For an arbitrary natural number n,we add 121to it when it is an odd number and have it divided by 2(接上)when it is an even number.This is one operation.Now if we perform this operation on 231 successively,whether 100 can appear or not in this pr

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For an arbitrary natural number n,we add 121to it when it is an odd number and have it divided by 2(接上)when it is an even number.This is one operation.Now if we perform this operation on 231 successively,whether 100 can appear or not in this pr

For an arbitrary natural number n,we add 121to it when it is an odd number and have it divided by 2(接上)when it is an even number.This is one operation.Now if we perform this operation on 231 successively,whether 100 can appear or not in this pr
For an arbitrary natural number n,we add 121to it when it is an odd number and have it divided by 2
(接上)when it is an even number.This is one operation.Now if we perform this operation on 231 successively,whether 100 can appear or not in this process?Why?(请翻译且讲解)

For an arbitrary natural number n,we add 121to it when it is an odd number and have it divided by 2(接上)when it is an even number.This is one operation.Now if we perform this operation on 231 successively,whether 100 can appear or not in this pr
对于任意自然数n,当它是奇数的时候,把它加上121,当它是偶数的时候,把它除以2.这是一种运算.当我们这样连续运算231次后,在这过程中,100是否会出现?为什么?
当然会出现,比如:100*2^231这个数就是啊