150.
100/ 6 = 16 2/3 so 17 x 6 = 102 is the first three digit multiple of 6
1000 / 6 = 166 2/3 so 166 x 6 = 996 is the last three digit multiple of 6
So there are 166 - 17 + 1 = 150 three digit multiples of 6.
Three of them.
There are 21.
31 + 62 + 93 + 13 + 26 + 39 = 264
There are 720 of them.
I believe there are 2 positive three-digit perfect cube numbers, that are even.
10 of them.
the range of three-digit integers is from 100 to 999. Therefore, there are 300 positive three-digit integers that are divisible by neither 2 nor 3.1 day ago
Three of them.
997
There are 21.
Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.
102 = 100 which is the first possible three digit number that is a perfect square. 312 = 961 which is the last possible three digit number that is a perfect square. So there are 22 three digit positive numbers that are perfect squares.
Any multiple of 24 has 2, 3 and 8 as factors which are all positive single-digit integers.
There are 898 three-digit even numbers. Nine of them are multiples of 55. That leaves 889 * * * * * There are 450 three-digit even numbers and 17 of them are multiples of 55. So that leaves 433.
The set of integers is divided into three subsets. One is the positive integers. Another is the negative integers. The last subset has one element -- zero. In sum, integers are composed of the positive integers, the negative integers, and zero.
The first three positive integers, 1, 2, and 3, satisfy this condition.
31 + 62 + 93 + 13 + 26 + 39 = 264