c**********e
发帖数: 2007 | 1 How to solve this problem?
7. Three nine digits number e,f,and g. If we replace f_i with e_i they the
new nine digites can be divided by 7, for all i. Same happens if we replace
g_i with f_i,prove that g_i-e_i==0 mod 7 for all i. | l*********t
发帖数: 89 | 2 同问。有没有牛人出来指点一下啊？
即使没有最终答案，说说思路也好，大家一起讨论。多谢了！
replace
【在 c**********e 的大作中提到】
: How to solve this problem?
: 7. Three nine digits number e,f,and g. If we replace f_i with e_i they the
: new nine digites can be divided by 7, for all i. Same happens if we replace
: g_i with f_i,prove that g_i-e_i==0 mod 7 for all i.
| w******g
发帖数: 313 | 3 这不显然么。。考虑f的第i位，和e的第i位交换以后f能被7整除，和j的第i位交换以后
f能被7整除，所以(e_i-g_i)*10e7是7的倍数，for all i | x******a
发帖数: 6336 | 4 e_i+\sum_{ j \ne i} f_i == 0 mod 7 for all i=1, 2, ..., 9,
f_i +\sum_{j \ne i} g_i == 0 mod 7 for all i=1, 2, ..., 9.
=> (sum up)
\sum e_i + \sum f_i == 0 mod 7.
\sum f_i + \sum g_i == 0 mod 7.
e_i== -\sum f_j + f_i == \sum g_j + f_i == g_i mod 7. | c**********e
发帖数: 2007 | 5 The original problem says "Same happens if we replace g_i with f_i". I think
it means the new G is divisable by 7.
I guess you misunderstood the original problem. Otherwise the problem was
too easy.
【在 w******g 的大作中提到】
: 这不显然么。。考虑f的第i位，和e的第i位交换以后f能被7整除，和j的第i位交换以后
: f能被7整除，所以(e_i-g_i)*10e7是7的倍数，for all i
| x******a
发帖数: 6336 | 6 e_i10^i + \sum_{j \ne i}f_j10^j==0 mod 7. ------1
f_i10^i + \sum_{j \ne i}g_j10^j==0 mod 7. ------2
==> (sum up 1 and 2 separately)
\sum e_i10^i +\sum f_i 10^i==0 mod 7 ------------3
\sum f_i10^i +\sum g_i 10^i==0 mod 7 ------------4
==>
e_i10^i== -\sum f_j 10^j +f_i 10^i (b/c 1)
==\sum g_j 10^j + f_i10^i (b/c 4)
==g_i10^i mod 7. (b/c 2)
It follows 7|e_i-g_i for each i since 7 is prime and 7 doesn't divide any 10
^i.
replace
【在 c**********e 的大作中提到】
: How to solve this problem?
: 7. Three nine digits number e,f,and g. If we replace f_i with e_i they the
: new nine digites can be divided by 7, for all i. Same happens if we replace
: g_i with f_i,prove that g_i-e_i==0 mod 7 for all i.
|
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