Q: Three consecutive numbers have a sum of 36?

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The numbers are 11, 12 and 13.

The numbers are 36, 37 and 38.

The numbers are 32, 34, and 36.

36 + 37 + 38 = 111

102 = 32 + 34 + 36

36, 37, 38 36+37+38=111

10, 12, 14 are the consecutive even #'s that equal 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.

1, 3, 5, 7, 9, 11 are six consecutive odd numbers whose sum is 36

11, 12, and 13

All the odd numbers.

(-11) + (-12) + -(13) = -36

You may mean: 32+34+36 = 102

11,12,13

Call the smallest number x, which makes the next two consecutive even numbers x+2 and x+4. The sum of these three numbers is 36, so: x + (x + 2) + (x + 4) = 36 3x + 6 = 36 3x = 36 - 6 = 30 x = 10 So the three numbers are 10, 12, and 14, which indeed total 36. So the largest number is 14. Note that the question was about consecutive even numbers. Had the question just been about consecutive numbers, the numbers would be x, x+1, and x+2, so you'd get x + (x+1) + (x+2) = 36, 3x + 3 = 36, 3x = 33, x=11. The three numbers would be 11, 12, and 13, so the answer would be 13.

36 / 3 = 13. So lets try 10, 12, 1410 +12+14 = 36

No, they are consecutive but sum means that if you add them together it will equal 36. Also 2 is not odd. The numbers you are looking for are 17 and 19.

10 12 14 are the consecutive even integers which add up to 36

36+37=73

The numbers are 36, 37 and 38.

The integers are -14, -12 and -10.

10 12 14

-13, -12, & -11.

35+36+37

33, 36, 39, 42

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