Question

英文数学问题！

How many four-digit numbers containing no zeros have the property that whenever any one of its four digits is removed , the resulting three-digit number is divisible by 3?
谢谢

Answer 1

One thing to remember is that a number is divisible by 3 if the sum of all its digits is divisible by 3. Since 0 is not allowed and removing any one of the 4 digits results in a number divisible by 3, there're types of possibilities:
1. all 4 digits are the same, there're 9 possibilities: 1111, 2222, ..., 9999
2. all digits are 3, 6, 9:
(a) the 4 digits are 3, 3, 3, 6: 4 possibilities (take 333, then insert 6 to it)
(b) the 4 digits are 3, 3, 3, 9: 4 possibilities (take 333, then insert 9 to it)
(c) the 4 digits are 6, 6, 6, 3: 4 possibilities
(d) the 4 digits are 6, 6, 6, 9: 4 possibilities
(e) the 4 digits are 9, 9, 9, 3: 4 possibilities
(f) the 4 digits are 9, 9, 9, 6: 4 possibilities
(g) the 4 digits are 3, 3, 6, 9: P₄/2 = 4/2 = 12 possibilities
(h) the 4 digits are 3, 6, 6, 9: P₄/2 = 4/2 = 12 possibilities
(i) the 4 digits are 3, 6, 9, 9: P₄/2 = 4/2 = 12 possibilities
(j) the 4 digits are 3, 3, 6, 6: P₄/(2*2) = 6 possibilities (3366, 3636, 3663, 6336, 6363, 6633)
(k) the 4 digits are 3, 3, 9, 9: similar to (j), 6 possibilities
(l) the 4 digits are 6, 6, 9, 9: similar to (j), 6 possibilities

Total: 9 + 4*6 + 12*3 + 6*3 = 87

Answer 2

198种（四位数中有零）一个数能不能被3整除，就要看其各位数字的和能否被3整除。又因为是随意去掉一个数字，所以可能的数字为0,3,6,9四个，（0不能为开头）。。。
87种（其中无零）。。。英语，我翻译的不太准，，，