数学题:在AB两点之间等距离安装路灯
- 时间:2020-06-05 20:41:20
- 分类:数学世界
- 阅读:99 次
数学题:在AB两点之间等距离安装路灯,AC点长560米,在C处及AC中心点必须安装一盏路灯,CB点长630米,在CB点中心处必须安装一盏路灯,问至少安装多少路灯?(求算式)
数学题解答:由于“在C处及AC中心点必须安装一盏路灯”,所以路灯的间距必须是280米的因数,因为:
560÷2=280(米)
又由于“在CB点中心处必须安装一盏路灯”,所以路灯的间距还必须是315米的因数。因为:
630÷2=315(米)
求出280和315的最大公因数是:
280=2×2×2×5×7
315=5×7×9
最大公因数是:5×7=35
最少要装的路灯数是:(630+560)÷ 35 + 1 =34+1 = 35(盏)
这个答案是考虑AB两点首尾都装的情况。
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