public static ArrayList<Integer> prime(int num){ ArrayList<Integer> primes = new ArrayList<>(); ArrayList<Integer> nul = new ArrayList<Integer>(); nul.add(-1); for(int i = 1; i < num; i++) { for(int j = 2; j < i - 1; j++){ if (i % j == 0) { return nul; } p...
89.-|||-Crossing off numbers that are not 1 more than a multiple of 4 (in other words,numbers that are 1 less than a multiple of 4,since all numbers are-|||-odd),we get:-|||-29,49,89.-|||-Noting that 49 is not prime,we have only 29 and 89,which give a sum of 118,...
回答:匿名 p1,牳九旧厦挥械总理的一组数字,其单位的所有数字为1 2013-05-23 12:23:18 回答:匿名P1,表示单位数字是1的套所有质数 2013-05-23 12:24:58 回答:匿名P1,用以表明所有素数的单位数字是 1 套 2013-05-23 12:26:38 回答:匿名P1,表示一套所有素数其单位数字是 1 2013-05-23 12:28:18 ...
Arrange all the prime numbers in the order from little to great,counting from least one,the N-th prime number is 53,then N=__ 相关知识点: 试题来源: 解析 大意是安排所有的质数顺序由小大、计数,他们从最小的一个,最后一个质数是53,n是几?排一下就知道了∴n=16...
1.Arrange all the prime numbers in the order from little to great,counting from least one,the N-th prime number is 53, then N=__2.For each pair of integer number A and B,define the operation*as A*B=AB+1,then(2*4)*2=___3.已知...
选C~20以下的质数有2.3.5.7.11.13.17.19他们乘积为9699690最接近10的7次方
A If the product is divided by 210, the remainder is 0 since 210×2×3×5×7= product of the first 4 crimes. 如果所有素数的乘积在 1 和210 之间,那么这个数除以 21 余数是多少( )? A.0 B.3 C.7 D.21 如果乘积除以210,则余数为0,因为210×2×3×5×7=前4个素数的乘积. 故选A....
百度试题 结果1 题目The product of all prime numbers between 1 and 10 is( ). A. 210 B. 105 C. 1890 D. none of the above相关知识点: 试题来源: 解析 A
一道英文数学题.Arrange all the prime numbers in the order from little to great,counting from the least one,the n-th prime number is 53.n=? 答案 素数啊~~一个一个列吧~~~相关推荐 1一道英文数学题.Arrange all the prime numbers in the order from little to great,counting from the least on...
2. Please list all the prime numbers in the interval [1, 1000]. A. a=1:1000; b=a(primes(a)) B. a=1:1000; b=b(primes(a)) C. a=1:1000; b=a(isprime(b)) D. a=1:1000; b=a(isprime(a)) 相关知识点: 试题来源: 解析 D ...