Prime Factors

so n = 1*2*3*4*5*6*7*8
prime factors of those are: 2,3,5,7
so 4 prime factors?

sfreeman wrote:if n is the product of the integers from 1 to 8, inclusive, how many different primer factors greater than 1 does n have?

product of n: 1*2*3*4*5*6*7*8

We can re-write this as:
1*2*3*(2^2)*5*(3*2)*7*(2^3)

See what I did? I reduced everything integer in the product to its factors.
Now its easy, we count all the factors greater than 1
# of 2s: 7
# of 3s: 2
# of 5s: 1
# of 7s: 1

total # of prime factors: 7+2+1+1 = 11 different prime factors. (Also note that since the question did not ask for DISTINCT or UNIQUE factors, we count all prime numbers, even as the prime numbers repeat themselves.

I hope that helped! Let me know if I am correct!

pathaniaus you gotta stop smoking man

The answer is 4. There are only 4 different prime factors: 2, 3, 5, and 7.

yes, the operant word being "different" primes :p

pathaniaus wrote:

sfreeman wrote:if n is the product of the integers from 1 to 8, inclusive, how many different primer factors greater than 1 does n have?

product of n: 1*2*3*4*5*6*7*8

We can re-write this as:
1*2*3*(2^2)*5*(3*2)*7*(2^3)

See what I did? I reduced everything integer in the product to its factors.
Now its easy, we count all the factors greater than 1
# of 2s: 7
# of 3s: 2
# of 5s: 1
# of 7s: 1

total # of prime factors: 7+2+1+1 = 11 different prime factors. (Also note that since the question did not ask for DISTINCT or UNIQUE factors, we count all prime numbers, even as the prime numbers repeat themselves.

I hope that helped! Let me know if I am correct!

1 is not a prime number

lol... man i got mixed up with the length of prime factors and different prime factors...crap.