题库搜索>问题求解PS-17864
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How many positive integers n have the property that both 3n and n/3 are 4-digit integers?
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首先从3n 和n/3共同的Min max的交集在999-10000的范围内（包含，得出为3000-3333是n的范围，进而，寻找到3的倍数，随便找0-10；10-20 差不多是1/3的概率为3的倍数，所以3333-3000=333 333/3=111 但是因为这个范围是包含了3000所以，最后需要再加1。
680461elv
漏掉条件了！n/3也必须是整数！所以得到范围后还要除以3
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2022-06-14 18:51:59
lele要750
数学题怎么这么多小细节…要疯了
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2022-05-13 12:32:52
lele要750回复lele要750
md我又错了
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2022-07-03 18:53:08
221555tvgl
第一步判断出：1000<= n/3 ; 3n<=9999;可得 3000<= n <=3333 ;第二步就是把这里面能整除3的数挑出来，那这些数里面有多少个呢？有一个等差数列公式，可以用来算个数：(最后一个能被N整除的数-第一个能被N整除的数)/N +1,就是这个区间有多少数是这个数的倍数，所以 （3333-3000）/3 +1 =111+1=112.
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2021-12-04 16:54:18
221555tvgl回复221555tvgl
用这个公式还可用来算 99-301之间有多少个偶数（我总是算不清），（300-100）/2 +1=100+1=101
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2021-12-04 16:56:33
屠G旦
必须是3的倍数 没看到选了334
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2021-10-20 17:18:54
lavin.Z
眼瞎没看到得整除（both are digit integers），还考虑到了3000，美滋滋的选了334。。。给我整无语了
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2021-10-14 19:02:42
贾思敏
pace慢的题6反思：算范围那里有点卡壳 因为3n是 4-digits integers， 所以3n的取值范围是【1000,9999】，n最大能取3333 又因为n/3也是四位数，n/3得大于等于1000, 所以n大于等于3000 综上，n的取值范围是【3000,3333】 还有一个条件是要满足n/3的话，那n必须是3的倍数，而一组连续的整数中每3个就会有1个是3的倍数，所以（3333-3000）/3=111，最后还有一点很阴险的地方就是3000也满足条件，所以111+1=112
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2021-10-07 12:36:04
162466yocym
这道题我有个简单的思路，3n和n/3差了九倍，且都为四位数。所以n/3最小取1000，最大取1111，因为1111的9倍是最大的四位数。所以从1000-1111有112个数字，选B。
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2021-07-12 23:48:26
734893yczj回复162466yocym
好思路。
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2022-01-04 13:17:54
626617yh
太阴了吧，必须能整除
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2021-03-27 23:24:01
William(GMAT 750+)
注意必须得是3的倍数
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2021-03-06 17:45:23
Go800
先算出来n的范围为3000<=n<=3333。这里有个简单方法：试想0到333中，有多少个可以整除3的，就用333/3=111，再加上300这个数，为112。
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2021-02-20 17:58:25
clark
n在3000-3333之间，inclusively，但需要是3的倍数
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2019-10-19 14:05:48
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