If $n! = 3! \times 5! \times 7!,$ what is $n?$

The answer is 10.

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Good! Since you already know that the answer is $10$ , to show less work, can you prove that $3! \times 5! = 8 \times 9 \times 10$ ?

(3!)(5!)(7!)=10! (3!)(5!)=10!/7!=8x9x10

Brigham Lucero
- 6 years, 1 month ago

However this is a very specific case in which the general equation "n!=a! * b! * c!" has an actual integer solution.

For example, try to solve it for n! = 4! * 5! * 6! or for n! = 2! * 4! * 6!, I think you can't

Roberto Carrasco
- 5 years, 12 months ago

no factorial identities?

Eka Kurniawan
- 6 years, 1 month ago

By rearranging the prime factors of $3!\times5!$ , we get $(1\times1)\times(2\times2\times2)\times(3\times3)\times(5\times2) = 8\times9\times10$

Feathery Studio
- 6 years, 1 month ago

1
*
2
*
3
*
1
*
2
*
3
*
4
*
5 = (5
*
2)
*
(4
*
2)
*
(3
*
3
*
1
*
1)=8
*
9
*
10

Akash Taru Das
- 6 years, 1 month ago

what's that ?
what's the meaning of !
and how 3! = 1
*
2
*
3 in your example ? how get this ?

marina rasmy
- 6 years, 1 month ago

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$n!$ refers to factorials, where positive integer $n$ is multiplied by all integers from $1$ to $n$ .

Feathery Studio
- 6 years ago

He's omitted the × signs. By 1
*
2
*
3
*
1
*
2
*
3
*
4
*
5
*
= (
*
5
*
2)(
*
4
*
2)(
*
3
*
3
*
1
*
1)=8
*
9
*
10 he means 1×2×3 × 1×2×3×4×5 = (5×2)(4×2)(3×3×1)=8×9×10. It's a little difficult to see, but he uses alternating italics to separate the terms, except for the It confused me too at first, not least because the notation is a bit inconsistent, ie the inclusion of (33
*
1
*
1). As 1 is the identity it can easily be omitted, but if you're going to include it, you should always include it.

Nicholas Pretzel
- 6 months, 2 weeks ago

! Means factorial let m! Then m! =1×2×3×4×.........(m-2)×(m-1)×m

abhishek sahoo
- 6 years ago

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, $\boxed{n = 10}$ .

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Great work of bounding!

3! * 5! * 7!

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

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

= 7! * 8 * 9 * 10

= 10!

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True. Alternatively if we remember the 10th row of the pascal triangle, we see that $\dbinom{10}{3} = 120 = 5!$ , then the answer is immediate.

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

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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

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

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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.

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So n= 10

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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.

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

now

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

$\boxed{n=10}$

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$3!=1\times2\times3$

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

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

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

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

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

$n!=10!$

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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

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$3! \times 5! \times 7! = n!$ , 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 $10! = n! \Rightarrow \boxed { n = \boxed { 10 } }$ .

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

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$3! = 1×2×3 = 6 \therefore$

$3! × 5! = 5! × 6 = 6!$ 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:

$7! × 8 × (3 × 5 × 2 × 3) =$

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

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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

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

$n!=10!$

$n=10$