C*****x
发帖数: 109 | 1 show that for any N>=1,there exists an integer with n digits from {1,2} that
is divisible by 2^n.For example,for n=4,2112 is a four-digit number with
digits from {1,2} that is divisible by 2^4.
谢谢！ | c****l
发帖数: 88 | 2 fix n, there are 2^n such integers, lemma: no any two such integers a=b mod
2^n. lemma=> there is one which is divisible.
proof of lemma, suppose two such integers a=b mod 2^n or a-b|2^n. then a-b=\
sum_{i=1}^n c_i 10^(i-1), where c_i=0,1,-1. if c_1=1 or -1, a-b is odd,
contradict. if c_i=0, use induction for (a-b)/10.
广告帖，包子酬谢
http://www.mitbbs.com/article_t/Mathematics/31181125.html | B****n
发帖数: 11290 | 3 可以考慮用數學歸納法 假設n=k時成立 也就是有一個k位數的x可以整除2^k
那只有兩種情況一種是x可以整除2^(k+1) 那就2*10^k+x可以整除2^(k+1)
或是x不能整除 那10^k+x可以整除2^(k+1)
that
【在 C*****x 的大作中提到】
: show that for any N>=1,there exists an integer with n digits from {1,2} that
: is divisible by 2^n.For example,for n=4,2112 is a four-digit number with
: digits from {1,2} that is divisible by 2^4.
: 谢谢！
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