2007-04-03 06:42:56 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
What a load of rubbish answers so far.
(x^n - x) is always divisible by n if n is prime. This is called Fermat's Little Theorem, and it has been known since about 1640.
(x^n - x) is very seldom divisible by n if n is composite. However, it does happen, and in that case n is called a "pseudoprime to base x".
2007-04-03 07:09:55 · answer #1 · answered by Anonymous · 0⤊ 0⤋
No.
Suppose x=3 and n=2
3^2=9, which is not divisible by 2 or n.
2007-04-03 13:52:27 · answer #2 · answered by Bubblez 3 · 0⤊ 0⤋
No. Try x=3, n=4.
â x^n - x = 81 - 3 = 78 â not divisible by 4.
2007-04-03 13:50:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
No. Log x to the (power n) or 10 raised to 2nd power is 1 over 10 to the minus 2 is .001. Such that X to the n times X to the m=X to the plus (m+n); further: 1 over X to the n times X to the m=1 over X to the n minus m (m-n). Note: provided coefficient of the denominator does not equal zero. Division by zero is not undefined.
In other words: when multiplying coefficients we always add Exponents; when dividing coefficients we subtract exponents. Provided that there is no division by zero.
2007-04-03 14:00:19 · answer #4 · answered by Ke Xu Long 4 · 0⤊ 1⤋
there is no reason I can see where x^n - x must be divisible by n, although it is certainly divisible by x (as x is in each term).
how about the counter-example of x=7 and n=3?
2007-04-03 13:47:58 · answer #5 · answered by emptydoubleyou 2 · 0⤊ 1⤋
its always divisible by x , for n i am not sure yet
2007-04-03 13:48:53 · answer #6 · answered by gjmb1960 7 · 0⤊ 1⤋