} \color{#333333} {\text {and 3 times the number}} \color{#D61F06}{\;2\;}$

LaTeX: $\text {We stop when we have found 3 times the digit 5. We notice that }$ LaTeX: $\text {the first one appears in 5!, the second one in 6.5! and the last one appears in 7.6.5!}$

LaTeX: $\text {Let's simplify :} \; \huge \color{#3D99F6}{\frac{1!.2!.3!.4!.5!.6!.7!}{1000} = \frac{1\times2\times6\times24\times120\times720\times5040}{1000}}$

LaTeX: $\text {Finally we get :} \; \Large \color{#3D99F6}{ 1\times2\times6\times24\times12\times72\times504 = } \boxed{\color{#69047E}{125.411.328 }}$

LaTeX: $\text {So the answer is :} \; \Large \boxed{\color{#D61F06}{n = 7 }}$

125,411,328,000 must be the least double factorial divisible by 1000 to account for the three 0's in the low order digits. That means a factor of 5 exactly 3 times as there are abundant 2's to make factors of 10. These come in exactly at 5, 6, and 7. Moving on to 8 or more would add additional 5's and would make a multiple of 10,000, 100,000 etc with more low order zeros. Thus 7 is the answer.

Please excuse the use of plain English and lack of symbols.

$N=125,411,328 × 1000 = q × 10^3$ .
$N=q × (2 × 5)^3 = q × 2^3 × 5^3$ .

5 occurs in 5! (the first time).
$5^2$ occurs in 6! and.
$5^3$ occurs in 7!

Then the product stops at 7!

$N$ has $3$ trailing zeros that means it has $5^3$ as factor. The first time that appears a $5$ factor is when it's reached $5!$ then the second $5$ when $6!$ and the third $5$ when $7!$ . Therefore $\boxed{n=7}$

Just focus on number of 0's.

Methods to get 0 = 1x10, 2x5

so if we take 0's by 1 x 10 which we will get only in 10!, it will have already extra 0's created by 2 x 5 in 5!, 6!, 7!, 8!, 9! and 10!

so we will take 0's by only 2 x 5 which we will get in 5! and above it

So we need three 0's, i.e, three 5's are needed which will be easily given by 5!, 6! and 7!

Hence n = 7

Look out for the number of zeroes. Each zero must be contributed by $i!$ for $i>=5$ so since we have three zeroes, we must have:

$n-5+1=3$

$n-4=3$

$n=\boxed{7}$

I've seen many complex solutions below, what I did was check how many 10s the final number had.

It had 3 zeros, so 3 tens.

Now, to get 3 tens, it has to have 3 pairs of 5 and 2

And to get 3 pairs of 5x2(=10) the expression has to have 5!, 6! and 7!

Thus the value of n is 7.

Any more and the number of zeros increases, any less and it decreases.

Prime numbers are the atoms of numbers: 125,411,328,000 has to be dividable by 1000: If we analyze in primes $1000 = 2\times5\times2\times5\times2\times5$ Thus 2 and 5 should appear at least 3 times in 125,411,328,000

125,411,328,000 should not be dividable by any number >1000 Thus at least one of 2 or 5 should not appear more than 3 times in 125,411,328,000.

To get three trailing zeroes, we need three factors of 10 (which means using prime factors, three 2s and three 5s). The first 5 shows up in 5!. We have lots of 2s at this point, so no need to count those. By 6! we have one more 5 and by 7! we have three 5s and more than three 2s, so the answer is 7!.

The product has 3 zeroes. Therefore, to have 3 zeroes you must have 3 2s and 3 5s. We get the first 3 5s in 5!, 6! and 7!. So, n=7.