64 of 100: Prime Octet

We know that $\color{#333333} 8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8$ , so we can write these numbers as the following:

$\displaystyle \begin{aligned} {\color{#D61F06}{A}} & {\color{#333333}{= 3 \times \bigg[(1 \times 2 \times 4 \times 5 \times 6 \times 7 \times 8) + 1\bigg]}} \\ {\color{#EC7300}{B}} & {\color{#333333}{= 4 \times \bigg[(1 \times 2 \times 3 \times 5 \times 6 \times 7 \times 8) + 1\bigg]}} \\ {\color{gold}{C}} & {\color{#333333}{= 5 \times \bigg[(1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8) + 1\bigg]}} \\ {\color{#20A900}{D}} & {\color{#333333}{= 6 \times \bigg[(1 \times 2 \times 3 \times 4 \times 5 \times 7 \times 8) + 1\bigg]}} \\ {\color{teal}{E}} & {\color{#333333}{= 7 \times \bigg[(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 8) + 1\bigg]}} \\ {\color{#3D99F6}{F}} & {\color{#333333}{= 8 \times \bigg[(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7) + 1\bigg]}} \\ {\color{#BA33D6}{G}} & {\color{#333333}{= 9 \times \bigg[(1 \times 2 \times 4 \times 5 \times 2 \times 7 \times 8) + 1\bigg]}} & \small \color{#3D99F6} 9 = 3 \cdot 3 = 3 \cdot \dfrac 62 \\ {\color{#69047E}{H}} & {\color{#333333}{= 10 \times \bigg[(1 \times 3 \times 4 \times 6 \times 7 \times 8) + 1\bigg]}} & \small \color{#3D99F6} 10 = 2 \times 5 \end{aligned}$

It is clear that $\boxed{0}$ of them are prime.

Note that this a specific case of the following theorem:

If $a \mid M$ and $a \mid N$ , then $a \mid M \pm N$ .

8! is divisible by 3,4,5,6,7,8,9 and 10

So 8! plus any of those numbers is also divisible by 3,4,5,6,7,8,9 and 10

8! is divisible by every one the numbers added when the result can be factored n( x + 1) where nx = 8!. Clearly not prime.

3..8 are obviously direct factors of 8!.

9 is found in 3 x 6 , 10 is found in 2 x 5.

Just thinking through this logically, we know that 8! is a multiple of 3, 4, 5, 6, 7, and 8. So A, B, C, D, E, and F cannot possibly be prime, since they are just adding one more of something that is already a factor of 8!.

Utilizing that same logic, we can rule out G and H. 9 is a multiple of 3, so adding 9 to a number that is already a multiple of 3 can't be prime. 10 is a multiple of 5, so adding 10 to a number that is already a multiple of 5 can't be prime.

Because 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8, we can add the numbers 1 through 8, or any multiple thereof, to 8! and still have it be a multiple of that specific number. 9 is simply adding 3 to 8! three times whereas 10 is simply adding 2 to 8! 5 times, both of which work according to the prior sentence. Thus none of them are prime.

So we know that $8!=8\times7\times6\times5\times4\times3\times2\times1$ . This can be rewritten as $8!=2^6\times3^2\times5\times7$ . Also, $8!$ is an even number. We remove the letters that add an even number as Even + Even = Even, therefore, it's divisible by $2$ . The letters removed are B, D, F and H. The remaining $4$ letters aren't prime as you can take out the respective factor. The general equation is $x(\frac {8!}{x}+1)$ for the remaining letters. For A, C, E and G you make $x$ equal to $3$ , $5$ , $7$ , $3^2$ respectively.

• Every factorial number after 5! ends with digit zero, therefore C and H must not be primes, because adding 0 with 5 would end up with a number divisible by 5 (not prime), and adding 0 with 0 would end up with a number divisible by 10 (not prime).

• Therefore B, D, and F are definitely even numbers, divisible by 2 (not prime).

• A, E, and G remain. But, since 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, A is not prime (because 3 is one of its factor), E is not a prime as well (because 7 is one of its factor too), and then G is not a prime too (because 3 and "2x3" or 6 are two of its factor, therefore it's divisible by 9).

After these three steps: None of them are prime numbers.

Each one of them can be factorised easily enough. Up to $8! + 8$ , this is done by taking out the number added. For example:

$8! + 3 = 1\times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 + 3 = 3(2 \times 4 \times 5 \times 6 \times 7 \times 8 + 1)$

In the final two, this can be done by factoring out factors of the numbers added.

$8! + 9 = 1\times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 + 9 = 3(2 \times 4 \times 5 \times 6 \times 7 \times 8 + 3)$

and

$8! + 10 = 1\times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 + 3 = 10(4 \times 6 \times 7 \times 8 + 1)$

So none of the numbers are prime.

Incidentally, for the next number in the sequence $8! + 11$ , the same arguments can't be made as 11 does not feature in the factorial and is prime.

Use the fact, if d | a and d | b, then d | (a + b). Since 3, 4, 5, 6, 7, 8 | 8!, it follows that 3 | A, 4 | B, 5 | C, 6 | D,

7 | E, 8 } F. So A, B, C, D, E, F are composite. Note 3 | 8!, 3 | 9, then 3 | G, also 2 | 8!, 2 | 10, then 2 | H.

So also G and H are also composite.

Let us check the statement that: for any positive integer $n$ from $2$ to $10$ , $8!+n$ is divisible by $n$ .

