Related to least common multiple:least common denominator,greatest common factor least common multiple n.Abbr.lcm The smallest quantity that is divisible by two or more given quantities without a remainder:12 is the least common multiple of 2, 3, 4, and 6.Also calledlowest common multiple. ...
Related to least common multiples:lowest common multiple least common multiple n.Abbr.lcm The smallest quantity that is divisible by two or more given quantities without a remainder:12 is the least common multiple of 2, 3, 4, and 6.Also calledlowest common multiple. ...
least common multiple ▾ 英语-中文正在建设中 leastcommonmultiple 最小公倍数() 也可见: least— 至少形 · 最少形 · 最少的形 · 最差 · 丝毫的形 · (程度)最小地副 · 微乎其微的形 · (某一特性)最少副 公倍数 multiple形—
What is the least common multiple of 84 and 112( ). A. 28 B. 196 C. 336 D. 9408 相关知识点: 试题来源: 解析 C 本题题意为“84和112的最小公倍数是多少?”,本题考查因数分解,84与112的最小公倍数为2×2×7×3×4=336. 故选C....
least common multiple 英文least common multiple 中文最小公倍数 【计】 最小公倍数
My code below works for small ranges only, not something like [1,13] (which is the range 1,2,3,4,5,6,7,8,9,10,11,12,13), which causes a stack overflow. How can I efficiently find the least common multiple of a range? function leastCommonMultiple(arr) { var minn, max; ...
The multiples of 2 are 2, 4, 6, etc. The multiples of 4 are 4, 8, 12, etc. The smallest multiple they have in common is the 4. So 4 is the LCM. What is the LCM of 12 and 8? Listing the multiples of each number will work for these two small numbers. The 12 has ...
Least Common Multiple (最小公倍数,先除再乘) 思路: 求第一个和第二个元素的最小公倍数,然后拿求得的最小公倍数和第三个元素求最小公倍数,继续下去,直到没有元素 注意:通过最大公约数求最小公倍数的时候,先除再乘,避免溢出 1#include <iostream>2#include <cmath>3#include <cstdio>4#include <...
the least common multiple of 青云英语翻译 请在下面的文本框内输入文字,然后点击开始翻译按钮进行翻译,如果您看不到结果,请重新翻译! 翻译结果1翻译结果2翻译结果3翻译结果4翻译结果5 翻译结果1复制译文编辑译文朗读译文返回顶部 的最小公倍数 翻译结果2复制译文编辑译文朗读译文返回顶部...
The Least Common Multiple of 12, 8 and 10 is 120 because 120 is the smallest number that is a multiple of all 3 numbers. Simply type in your numbers into the space below with a SPACE after each one and the calculator will do the rest!