Therefore, M=S(T+A2003)=S×T+A2003×S, n = (s+a2003) t = s× t+a2003× t.

Because s = t+a1>; T, so M> is ordinary.

Second, because 3×6= 18, 2a×2b=2c.

That is, c=2ab.

( 1) 1.(x+7)(x+9)= x2+ 16x+63；

2.(x- 10)(x+20)=x^2+ 10x-200；

3.(x-3)(x-2)=x^2-5x+6

(I wonder if the title means this)

(2).(x+a)(x+b)=x^2+(a+b)x+ab

(3).(t+ 1\6)(t- 1\7)=t^2+ 1/42t- 1/42

(4) Because ab=36 and A and B are integers.

So ab =1× 36 = 2×18 = 3×12 = 4× 9 = 6× 6.

So m=a+b= 1+36 or 2+ 18 or 3+ 12 or 4+9 or 6+6.

M=37 or 20 or 15 or 13 or 12.

(1) Original number:11x-10; New number: 1 1x- 1.

(2)from( 1 1x- 1)( 1 1x- 10)=( 1 1x- 10)2+405。

Solution: x=5

The original number was 45.