79 and 2 if you add them.
1 3 9 27 81
You have a few choices: > 1 and 81 > 3 and 27 > 9 and 9 > 3 and 3 and 3
3 x 3 x 3 x 3 = 81
(x) + (x + 2) + (x + 4) = 81 3x + 6 = 81 3x = 75 x = 25 The three numbers are 25, 27, and 29
They are: 9 times 9 = 81
There is only one number that's equal to 81. That number is 81. There are many other sets of two or more numbers that you can manipulate with an arithmetic operation and produce 81 as the result. For example: Addition: 27 + 54 = 81 Subtraction: 175 - 94 = 81 Multiplication: 3 x 27 = 81 Division: 567 / 7 = 81 But none of those other numbers is "equal" to 81.
1 x 81, 3 x 27, 9 x 9.
81 = 1 x 81, 3 x 27, 9 x 9, 27 x 3, 81 x 1.
Techcially all numbers are divisible by 81, however many would result in a decimal answer. If you mean how many numbers are divisible by 81 and have an integer answer, then it is "all multiples of 81" - e.g 81 x 2, 81 x 3 etc...81, 162, 243 and keep adding 81 forever.
1 x 81, 3 x 27, 9 x 9
1 x 81 3 x 27 9 x 9
1 3 9 27 81
3 * 27
There are no two real numbers that will add to -9 and multiply to 81. Using complex numbers, the two numbers are: (-9/2 + i9/2√3) and (-9/2 - i9/2√3)
170 * * * * * No. These are numbers of the form AB00 (9*9 = 81 of them) C0D0 (81 of them), and E00F (81 of them). 243 in all.
Yes. The square root of 81 is 9 - a natural number and all natural numbers are rational numbers.