If the given number is,say N, and if its prime factors are k1,k2,k3.... raised to powers r1,r2,r3 respectively. N can be written as,
N = (k1^r1)*(k2^r2)*(k3^r3)......
Now can anybody off you give me the formula for above mentioned question ?
2007-03-09 16:32:42 · 4 answers · asked by megh 1 in Science & Mathematics ➔ Mathematics
hi bh8153.The formula you gave is for Number factors of a number. Read my question carefully.
I had derived a formula for this,but now i dont remeber what i had done
2007-03-11 04:34:47 · update #1
Any factor of N is (k1^e1)*(k2^e2)*(k3^e3)... where 0 <= e1 <= r1, 0 <= e2 <= r2, etc. So the total number of factors is f = (r1+1)*(r2+1)*(r3+1)... Every 2 factors multiply to N, so the product of all the factors is N^(f/2).
If all the r's are even, then N is a perfect square, and f/2 comes out to be "half-integral" because the square root of N is a factor only once, not twice.
2007-03-10 04:21:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
That's the formula, as it were -- the prime factorization.
2007-03-10 00:41:06 · answer #2 · answered by Curt Monash 7 · 0⤊ 1⤋
buddy, there isn't even a formula to tell you all the factors of a given number...how can we have one to tell you the product of them all?
pretty sure you have to figure it out manually.
2007-03-10 00:42:28 · answer #3 · answered by sage and wise 3 · 0⤊ 1⤋
The formula for what?
2007-03-10 00:38:26 · answer #4 · answered by AnyMouse 3 · 0⤊ 1⤋