10a/ 10b=a/b=3...9
a+b+3+9= 10 ( 1)
a=3b+9 (2)
Bring (2) into (1): 3b+9+b+12 =10b =-4/1a = 3b+9 = 73/9.
2. It is to find the allocation method that the sum of one or more numbers can reach 1, 2,3. . . Every number in 3 1.
The first bag contains 1, so the second bag contains 1 or 2. The classification analysis shows that:
(1). If the second bag contains 1, the third bag can only contain 3; 1+ 1+3=5, and the fourth bag can only hold 6;
1+1+3+6 =11,the fifth bag can only hold 12, while1+1+3+6+1.
(2). If there are two in the second bag, only four in the third bag and only seven in the fourth bag;
1+2+4+7= 14, so the fifth bag can only hold 15, and 1+2+4+7+ 15=29. Therefore excluded;
So the proposition doesn't hold! !