F.4 maths~~~15marks

1. Let the first number be x, where x is integer
So, the second number is x+1;
third number is x + 2
Then, sum = x + (x + 1) + (x + 2) = 3x + 3 = 3(x+1)
which is multiple of 3, where the quotient is x+1

2. Let the first number be 2x , where x is an integer.
So, the second, third, and fourth number should be:
2x+2, 2x+4, 2x+6
sum = 2x + (2x+2) + (2x+4) + (2x+6) = 8x + 12
Assume sum = 8x + 12 = 56,
then 8x = 44
x = 5.5 which is non-integer
But x is integer.
So, sum can not be 56.
3. Let the first number be 2x+1 , where x is integer
then, the second and third number should be 2x+3 and 2x+5 respectively.
sum = (2x+1) + (2x+3) + (2x+5) = 6x+9 = 3(2x+3)
which is multiple of 3.

2009-08-21 23:59:37 補充：
For question 2), we have applied the set theory:
If P is true, then Q is true.
In other words,
If Q is false, then P is false.

That is, the statement P -> Q
is equivalent to the statement
~Q -> ~P

2009-08-21 23:59:47 補充：
P refers to the statement :
x is integer

Q refers to the statement :
sum of four consecutive even numbers = 8x + 12 != 56

That is, prove that if P is true, then Q is true.

2009-08-21 23:59:52 補充：
To prove that 8x+12 != 56 if x is integer,
we just need to show that x is non-integer if 8x + 12 = 56
That is, show that ~P is true is ~Q is true.

2009-08-22 00:01:45 補充：
Sorry... Typing mistake:

referring to 2009-08-22 00:00:02 補充 :
To prove that 8x+12 != 56 if x is integer,
we just need to show that x is non-integer if 8x + 12 = 56
That is, show that ~P is true is ~Q is true.

The last statement should be:
That is, show that ~P is true if ~Q is true.