標題:
數學兩題(關於次方)
4^11213次方的首位數是什麼?? 47^2007次方的最後兩位數是什麼??? 請詳細回答....... 更新: 在不用計算機的情況下解決此問題........ 問題2改為......... 57^2006次方的最後兩位數是什麼??? 更新 2: 用中文回答.........
最佳解答:
4^11213次方的首位數是什麼?? 4^1 = 4 => 4 4^2 = 16 => 1 4^3 = 64 => 6 4^4 = 256 => 2 4^5 = 1024 => 1 4^6 = 4096 => 4 4^7 = 16384 => 1 4^8 = 65536 => 6 4^9 = 262144 => 2 4^10 = 1048576 => 1 ... the pattern repeats every factor of 5 Therefore, 4^11213次方的首位數是6 (11213 mod 5 = 3). 47^2007次方的最後兩位數是什麼??? Since only the last 2 digits are required, we only need the last 2 digits of the product. By using EXCEL, it is found that the last 2 digits repeats itself every factor of 20 (it is also interesting to find that the last digit repeats every factor of 4...):- 1: 1 * 47 = 47 => 47 2: 47 * 47 = 2009 => 09 3: 9 * 47 = 423 => 23 4: 23 * 47 = 1081 => 81 5: 81 * 47 = 3807 => 07 6: 7 * 47 = 329 => 29 7: 29 * 47 = 1363 => 63 8: 63 * 47 = 2961 => 61 9: 61 * 47 = 2867 => 67 10: 67 * 47 = 3149 => 49 11: 49 * 48 = 2303 => 03 12: 3 * 47 = 141 => 41 13: 41 * 47 = 1927 => 27 14: 27 * 47 = 1269 => 69 15: 69 * 47 = 3243 => 43 16: 43 * 47 = 2021 => 21 17: 21 * 47 = 987 => 87 18: 87 * 47 = 4089 => 89 19: 89 * 47 = 4183 => 83 20: 83 * 47 = 3901 => 01 Therefore, 47^2007次方的最後兩位數是63 (2007 mod 20 = 7). 2007-05-13 00:22:47 補充： 唔好意思，冇睇番你的補充﹗ 57^2006次方的最後兩位數是什麼??? 其實原理一樣： 由於只要最後兩位數，我們可把積的最後兩位乘57，直至積的最後兩位等於01，如下： 1: 1 x 57 = 57 = 57 2: 57 x 57 = 3249 = 49 3: 49 x 57 = 2793 = 93 4: 93 x 57 = 5301 = 01 所以，57^2006次方的最後兩位數是01(2006 / 4 的餘數是0)。
其他解答:
數學兩題(關於次方)
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發問:4^11213次方的首位數是什麼?? 47^2007次方的最後兩位數是什麼??? 請詳細回答....... 更新: 在不用計算機的情況下解決此問題........ 問題2改為......... 57^2006次方的最後兩位數是什麼??? 更新 2: 用中文回答.........
最佳解答:
4^11213次方的首位數是什麼?? 4^1 = 4 => 4 4^2 = 16 => 1 4^3 = 64 => 6 4^4 = 256 => 2 4^5 = 1024 => 1 4^6 = 4096 => 4 4^7 = 16384 => 1 4^8 = 65536 => 6 4^9 = 262144 => 2 4^10 = 1048576 => 1 ... the pattern repeats every factor of 5 Therefore, 4^11213次方的首位數是6 (11213 mod 5 = 3). 47^2007次方的最後兩位數是什麼??? Since only the last 2 digits are required, we only need the last 2 digits of the product. By using EXCEL, it is found that the last 2 digits repeats itself every factor of 20 (it is also interesting to find that the last digit repeats every factor of 4...):- 1: 1 * 47 = 47 => 47 2: 47 * 47 = 2009 => 09 3: 9 * 47 = 423 => 23 4: 23 * 47 = 1081 => 81 5: 81 * 47 = 3807 => 07 6: 7 * 47 = 329 => 29 7: 29 * 47 = 1363 => 63 8: 63 * 47 = 2961 => 61 9: 61 * 47 = 2867 => 67 10: 67 * 47 = 3149 => 49 11: 49 * 48 = 2303 => 03 12: 3 * 47 = 141 => 41 13: 41 * 47 = 1927 => 27 14: 27 * 47 = 1269 => 69 15: 69 * 47 = 3243 => 43 16: 43 * 47 = 2021 => 21 17: 21 * 47 = 987 => 87 18: 87 * 47 = 4089 => 89 19: 89 * 47 = 4183 => 83 20: 83 * 47 = 3901 => 01 Therefore, 47^2007次方的最後兩位數是63 (2007 mod 20 = 7). 2007-05-13 00:22:47 補充： 唔好意思，冇睇番你的補充﹗ 57^2006次方的最後兩位數是什麼??? 其實原理一樣： 由於只要最後兩位數，我們可把積的最後兩位乘57，直至積的最後兩位等於01，如下： 1: 1 x 57 = 57 = 57 2: 57 x 57 = 3249 = 49 3: 49 x 57 = 2793 = 93 4: 93 x 57 = 5301 = 01 所以，57^2006次方的最後兩位數是01(2006 / 4 的餘數是0)。
其他解答:
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