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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
79 and 2 if you add them.
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)
You have a few choices: > 1 and 81 > 3 and 27 > 9 and 9 > 3 and 3 and 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.