- #1

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**[P! + P]/P^2 = INTEGER**

if and only if P is a prime number

P! is P factorial, e.g. 3*2*1 , 5*4*3*2*1 , 7*6*5*4*3*2*1, etc...

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- Thread starter Russell E. Rierson
- Start date

- #1

- 384

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if and only if P is a prime number

P! is P factorial, e.g. 3*2*1 , 5*4*3*2*1 , 7*6*5*4*3*2*1, etc...

- #2

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- #3

- 384

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3*2 = 6 , 6 = 6

5*3*2 = 30 , 3+0 = 3

7*5*3*2 = 210 , 2+1+0 = 3

11*7*5*3*2 = 2310 , 2+3+1+0 = 6

13*11*7*5*3*2 = 30030 , 3+3 = 6

17*13*11*7*5*3*2 = 510510 , 5+1+0+5+1+0 = 12 , 1+2 = 3

etc...

etc...

23 ---> 6

29 ---> 3

31 ---> 3

37 ---> 3

41 ---> 6

Multiply the sequence of prime numbers then sum the digits[omit 1] of

the product until it is a single integer. It appears to be 3 or 6.

- #4

- 384

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[P! +P^2 +P]/P^2 = integers >=1 when P is 1 or prime?

[P!+P^3+P^2+P]/P^2 = integers >=1 when P is 1 or prime?

[P! + P^4 + P^3 + P^2 +P]/P^2 = integers >=1 when P is 1 or prime?

[P! + P^n + P^(n-1) + ...+ P]/P^2 = integers >=1 when P is 1 or prime?

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