27.
1 ÷ 3 = 0 r 3 → first multiple of 3 in the range 1-81 is 1 x 3 = 3
81 ÷ 3 = 27 → last multiple of 3 in the range 1-81 is 27 x 3 = 3
→ there are 27 - 1 + 1 = 27 multiples of 3 in the range 1-81, which are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81.
They are multiples of 1, 3, and 9.
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.
81, 84, 87.
81 84 87
There are infinitely many different choices of 3 numbers that add up to 81. 50 + 30 + 1 = 81 40 + 40 + 1 = 81 10 + 11 + 60 = 81 etc.
1, 3 and 9 are common factors of 27, 45 and 81. 27, 45 and 81 are multiples of 1, 3 and 9.
They are multiples of 1, 3, and 9.
The multiples of 243 is 1, 3, 9, 27, 81, and 243.
The multiples of 81 are infinite. The factors of 81 are: 1, 3, 9, 27, 81.
They are multiples of 1, 3 and 9.
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.
81 and 162 are both multiples of 3. In mathematics, a number that can be divided by another number without leaving a remainder is said to be divisible by that number. Both 81 and 162 can be evenly divided by 3, making them multiples of 3.
Incorrect. Instead they are factors of I, 3 and 9.
81, 84, 87.
3 x 3 x 3 x 3
1, 3, 29, 87.
The factors of 81 are 1, 3, 9, 27, and 81.The factors of 162 are 1, 2, 3, 6, 9, 18, 27, 54, 81, and 162.81 is the first common multiple over 80 of 81 and 162.