The coordinates of point P are (1, - 1), and point q is a point on the straight line x + y-4 = 0. When vector PQ is perpendicular to vector a = (1. - 1), module PQ|=

The coordinates of point P are (1, - 1), and point q is a point on the straight line x + y-4 = 0. When vector PQ is perpendicular to vector a = (1. - 1), module PQ|=

Point q is a point on the straight line x + y-4 = 0, so let the coordinates of point Q be: (x, 4-x),
Vector PQ = (x-1,4-x + 1) = (x-1,5-x),
Vector PQ is perpendicular to vector a = (1, - 1), so
1*(x-1)-1*(5-x)=0,
x=3,
So the vector PQ = (2,2),
|PQ|=√(2^2+2^2)=2√2.

If P and P + 2 are prime numbers, find the remainder of P divided by 3 (P > 3)
Process!!!

P and P + 2 are prime numbers,
The remainder of any number divided by 3 is only 0, 1 and 2
If the remainder = 0,
Then p is a multiple of 3,
Because P is prime, P can only be 3
If the remainder = 1,
Then p + 2 is a multiple of 3, and P + 2 is a prime number, it can only be 3
But p can't be 1, it doesn't hold
If the remainder = 2,
The remainder of P + 2 divided by 3 is 1
So there are two answers:
If P = 3, the remainder = 0
If P is not 3, the remainder is 2

If P and Q are prime numbers and the root of the equation PX + 5q = 97 is 1, then p2-q=______ ．

Because 1 is the root of the equation PX + 5q = 97, one of P + 5q = 97, P and 5q must be odd and the other even. If P is odd and 5q is even, only Q can be even prime 2. In this case, P = 97-5 × 2 = 87 = 3 × 29, which is inconsistent with the condition that P is prime. Therefore, only P can be even prime 2, 5q = 95, q = 19.. p2-q = 4-19 = - 15, so the answer is: - 15

If we define P ￣ q = 5p + 3q, P ￣ q = 3p-5q, then (3 ￣ 23) ￣ 0.6=______ ．

(3 · 23) [0.6, = (3 × 3-5 × 23) [0.6, = (9-103) [0.6, = 173 [0.6, = 5 × 173 + 3 × 0.6, = 853 + 95, = 45215; so the answer is: 45215