Which of the following is the product of 2 consecutive integers?( ) A.182 B.195 C.208 D.221相关知识点: 试题来源: 解析 A 本题题意为“以下哪个是连续2个整数的乘积?”, A选项:182=13×14; B选项:195=13×15; C选项:208=13×16. 因此只有A选项是连续两个整数相乘. 故选A....
The product of two consecutive positive integers is always .A. odd B. even C. prime D. composite 相关知识点: 试题来源: 解析 B 两个连续的正整数相乘得到的数总是什么数? product 结果,consecutive 连续的,positive integer正整数,odd奇数, even偶数, prime质数, composite合数,选 B....
两个连续正整数的积不能是: 解决这类数字规律题一般情况下应采用排除法,通过找反例的途径来排除错误选项。(A)不对,因为惟一的例外是1×2=2;(B)显然不对,如10与11,11与12等;(C)同理也不对;(D)两个连续整数的积是一个偶数是肯定的,但完全可以大于10,所以不对。(E)找不到反例.两个连续正整数的积的...
Answer to: The product of two consecutive positive integers is five more than their sum. Find the integers. By signing up, you'll get thousands of...
A. −1 B. 0 C. 1 D. 1989 相关知识点: 试题来源: 解析 B The sum of two consecutive integers is always an odd number. 下列哪一个不能表示为两个连续整数的和( )? A.−1 B.0 C.1 D.1989 两个连续正数的和等于一个奇数. 故选B.反馈 收藏 ...
A.Less than 1009 B.1009 C.bigger than 1009 and less than 2018 D.2018相关知识点: 试题来源: 解析 B 本题题意为“两个连续偶数的和是2018,那么他们的平均值为多少?”由题意可得,这两个连续偶数的和为2018,因此平均数为:2018÷2=1009. 故选B.反馈 收藏 ...
which of the following CANNOT be expressed as the product of exactly 2 consecutive integers?A:(2)(3)(7)B:(2)(3)(7)(11)C:(3的平方)(11)(13)D:(2)(5的平方)(13)E:(2的平方)(3)(5)(7)题目就是这样,由于我不会打平方,所以用文字表示.答案是选C, 扫码下载作业帮搜索答疑一搜即得 ...
题目 The product of three consecutive integers must be Ⅰ. Divisible by 3 Ⅱ. Divisible by 5 Ⅲ. Divisible by 6 A.Ⅰ onlyB.Ⅱ onlyC.Ⅰand Ⅱ onlyD.Ⅰ and Ⅲ onlyE.(E) Ⅰ, Ⅱ, and Ⅲ 相关知识点: 试题来源: 解析 D 反馈 收藏 ...
CWrite out the product of five consecutive integers and call their product P . Then P=(n)(n+1)(n+2)(n+3)(n+4) . Note that exactly one of these numbers will be divisible by 5 , and at least one of these numbers will be divisible by 3 and 4. Either two or three of these ...
conjecture.Thus weshallprove THEOREM1. Theproductoftwoormoreconsecutivepositiveintegersis neverapower. Infactweshallproveastrongerresult: THEOREM2. Letk,1,nbeintegerssuchthatk >_3,1>_2andn+k>_p(k), wherep(k)istheleastprimesatisfyingp(k) >_k.Thenthereisaprimep>_kfor whichap#0(mod 1),...