1, in 64- 127, can be expressed as: 2 6+2 5+2 4+2 3+2 2 1+2 0 (* * has 7 items), and the coefficient of each item is 65433. The numbers with 5-0 terms and a coefficient of 0 are C6 and 5 respectively; C6，4； C6，3； C6，2； C6， 1； C6，0 .
2. In 32-63, it can be expressed as: 2 5+24+23+22+2 1+20 (* *), and the coefficient of each item is1or 0, except "25". C5，4； C5，3； C5，2； C5， 1； C5，0 .
3. Similarly, in 16-3 1, the coefficients of 4-0 are C4 and 4 respectively; And 4; C4，3； C4，2； C4， 1； C4，0 .。
In 8- 15, the coefficient of 3-0 is 0: C3, 3; C3，2； C3， 1； C3，0 .。
In 4-7, the coefficient of 2-0 is 0: C2, 2; C2， 1； C2，0 .
In 2-3, the coefficient of 1-0 is 0: C 1,1; C 1，0 .
In 1, the term with a coefficient of 0 is 0.
In combination with the above,
The number of six numbers with coefficient 0 is C6, 6 =1;
The number of five numbers with a coefficient of 0 is C6, 5+C5, and 5 = 7;
The number of four numbers with coefficient 0 is C6, 4+C5, 4+C4, 4 = 21;
The number of three numbers with coefficient 0 is C6, 3+C5, 3+C4, 3+C3, 3 = 35;
The number of numbers with binomial coefficient of 0 is C6, 2+C5, 2+C4, 2+C3, 2+C2, 2 = 35;
The number of the term 1 with a coefficient of 0 is C6, 1+C5, 1+C4, 1+C3, 1+C2, 1+C 1,/kloc.
The number of numbers with a coefficient of 0 is C6, 0+C5, 0+C4, 0+C3, 0+C2, 0+C 1, 0+C0, 0 = 7.
The sum of the original problems is: 26x1+25x7+24x21+23x35+22x35+21x 21+20x7 = 64+224+336+24.