Consider the divisors of $6!$ , or $720$ . It can be factorised into $(2^4)(3^2)(5)$ . From this, we can conclude that for the integers from $2$ to $10$ , only $7$ is not a factor. We can verify this by checking the powers of $2$ that are $≤4$ : $1,2,4,8$ , powers of $3$ that are $≤2$ : $1,3,9$ , as well as $2*3$ and $2*5$ .

Now, $7*6!$ or $7!$ has the same divisors as $6!$ , but with $7$ added (and a few more others). Therefore, $7!$ is divisible by the numbers from $2$ to $10$ , and it follows that any multiple of $7!$ , including $8*7!$ or $8!$ also satisfy this property.

Since $8!+n$ is divisible by an $n$ which are not $1$ or $8!+n$ , it doesn't satisfy the definition of a prime. Therefore, there are no numbers in the list which are prime.

$A=8!+3=40320+3=40323$ ............[ $\text{this is not a prime as summation of its digits(4+0+3+2+3=12) is divided by 3}$ ]

$B=8!+4=40320+4=40324$ ............[ $\text{simply divided by 2, so this is not a prime}$ ]

$C=8!+5=40320+5=40325$ ............[ $\text{simply divided by 5, so it is not a prime}$ ]

$D=8!+6=40320+6=40326$ ...........[ $\text{simply divided by 2, so this is not a prime}$ ]

$E=8!+7=40320+7=40327$ ...........[( $\frac{40327}{7}=5761$ , $\text{so, it is not a prime also}$ ]

$F=8!+8=40320+8=40328$ ...........[ $\text{simply divided by 2, so this is not a prime}$ ]

$G=8!+9=40320+9=40329$ ..........[ $\text{this is not a prime as summation of its digits(4+0+3+2+9=18) is divided by 3}$ ]

$H=8!+10=40320+10=40330$ .......[ $\text{simply divided by 2, so this is not a prime}$ ]

$\text{so, there is no prime number}$

8! is divisible by all the numbers 1 to 8 Not prime(duh) Divisible by 9 because 3*6 is divisible by 9 Divisible by 10 since 285 is 10 if you add anything between 1 and 10 to it it is still divisible by that number

(think about it)

8!=41320.

Then,

8!+3=41323

8!+4=41324

8!+5=41325

8!+6=41326

8!+7=41327

8!+8=41328

8!+9=41329

8!+10=41330

+4, 6, 8, 10 have to be composite because if you add an even number to an even number, it the sum is also an even number; so it is divisible by 2.

+3 is not prime because 41323 is divisible by 31.

+5 is composite because 41325 is divisible by 5.

+7 is also composite because it is divisible by 11.

Finally, +9 is not prime because divisible by 37.

So, ALL of the numbers are composite;

NONE of them are prime.

Suppose the answer is not $0$ . Then we have to show that at a number greater than $8!$ is prime. There is no elegant way to do this other than by brute force. Hence from a question setter's point of view, no answer other than $0$ makes sense.

In our problem $n$ and $k$ are non-negative integers :

N = $n$ ! + $k$ where $n$ = 8 and $3 \leq k\leq 10$

n ! = 8 ! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

In eqns B, D, F and H, $k$ is even, multiply by 2 / 2

N = 2 [8 ! + $k$ ] / 2

2 is a factor of N, 8 ! and $k$ Therefore N is non-prime where $k$ is even

In eqns A, C, E and G, $k$ is odd, multiply by corresponding $k$ / $k$

N = $k$ [8 ! + $k$ ] / $k$

since $k$ is a factor of N, 8 ! and $k$ , therefore N is non-prime where $k$ is odd

In general,

N = $n$ ! + $k$ where $n$ = 8 and $3 \leq k\leq 10$

since $k$ is a factor of N, n ! and $k$ , therefore N is non-prime for $n$ = 8 and $3 \leq k\leq 10$

The answer is 0

Each expression has a factor of 3,4,5,6,7,8,9,10 respectively In some sense factorials are extremely composite

8!+N= 8!/N*(1+N) where N is 3 to 10. 8!/N is an integer number, therefore the result is dividable by (1+N) and not a prime. 8! is dividable by 9 and 10 (not included in 8!)

All primes (except 2 and 3) are equal to 6n + 1 so the only candidate that meets this criteria in E = 8! + 7. But 7 is already a factor of 8! so adding seven to it yields a number also divisible by 7.

Let's generalize the problem a little bit. Let $n$ and $k$ be non-negative integers.

I claim that $n! + k$ cannot be a prime number if $k$ has a divisor $d$ such that $1 < d \leq n$ .

Proof is simple because by definition $d \mid k$ and since $n! = 1 \cdot 2 \cdot \ldots \cdot d \cdot \ldots \cdot n, \quad d \mid n!$ , meaning that $d \mid n! + k$ , but clearly $d < n! + k$ , so there exists a positive integer $a$ greater than $1$ such that $da = n! + k$ .

Now, $n! + k$ cannot be prime because it is divisible by $1, a, d, ad$ and since $1 < a, d < ad$ it has at least 3 different positive divisors meaning that it cannot be a prime.

Using the theorem above we can easily conclude that none of the numbers are prime because the corresponding $k$ all have at least one divisor $d$ such that $1 < d \leq 8$ .

Hence, the answer is $\boxed{0}$ .