There are 45×2000=1600 numbers that are not multiples of 5. 23×34×1600=800 are not multiples of 3 or 4, so 800 numbers are. 800+1=801. Take a good−sized sample of consecutive integers; for example, the first 25 positive integers. Determine that the numbers 3, 4, 6, 8, ...
4×1=44×2=84×3=124×4=16… and so on.Take a look at the multiplication tables (times tables) which represent the first ten multiples of numbers from 1 to 10.So, what are common multiples? Let us try and understand with an example. It is quite simple. Let’s find the common...
1) 3, 4 Multiples of 3 : Multiples of 4 : Common multiples : and 2) 2, 5 Multiples of 2 : Multiples of 5 : Common multiples : and 3) 12, 8 Multiples of 12 : Multiples of 8 : Common multiples : and 4) 4, 6 Multiples of 4 : Multiples of 6 : Common multiples : and 5)...
we get 3, 5, 6 and 9. The sum of these multiples is 23.2--Find the sum of all the multiples of 3 or 5 below 1000.34--set up iterator5functionlist_iter()6locali =1;7localn =100;8returnfunction()9i = i +1;10ifi < nthen11returni;12end13end14end1516localsum =0;17foriinlis...
The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. 我们列举10以下的自然数且是3和5的倍数,可以得到3、5、6、9,它们的和是23。请找出1000以下的所有的3和5的倍数。 最简单的方法: class Untitled { public static int multiplesSum(int n) { int i...
Find the sum of all the multiples of 3 or 5 below 1000. 这个题目提供两种方法来计算。 方法一:采用最笨拙的方式,直接从1到1000,每一个数都去看看是否可以被3或者5整除。Java实现代码如下: 方法一 方法二:利用求和公式,我们知道1000之内被3整数的最大数,应该是999/3=333,所以所有可以被3整除的数的总和...
Multiples of 3 and 5 文章标签Project EulerJAVAJava后端开发 1--If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.2--Find the sum of all the multiples of 3 or 5 below 1000.34--set up iterator5...
How many positive integer factors of 72 are also multiples of 4 A. 2 B. 3 C. 4 D. 5 E. 6 相关知识点: 试题来源: 解析 E72=3^2*2^3=4(3^2*2) . The number of positive integer factors of 72 that are also multiples of 3 is: (2+ 1)(1+1)= 6. ...
4. Write the first five multiples of the following:(i)4(ii)3(ii)7(iv)9(v)5(vi)8(vii) 12(viii) 15 相关知识点: 试题来源: 解析 4.(i)4,8,12,16,20(ii) 3,6,9,12,15(iii)7,14,21,28,35(iv)9,18,27,36,45(v) 5,10,15,20,25(vi)8,16,24,32,40(vii)12,24,36,48,...
Find the sum of all the multiples of3or5below1000. 和各大OJ网站一样,第一题通常都很简短(题目描述也很简短,英语渣渣的我居然看懂了)。 求小于1000,且能被3或5整除的所有数之和。 一个循环+一个分支解决,代码如下: 1#promble 1: Multiples of 3 and 52RANGE = 10003sum =04num = 35whilenum <...