- 在線時間
- 205 小時
- 最後登錄
- 22-8-2
- 國民生產力
- 65
- 附加生產力
- 2084
- 貢獻生產力
- 0
- 註冊時間
- 10-2-26
- 閱讀權限
- 10
- 帖子
- 1166
- 主題
- 16
- 精華
- 0
- 積分
- 3315
- UID
- 588306
|
A more complicated algorithm for testing divisibility by 7 uses the fact that 100 ≡ 1, 101 ≡ 3, 102 ≡ 2, 103 ≡ 6, 104 ≡ 4, 105 ≡ 5, 106 ≡ 1, ... (mod 7). Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is).[8]
This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorize the sequence above (132645...), and to add and subtract, but always working with one-digit numbers. |
|