有關一個數字能否被3除的疑問

如果一個數字 N = AnAn-1An-2 ... A3A2A1A0
注意 Ax 是該數位之數字

如 A0 指個位數字

咁 N 的數值為:
N
= An x 10^n + An-1 x 10^(n-1) + ... + A1 x 10 + A0
= An x (1+9)^n + An-1 x (1+9)^(n-1) + ... + A1 x (1+9) + A0

考慮 (1+9)^n
= 1 + nC1 (9) + nC2 (9^2) + ...
= 1 + 9M

所以 N
= (An + An-1 + ... + A2 + A1 + A0) + 9M

由於 9M 必可給 3 整除，
所以只須考慮 (An + An-1 + ... + A2 + A1 + A0) 可否給 3 整除便行了

I don't know that
I only know some things about it.

If the sum like this365=3+5+6=14,14/3=4...2
All the remainder that (sum/3=x...1)=y.333...
All the remainder that (sum/3=z...2)=a.666...

Let a 4 digit number be abcd, so its value
= 1000a + 100b + 10c + d
= 999a + 99b + 9c + ( a + b + c + d).
999a is divisible by 3,
99b is divisible by 3,
9c is divisble by c and
a + b + c + d is divisible by 3.
So the number abcd is divisible by 3.
The same reasoning can apply to number with any digits.




2009-03-01 21:46:56 補充：
Using the same reasoning, if the sum of digits is divisible by 9, the number is divisible by 9

2009-03-01 21:48:46 補充：
Correction: 9c is divisible by 3, not only c.