The integers are 4 and 9, so the equation would be 16+81=? and 16+81=97
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.
35,13,33
The first three positive integers, 1, 2, and 3, satisfy this condition.
25, 27, 29
27 and 81; 27 + 81 = 108, and 3 x 27 = 81
The integers are 4 and 9, so the equation would be 16+81=? and 16+81=97
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.
35,13,33
The first three positive integers, 1, 2, and 3, satisfy this condition.
25, 27, 29
It may be either. If any of the integers is zero, the product will be zero. Else, if one or three of the integers is negative, the product will be negative. Otherwise, it will be positive.
The numbers are 81, 82 and 83. Also, 80, 82 and 84
If a, b, c, and d are positive integers that add up to 20 they each must be less than 20. If abcd=81 then a, b, c, and d must be factors of 81. The factors of 81 are 1,3,9 then 27, so there is no solution to this problem.
27,28,29The sum of which is 84.
81
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