UC知道数学 二进制 ！！
来源：百度知道 编辑：UC知道 时间：2024/06/14 01:26:01
数学 二进制
Let k be a positive integer. Show that the number of powers of 2 that have k digits in decimal notation is at least three and at most four. Given that the largest power of 2 which is less than 10^2009 is 2^6673 , for how many k between 1 and 2009 inclusive, are there four powers of 2 that have k digits?
Let k be a positive integer. Show that the number of powers of 2 that have k digits in decimal notation is at least three and at most four. Given that the largest power of 2 which is less than 10^2009 is 2^6673 , for how many k between 1 and 2009 inclusive, are there four powers of 2 that have k digits?
二进制的1011，转化为十进制
=1×2^0+1×2^1+0×2^2+1×2^3
=1+2+0+8
=11
1011从左到右，权依次是2的0次方，2的1次方，2的2次方，2的3次方。系数依次是1,1,0,1
个人翻译，希望没错：
令k为正整数，试证明：位数为k（十进制）且为2的整数幂的数最少有3个，最多有4个。若其中最大的2的整数幂小于10^2009 即2^6673 ，对于1和2009之间的k，有没有4个k位数的2的整数幂？
您这个题貌似不是二进制那么简单了，应该涉及数论的知识。当然，第二个问题，由于是有限数，可以用编程来解决。