4 is an aliquot part of 12.
4 能12。
This paper studies the numeral divisibility characteristic of round number, and solves the problem in theory about the divisibility characteristic of any round number.
摘要针对论进行了研究,从理论上解决了任意问题。
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I will switch every light bulb that's divisible by two.
我会按下编号2整除的那些灯泡。
I know that because two is divisible into eight.
我知道这个 因为82整除。
2,4,6, and 8 are even numbers and can be divided exactly by 2.
2、4、6、8是偶数,且可2整除。
A quantity of pearls that can appease the sea monsters must be divisible by 7,11, and 13.
平息海怪的珍珠数量必须 7、11、 13 整除。
Being divisible by 7,11, and 13 means that our number must be a multiple of 7,11, and 13.
7、11、 13 整除,说明这个数字必须同时是 7、11、 13 的倍数。
A lot of numbers are divisible by 3.
很多数字可以3整除。
The lowest common denominator, for example, is the smallest number that can be divided evenly into a set of fractions.
例如," 最小公分母(The lowest common denominator)" 是指一组分数整除的最小的数字。
12 is a whole, complete, perfectly divisible number.
12 是一个完整的、完整的、可以完全整除的数字。
This covers every number which divides evenly into 45, once, and only once.
这涵盖了所有 45 整除的数字,一次且仅一次。
This diagonal is also out, as they’re divisible by 11.
这条对角线也不对, 因为它们可以 11 整除。
And, when it's divisible by both 3, and 5. you print fizzbuzz.
并且,当它可以 3 5 整除时,您将打印 fizzbuzz。
You might remember this from elementary school math, but a prime number is only divisible by itself and 1.
你可还记得小学数学中的这个,但质数只它自己 1 整除。
We can eliminate anything that ends in 2 or 5, as they’ll always be divisible by 2 and 5.
我们可以消除任何以 2 或 5 结尾的东西,因为它们总是可以 2 5 整除。
Any one-gem advantage where your starting total is divisible by three will lead to victory by the same logic.
如果您的起始总数可以三整除,任何一颗宝石的优势都会按照相同的逻辑导致胜利。
Well, the slime-lined are in their prime— which is to say they’re prime numbers, divisible only by 1 and themselves.
好吧, 粘液衬里处于质数状态——也就是说它们是质数,只 1 它们自己整除。
Heir 3 will change the status of every third locker, specifically if it's open, she'll close it, but if it's closed, she'll open it.
3号改变整除3的储藏柜的状态,如果储藏柜开着,那就关上,如果储藏柜关着,那就打开。
To simplify this, both of these equations, actually the top one and the bottom one, both sides are divisible by 2n.
为了简化这一点, 这两个方程式,实际上是上面的方程式下面的方程式, 两边都可以 2n 整除。
Even though you don't know how to read the numbers on the chests, you can read which pattern of digits represents a number divisible by 1001.
因此,即使你看不懂宝箱上的古数字,你仍然找出符合1001 整除规律的那一组。
I guess there are more of them that work nicely in base 12 because there are more numbers that divide exactly into 10.
我猜它们中有更多的以 12 为基数工作得很好,因为有更多的数字可以整除 10。
Negative 20 over 84, which is the same thing, they're both divisible by 4, the numerator divided by 4 is negative 5, over 21.
负 20 大于 84, 这是一回事,它们都可以 4 整除, 分子除以 4 是负 5, 大于 21。
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