英文数学题 高分求强人!Starting with any positive integer n, we produce
英文数学题 高分求强人!
Starting with any positive integer n, we produce another number m as follows.
If n is odd, then m = 3n -1
If n is even, then m = n/2
By repeating this process, n generates a sequence called a snowstorm, the numbers of which are called snowflakes of n.
For example the snowstorm of 4 is: 4, 2,1,2,1 . and so on. Sometimes we arrange the snowstorm in a diagram to indicated cycles.
4⇒ 2 ⇒ 1 (and then an arrow goes from the end (1) and back to the (2) to form a cycle.
Here we have a cycle of length 2.
Question: Explain why no cycle can contain a number that is a multiple of three.
Starting with any positive integer n, we produce another number m as follows.
If n is odd, then m = 3n -1
If n is even, then m = n/2
By repeating this process, n generates a sequence called a snowstorm, the numbers of which are called snowflakes of n.
For example the snowstorm of 4 is: 4, 2,1,2,1 . and so on. Sometimes we arrange the snowstorm in a diagram to indicated cycles.
4⇒ 2 ⇒ 1 (and then an arrow goes from the end (1) and back to the (2) to form a cycle.
Here we have a cycle of length 2.
Question: Explain why no cycle can contain a number that is a multiple of three.
赞 莫优非游 幼苗
共回答了24个问题采纳率：87.5% 举报
我们可以分类讨论一下：
如果n是偶数,则m=n/2是偶数,一直循环下去得到的m都是更小的偶数,最后直到2,即得到了例子中的循环；
如果n是奇数,则m=3n-1又是偶数,偶数之后的下一个m还是更小的偶数,最后直到2又即得到了例子中的循环.（不理解,
if n is odd,then m=3n-1,so m0 is even.then the next m1=m0/2 ,m1 is even too and smaller then m0,so do this until we get the smallest even positive integer 2,then the next m equals 1,so we get the circle as showed above.
if n is even ,then m=n/2,so m1 is even ,then the next m2=m1/2,m1 is even too,then do the circle until we get the smallest even positive integer 2,then the next m equals 1,so we get the circle as showed above.
so no cycle can contain a number that is a multiple of three.
如果n是偶数,则m=n/2是偶数,一直循环下去得到的m都是更小的偶数,最后直到2,即得到了例子中的循环；
如果n是奇数,则m=3n-1又是偶数,偶数之后的下一个m还是更小的偶数,最后直到2又即得到了例子中的循环.（不理解,
if n is odd,then m=3n-1,so m0 is even.then the next m1=m0/2 ,m1 is even too and smaller then m0,so do this until we get the smallest even positive integer 2,then the next m equals 1,so we get the circle as showed above.
if n is even ,then m=n/2,so m1 is even ,then the next m2=m1/2,m1 is even too,then do the circle until we get the smallest even positive integer 2,then the next m equals 1,so we get the circle as showed above.
so no cycle can contain a number that is a multiple of three.
1年前
1
nongmin2007 幼苗
共回答了19个问题采纳率：84.2% 举报
....题目很多。
第一个题：k是一个正整数，找到最小的正整数k使得方程x^2 kx=4y^2-4y 1有整数解。
这个题比较轻松：
两边因式分x(x k)=(2y-1)^2 注意到右边是奇数的完全平方数，所以左边的x和x k都是奇数，并且都是完全平方数，或者是完全平方*a，于是k=(x k)-x至少是3^2-1^2=8。
第二题：在一个有p个0和q个1的数列里，...
第一个题：k是一个正整数，找到最小的正整数k使得方程x^2 kx=4y^2-4y 1有整数解。
这个题比较轻松：
两边因式分x(x k)=(2y-1)^2 注意到右边是奇数的完全平方数，所以左边的x和x k都是奇数，并且都是完全平方数，或者是完全平方*a，于是k=(x k)-x至少是3^2-1^2=8。
第二题：在一个有p个0和q个1的数列里，...
1年前
2
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