1, 3, 5, 7, 9, 11 are six consecutive odd numbers whose sum is 36
Let's represent the four consecutive numbers as x, x+1, x+2, and x+3. The sum of these numbers is x + (x+1) + (x+2) + (x+3) = 4x + 6. We know that this sum equals 36, so we can set up the equation 4x + 6 = 36. Solving for x, we find x = 7. Therefore, the four consecutive numbers are 7, 8, 9, and 10.
36 / 3 = 13. So lets try 10, 12, 1410 +12+14 = 36
The numbers are 11, 12 and 13.
The numbers are 11, 12 and 13.
11+12+13=36
1, 3, 5, 7, 9, 11 are six consecutive odd numbers whose sum is 36
Assuming you mean how to solve what three consecutive numbers sum to some value: Divide the sum by three to get the middle number. The other two numbers are one less than this and one more than it. example: Which three consecutive numbers sum to 108? 108 ÷ 3 = 36 → The other two numbers are 36 - 1 = 35 and 36 + 1 = 37 → The three numbers are 35, 36, 37. 35 + 36 + 37 = 108 as required.
36
Let's represent the four consecutive numbers as x, x+1, x+2, and x+3. The sum of these numbers is x + (x+1) + (x+2) + (x+3) = 4x + 6. We know that this sum equals 36, so we can set up the equation 4x + 6 = 36. Solving for x, we find x = 7. Therefore, the four consecutive numbers are 7, 8, 9, and 10.
33, 36, 39, 42
36 / 3 = 13. So lets try 10, 12, 1410 +12+14 = 36
114
The numbers are 11, 12 and 13.
The numbers are 11, 12 and 13.
All the odd numbers.
36 and 45