百度试题 结果1 题目Look at the numbers: 12, 17, 18. Which number is odd? A. 12 B. 17 C. 18 相关知识点: 试题来源: 解析 B。解析:17 不能被 2 整除,所以是奇数。反馈 收藏
E. The sum of two even numbers. 相关知识点: 试题来源: 解析 C Choice C is correct. The sum of an odd number and an even number can be expressed as (2n+1)+(2m), where n and m are integers. (2n+1 must be odd, and 2m must be even.) Their sum is equal to 2n+2m+1, o...
How many five-digit numbers are there with at least 2 digits which are odd numbers? 相关知识点: 试题来源: 解析 74375. 结果一 题目 【题目】问有多少个五位数有至少2个数位是奇数?How many five-digit numbers are there with at least2 digits which are odd numbers? 答案 【解析】74375.相关推荐...
(a) m is an odd number.Which of the numbers below mustbe even, and which must be odd?Write ‘odd’ or ‘even’ undereach one.2mm23m-1(m-1)(m+1)x_'___ⅵ-Ⅰ-4864+1)(b) m is an odd number.Is the number odd, or even, oris it not possible to tell?Tick (_) thecorre...
Why did Gregory Belenky restrict his subjects to odd numbers of sleep hours?A.Because he could acquire more precise information.B.Because he could provide different tests.C.Because he could find out the utmost effects of sleep.D.Because he could observe the relationship between sleep and cognitio...
Answer to: Find four consecutive odd numbers which add to 80. Give the numbers smallest to largest. By signing up, you'll get thousands of...
Which of the following numbers is odd for every integer n? A. 2003n B. n^{2}+2003 C. n^{3} D. n+2044 E. 2n^{2}+2003 相关知识点: 试题来源: 解析 E The numbers 2003n, n^3 and n + 2004 are all even when n is even, and n^2 + 2003 is even when n is odd. But ...
,…). One of the odd numbers has been rubbed out, and the sum of the remaining odd numbers becomes 2018. The odd number that has been rubbed out is . 黑板上写有从1开始的一些连续奇数:1,3,5,7,⋯,擦去其中一个奇数后,剩下的所有奇数的和是2018,那么擦去的奇数是 ....
EDIT: I'm using a lemma: the sum of two squareroots of natural numbers can be rational only when they both are in fact natural numbers (see here) Let a≥ba≥b (due to symmetry we can assume this, then all solutions will be permutations of the ones we find)...
odd numberssums of prime powersIn this paper, we find two integers k(0), m of 159 decimal digits such that if k k(0) (mod m), then none of five consecutive odd numbers k, k - 2, k - 4, k - 6 and k - 8 can be expressed in the form 2(n) +/- p(alpha), where p ...