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?
分析A选项
分析B选项
分析C选项
分析D选项
分析E选项
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。
爱琴海底火山
n/3为integers,3000<=n=<3333,1000<=n/3=<1111,1111-1000+1=112
学习才能配白敬亭
n/3是整数!!!隐含条件,救命这都没想到。。。。。
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
lele要750
数学题怎么这么多小细节…要疯了
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.
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
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
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
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。
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
William(GMAT 750+)
注意必须得是3的倍数
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。
clark
n在3000-3333之间,inclusively,但需要是3的倍数
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