Puggy, Whats the smallest prime factor of 100! + 1?

Hi everyone,





If c (vector b)- (vector a) = -3/4d (vector a) + 1/4d (vector b) , where


c and d are constant unknowns , can you conclude that





d=4/3 and c=1/3 through direct comparison of the coefficients of vector


a on the right hand side and the coefficients of





vector a on the left hand side to find d and then do the same for the


coefficients of vector b?





Is there any rule for using direct comparison? I feel that the vectors


must intersect before you can use direct comparison





because if you use the i, j , k in column form to replace vectors b and


a, they must intersect in order for all three levels of the





right and left hand side to have the same c and d. I'm getting confused


so anyone has any explanation?





I usually move all vector b to one side and all vector a to one side and


use the rule that if (c-1/4d)(vector b)= (1-3/4d)(vector





a), since vector a and vector b are not parallel, for vector b to be


equal to vector c, c-1/4d=0 and 1-3/4d=0.

Puggy, Whats the smallest prime factor of 100! + 1?
I hope you don't mind, I am just interested in the initial question. The answer is 101. If you work mod 101, then 100 = -1 So to prove that 100! +1 = 0, you need to prove that 99! = 1. For that it is enough to show that the numbers between 2 and 99 can be put in 49 pairs (p,q) such that pq=1. But Z/101Z is a field and this reduces to showing that no number is such that p^2=1. This can be checked easily because 102,203, 304 up to 2425 are not squares and that's it.
Reply:I only can tell that it is greater than 97 because the given number is not divided by any prime factor minor or equal to 97


a) As you write that vectors a and b are not paralel


the last equations you wrote are true becaus if they are NOT zero that means that both vector are paralel

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