how to you get the answer?

## The sum of three consecutive integers is 378. Find the three integers.

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# 4 Answers

Hi Ericka -- the middle number of the 3 must be 1/3 of 378 or 126 => 125,126,127 :)

Let's assume, that first number is "x", than second number will be "x + 1" and third one is "x + 2"

(For example if first number is 2, second will be 2 +

x + (x + 1) + (x + 2) = 378

x + x + 1 + x + 2 = 378

3x + 3 = 378

3(x + 1) = 378

÷ 3 ÷ 3

x + 1 = 126

- 1 - 1

x = 125

The first number is

125 + 126 + 127 = 378

(For example if first number is 2, second will be 2 +

**= 3 and third 2 +***1***= 4)***2*x + (x + 1) + (x + 2) = 378

x + x + 1 + x + 2 = 378

3x + 3 = 378

3(x + 1) = 378

÷ 3 ÷ 3

x + 1 = 126

- 1 - 1

x = 125

The first number is

**, second number is***125***and third number is***126***.***127*125 + 126 + 127 = 378

Suppose the least of the integers is n. Then, the three integers are n, n+1, and n+2. Hence, n + (n+1) + (n+2) = 378. This implies 3n + 3 = 378, and 3n = 375. Thus, n = 375/3 = 125. then, the three integers are 125, 126, and 127. Let's check: 125 + 126
+ 127 = 3(125) + 3 = 375 + 3 = 378.

Let x be the smallest one. These three numbers are x, x+1 and x+2.

x+(x+1)+(x+2) = 378

Solve for x,

3x+3 = 378

3x = 375

x = 125

Answer: 125, 126, 127