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$3!=1\times2\times3$

$5!=1\times2\times3\times4\times5$

$7!=1\times2\times3\times4\times5\times6\times7$

$3!\times5!\times7! = 1^{3}\times2^{3}\times3^{3}\times4^{2}\times5^{2}\times6\times7$

We can rearrange the expanded terms to get

$1\times2\times3\times4\times5\times6\times7\times(4\times2)\times(3\times3)\times(5\times2) = 10!$

$n!=10!$

$n=10$

We can see that $3! \times 5! = 6!$ as $3! = 6$ .

So, we need to find $6! \times 7!$ . Looking at the factors of this number, we see that $n < 11$ as there is no factor of 11 in $3!$ , $5!$ or $7!$ .

So, $n$ is 8,9 or 10. Now, we can see that $n!$ has 2 factors of 5 (one from $5!$ , one from $7!$ ).

Therefore, $n$ cannot be 8 or 9 as $8!$ and $9!$ only have one factor of 5. So, $\boxed{n = 10}$ .

3! * 5! * 7!

= (2 * 3) * (2 * 3 * 4 * 5) * 7!

= (2 * 4) * (3 * 3) * (2 * 5) * 7!

= 7! * 8 * 9 * 10

= 10!

$7!*5!*3!=7!*(4*2)*(3*3)*(5*2)=7!*8*9*10=10!$

3! * 5! * 7! Should be larger than 7! and already knowing that the result itself takes the form of n! we rearrange from the beginning.

7! * (3! * 5!).

P.S. Remember we have done no calculation yet so this step takes virtually no time at all (which is how basically human brain works)

Now factorials belong to only natural numbers the bracket part of expression has to anyhow take this shape :

7! * (3! * 5!) => 7! * (8 * 9 * 10 * ...) We only have to find out that what number is last and that will be our answer n.

P.S. no calculation yet.

3! * 5! = 6 * 120 = 720 Extract 8 from 720 720/8 = 90 Extract 9 from 90 90/9 = 10 Extract 10 from 10 10/10 = 1

So 10 is our last number, thus, our expression becomes 7! * 8 * 9 * 10 = 10! Answer n= 10

3! 5! 7!=10!=n! Then n=10

Feathery your solutions are good.I too did the same thing but what if the numbers were very big

By rearranging the values we get 1 2 3 4 5 6 7 (4 2) (3 3) (5 2) = 10! So, n!= 10!, implies n=10.

3!=6 3!x5!=6! factor 6! and try to make consecutive numbers after 7, 6!=(2)x(3)x(2x2)x(5)x(2x3) (2x4)x(3x3)x(2x5) 8x9x10 7!x8x9x10=10! n=10

3! * 5! * 7! = (2 * 3) * (2 * 3 * 4 * 5) * 7! = (2 * 4) * (3 * 3) * (2 * 5) * 7! = 7! * 8 * 9 * 10 = 10!

So n= 10

We know that $10!=7!6!$

Therefore on substituting the value of $7!$ ,

we get $3!5!7!=\frac{3!5!10!}{6!}$

Now proceed forward. This was just an innovative method.

$3!\times5!= 1\times2\times3\times1\times2\times3\times4\times5=8\times9\times10$

now

$3!\times5!\times7!=7!\times8\times9\times10=10!$

$\boxed{n=10}$

$3!=1\times2\times3$

$5!=1\times2\times3\times4\times5$

$\Rightarrow$ $Then$ $n!=3!\times5!\times7!$

$n!=7!\times3\times2\times1\times5\times4\times3\times2\times1$

$n!=7!\times(2\times4)\times(3\times3)\times(2\times5$ )

$n!=7!\times8\times9\times10$

$n!=10!$

W.K.T.. Factorial will have continoues number for 11 there are no factors and 11 does not exit in this example... so n cannot be greater than 10.. from 3!*5! we can generat following 8 = 4 * 2, 9=3 * 3, 10=5 * 2 , therefore we write, 7! * 8 * 9 * 10=10!

n!=3! x 5! x 7! =7! (3.2.1)(5.4.3.2.1) =7! (4.2)(3.3)(5.2)(1.1) =10! n=10

$3! \times 5! \times 7! = n!$ , then we get this.

$( 1 \times 2 \times 3 ) \times ( 1 \times 2 \times 3 \times 4 \times 5 ) \times ( 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 ) = 6 \times 120 \times 5040 = 720 \times 5040.$ .

$3! \times 5!$ equals to $6!$ , and $6! \times 7!$ equals to $10!$ .

Then $10! = n! \Rightarrow \boxed { n = \boxed { 10 } }$ .

$\boxed { \therefore \text { The answer is \boxed { 10 } } }$ .

$3! = 1×2×3 = 6 \therefore$

$3! × 5! = 5! × 6 = 6!$ So we have $n! = 6! × 7!$

This gives us the following factors:

$7! × 2 × 3 × 4 × 5 × 6 =$

$7! × 8 × 3 × 5 × 6 =$

$7! × 8 × 90 =$

$7! × 8 × 9 × 10 = 10!$

Alternatively, we can rewrite 7! × 8 × 3 × 5 × 6 as:

$7! × 8 × (3 × 5 × 2 × 3) =$

$7! × 8 × 9 × 10 = 10! \therefore n=10.$

3 + 5 + 7 = 15 5 is the common link, so it needs to be exterminated 3 + 7 = 10 or 15 - 5 = 10 therefore n! = 10!