Bunuel wrote:
zisis wrote:
I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....
so -
\(5! = 120\)
\(5!+3 / 3 = 123 / 3 =41\)correct
\(5!+4 / 4 = 124 / 4 =31\) correct
\(5!+5 / 5 = 125 / 5 =25\) correct
\(5!+6 / 6 = 126 / 6 =21\)correct
\(5!+7 / 7 = 127 / 7 =18.14\)INCORRECT !!!!
so...? what am i doing wrong???
You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here.
did it on excel and seems like you are right.....
15! 15!+x (15!+x)/x
1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00
1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00
1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00
1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00
1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00
1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00
now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b
?
It seems that you don't understand the explanation. No need to check in excel: no integer \(x\) satisfying \(15!+2\leq{x}\leq{15!+15}\) will be a prime number.
If \(x=15!+2\) then we can factor out 2, so \(x\) would be multiple of 2, thus not a prime;
If \(x=15!+3\) then we can factor out 3, so \(x\) would be multiple of 3, thus not a prime;
...
If \(x=15!+15\) then we can factor out 15, so \(x\) would be multiple of 15, thus not a prime.
also won't be a prime number as we can factor out this prime (15! has all primes less than or equal to 13).
Now, for \(x=15!+1\) or \(x=15!+17\) or \(x=15!+19\) we can not say for sure whether they are primes or not. In fact they are such a huge numbers that without a computer it's very hard and time consuming to varify their primality.