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

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4 Answers

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 + 1 = 3 and third 2 + 2 = 4)

     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, second number is 126 and third number is 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

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