

想法一
找出11的倍數且百位為2
209.220.231.242.253.264.275.286.297
加七
216.227.238.249.260.271.282.293
加三
219.230.241.252.263.274.285.296
能被八整除的為296,得解為293
想法二
A-7被11整除
則A-7-11-11也會被11整除
A+3被8整除
則A+3-8-8-8-8也會被8整除
代表A-29同時會被8和11整除
同時是8和11的倍數
又百位數為2的數=264
得解為264+29=293
想法一數字多
想法二對小五又有點難
請問有沒有更適合小五
但較快速的做法
謝謝各位大大
※ 編輯: bn11025 (1.172.198.251), 10/30/2015 23:06:44
推
10/30 23:34, , 1F
10/30 23:34, 1F
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10/30 23:37, , 2F
10/30 23:37, 2F
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10/30 23:38, , 3F
10/30 23:38, 3F
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10/30 23:38, , 4F
10/30 23:38, 4F
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10/30 23:38, , 5F
10/30 23:38, 5F
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10/30 23:38, , 6F
10/30 23:38, 6F
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10/30 23:40, , 7F
10/30 23:40, 7F
推
10/31 08:03, , 8F
10/31 08:03, 8F
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10/31 08:56, , 9F
10/31 08:56, 9F
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10/31 08:57, , 10F
10/31 08:57, 10F
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10/31 09:36, , 11F
10/31 09:36, 11F
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10/31 09:38, , 12F
10/31 09:38, 12F
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10/31 09:38, , 13F
10/31 09:38, 13F
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