960可以分解为2的6次方乘以3和5,这使得960可以分割成以下宽度的整数倍：（下面的整数倍的这些数值是怎么来的?不太理解）2,3,4,5,6,8,10,12,15,16,20,24,30,32,40,48,60,64,80,96,120,160,192,240,320,480共26种（

Question

960可以分解为2的6次方乘以3和5,这使得960可以分割成以下宽度的整数倍：（下面的整数倍的这些数值是怎么来的?不太理解）
2,3,4,5,6,8,10,12,15,16,20,24,30,32,40,
48,60,64,80,96,120,160,192,240,320,480
共26种（26 = 7 * 2 * 2 - 2,减去2是去掉1和960自身）,我们标记为：
N(960) = N(2^6 * 3 * 5) = 26
同理可以得到：
N(480) = N(2^5 * 3 * 5) = 6 * 2 * 2 - 2 = 22
N(750) = N(2 * 3 * 5^3) = 2 * 2 * 4 - 2 = 14
N(800) = N(2^5 * 5^2) = 6 * 3 - 2 = 16
N(1000) = N(2^3 * 5^3) = 4 * 4 - 2 = 14
N(1024) = N(2^10) = 11 - 2 = 9
N(1920) = N(2^7 * 3 * 5) = 8 * 2 * 2 - 2 = 30
证明整数倍的关系公式

Answer 1

我这样写你能明白么?
N(960) = N(2^6 * 3 * 5) = 7 * 2 * 2 - 2=（6+1）*2*(1+1）=26
N(480) = N(2^5 * 3 * 5) = = 7 * 2 * 2 - 2= = 7 * 2 * (1+1)- 2 = 22
N(750) = N(2 * 3 * 5^3) = 2 * 2 * 4 - 2 =(1+1)* 2 *（3+1)=14
N(800) = N(2^5 * 5^2) = 6 * 3 - 2 = (5+1)*（2+1）-2=16
N(1000) = N(2^3 * 5^3) = 4 * 4 - 2 =(3+1）*（3+1)= 14
N(1024) = N(2^10) = 11 - 2=（10+1)-2 = 9
N(1920) = N(2^7 * 3 * 5) = 8 * 2 * 2 - 2 = （7+1）*2*（1+1)=30