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Q: What are four consecutive integers with the sum of -66?

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65, 66, 67

x+(x+2)= 66.

There are two consecutive even integers of negative 130: -66 and -64.

If the range is three, then there are four whole numbers involved. The average value of these numbers is 66/4 = 16.5. Because the numbers are consecutive - evenly distributed - 16.5 is actually 'in the middle' of the four. From which we can infer that the numbers in question are 15,16,17 and 18.

I'll represent the smaller number with the variable 'n'. Now, how much larger must the second number be?The consecutive odd integers starting at 1:1,3,5,7,9,11,13...You can see that each number is two more than the last number. Therefore, we know that the larger number must be 2 more than 'n', or n+2. Finally, use the last piece of information that you are given to solve the problem.1st number + 2nd number = 130n + n+2 = 1302n + 2 = 1302n + 2 - 2 = 130 - 22n = 1282n/2 = 128/2n = 64This is the first number, and so we know that because the second number is 2 more than the second number, the second number must be 64+2, or 66.n+2 = 64 + 2n + 2 = 66The two integers are 64 and 66You can check to see if this is correct.64 + 66 =? 130130 = 130This is correct.

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21, 22, 23

The numbers are -18, -17, -16 and -15.

64 65 66 67

The numbers are -23, -22, and -21.

The average of three numbers which sum to 66 is 66/3 = 22. If 22 is the sum of three consecutive integers, then those three integers are 21,22,23. Kermit Rose

201

It's an imaginary number, one that exists only in fiction, if the author wants it to. There are no two consecutive integers whose sum is 66. 32 + 33 = 65 33 + 34 = 67 The sum of two consecutive integers is always an odd number.

The numbers are 21, 22 and 23.

65, 66, 67

-66 and -68

66

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