英语翻译Let N be the second smallest positive integer that is di
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英语翻译
Let N be the second smallest positive integer that is divisible by every positive integer less than7 .What is the sum of the digits of N 求解答T^T
N must be divisible by every positive integer less than7 ,1 or2,3,4,5 and6 .Each number that is divisible by each of these is is a multiple of their least common multiple.LCM(1,2,3,4,5,6,)=60,so each number divisible by these is a multiple of 60.The smallest multiple of 60 is clearly 60,so the second smallest multiple of 60 is 2*60=120 .Therefore,the sum of the digits of N is 1+2+=3
Let N be the second smallest positive integer that is divisible by every positive integer less than7 .What is the sum of the digits of N 求解答T^T
N must be divisible by every positive integer less than7 ,1 or2,3,4,5 and6 .Each number that is divisible by each of these is is a multiple of their least common multiple.LCM(1,2,3,4,5,6,)=60,so each number divisible by these is a multiple of 60.The smallest multiple of 60 is clearly 60,so the second smallest multiple of 60 is 2*60=120 .Therefore,the sum of the digits of N is 1+2+=3
让N是第二小的正整数,每一个正整数整除than7少.什么是数字的总和N吗?
护士必须每一个正整数整除than7少,or2 1、3、4、5 and6.每一个整除数是其中的每一篇都是是一个多的最小公倍式.LCM(1、2、3、4、5、6、)= 60,所以每一个整除数这些是一个多的60例的解除标准.最小的多个60显然是60,所以次之多60是2 * 60 = 120.因此,数字的总和N + 1 + 2 = 3
这是以为大学经验得出来的英语结果,希望能得到采纳,
护士必须每一个正整数整除than7少,or2 1、3、4、5 and6.每一个整除数是其中的每一篇都是是一个多的最小公倍式.LCM(1、2、3、4、5、6、)= 60,所以每一个整除数这些是一个多的60例的解除标准.最小的多个60显然是60,所以次之多60是2 * 60 = 120.因此,数字的总和N + 1 + 2 = 3
这是以为大学经验得出来的英语结果,希望能得到采纳,
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