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There are 33 consecutive integers that sum to 66. Which integers are they?\newline_\_, _\_, _\_

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Full solution

Q. There are 33 consecutive integers that sum to 66. Which integers are they?\newline_\_, _\_, _\_
  1. Define Integers: Let xx be the first integer. The three consecutive integers can be represented as xx, x+1x+1, and x+2x+2.
  2. Set Up Equation: Set up the equation to represent the sum of the three consecutive integers equal to 66: x+(x+1)+(x+2)=6x + (x+1) + (x+2) = 6.
  3. Simplify Equation: Simplify the equation: x+x+1+x+2=6x + x + 1 + x + 2 = 6 which becomes 3x+3=63x + 3 = 6.
  4. Isolate Variable: Isolate the variable term by subtracting 33 from both sides of the equation: 3x+3−3=6−33x + 3 - 3 = 6 - 3 which simplifies to 3x=33x = 3.
  5. Divide and Solve: Divide both sides of the equation by 33 to solve for xx: 3x3=33\frac{3x}{3} = \frac{3}{3} which results in x=1x = 1.
  6. Identify Integers: Now that we have x=1x = 1, the three consecutive integers are xx, x+1x+1, and x+2x+2, which are 11, 1+11+1, and 1+21+2. Therefore, the integers are 11, 22, and 33.

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