题目 How many two-digit multiples of 10 are also multiples of 12?A.4B.3C.2D.1 答案 D有多少个两位数是10的倍数也是12的倍数?[10,12]=60.故在两位数以内60的倍数只有60这一个.故选D.相关推荐 1How many two-digit multiples of 10 are also multiples of 12?A.4B.3C.2D.1 反馈 收藏 ...
百度试题 结果1 题目有多少个是的倍数的三位数? How many three-digit numbers, which are multiples of are there? 相关知识点: 试题来源: 解析 . .反馈 收藏
How many 4-digit multiples of 5 are there? How many 3-digit numbers can be written using the digits 0,2,2,4, and 6? How many perfect numbers are there? What is the value of the digit 3 in the number 3,725.89? How many 6-digit numbers can be made out of the digits 1, 2, ...
百度试题 结果1 题目How many -digit numbers whose sum of digit(s) is/ are or multiples of is/are there? 有多少个三位数的数位和是或的倍数? 相关知识点: 试题来源: 解析 . .反馈 收藏
How many of these three-digit numbers are multiples of4?从数字1,2,3,4,5,6,7,8,9中任取3个数组成三位数,所组成的数中,是4的倍数的三位数有___个. 相关知识点: 试题来源: 解析 133 要组成能被4整除的三位数,其后两位必须能被4整除。从给定的数字中,可以组成21个能被4整除的两位数。...
SAT数学题求简便算法How many of the positive divisors of 960 are also multiples of 6?A 960 B 160 C 80 D 10 E 12求简便算法,不要一个一个数的,要在考试时可以节省时间的算法,谢谢How many 3-digit numbers between 100 and 200 have a digit that is the average of the other 2 digits?A 10...
How many positive two-digit integers are multiples of 3? How many different four-digit even positive integers can be made using the digits 1, 2, 3, 4, and 5 if no digit can be used more than once? What is the smallest 4-digit odd number you can make?
How many positive integers less than 100 are neither multiples of 2 nor multiples of 3? Eliminating Integers: To find integers in a range that are not multiples of a specific number, or even two numbers, simply divide the number of integers in the range by the number of ...
How many 3 digit counting numbers are not multiples of 5? There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for ...
Each one contributes 2 valid permutations (the first digit can't be zero) so there are 9⋅2=18 permutations in this subcase. Case 3: All the digits are different Since there are 990−11011+1=81 multiples of 11 between 100 and 999, there are 81−8−9=64 multiples of 11 ...