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There are no such whole numbers.

The sum of three consecutive whole numbers must be a multiple of 3; as 68 is not a multiple of 3 (68 = 3 × 22 2/3) it cannot be the sum of three whole numbers.

Q: What are three consecutive numbers that add up to 68?

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They are: 68+69+70 = 20768, 69 and 70.

64 + 66 + 68 = 198

33 & 35.

Let the first even number be x. Then the second even number would be x + 2, and the third even number would be x + 4. The sum of these numbers is x + (x + 2) + (x + 4) = 3x + 6. We set this equal to 204 and solve for x: 3x + 6 = 204 --> 3x = 198 --> x = 66. Therefore, the three consecutive even numbers are 66, 68, and 70.

207

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They are: 68+69+70 = 20768, 69 and 70.

137

The three numbers will have a mean of 204/3 = 68. The two numbers are therefore 66, 68, 70.

66 + 67 + 68

5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 68

Answer: The even numbers are 62, 64, 66, and 68

64 + 66 + 68 = 198

The numbers are 33 and 35.

33 & 35.

Let the first even number be x. Then the second even number would be x + 2, and the third even number would be x + 4. The sum of these numbers is x + (x + 2) + (x + 4) = 3x + 6. We set this equal to 204 and solve for x: 3x + 6 = 204 --> 3x = 198 --> x = 66. Therefore, the three consecutive even numbers are 66, 68, and 70.

The idea is to add all three numbers, then divide the result by 3 (because there are 3 numbers).

Here's how you can work it out: First, let "x" be the first of those three numbers. We know that those numbers are three consecutive even integers, which tells us that the other numbers are "x + 2" and "x + 4". We also know that the three of them added together give us 210, so we can say: x + (x + 2) + (x + 4) = 210 ∴ 3x + 6 = 210 ∴ 3x = 204 ∴ x = 68 We now know that the lowest number is 68, which tells us that the other two must be 70 and 72. To check that, we simply add them together: 68 + 70 + 72 = 210 which confirms our answer.