Question

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say \(m\), of these three integers?

• A. \(7 \le m \le 9\)
• B. \(4 \le m \le 6\)
• C. \(1 \le m \le 3\)
• D. \(10 \le m \le 12\)
• E. \(13 \le m \le 15\)

Hint:

Write the three integers as \(m,\ m+1,\ m+2\), form the sum \(m+(m+1)^2+(m+2)^3\) and set it equal to the square of \(m+(m+1)+(m+2)\).

Answer 1

The Correct Option is C

Step 1: Set up the equation:
The integers are \(m,\ m+1,\ m+2\). Their total is \(3m+3\). The condition is
\(m+(m+1)^2+(m+2)^3=(3m+3)^2\).

Step 2: Expand:
Left side: \(m+(m^2+2m+1)+(m^3+6m^2+12m+8)=m^3+7m^2+15m+9\).
Right side: \(9m^2+18m+9\).

Step 3: Solve:
\(m^3+7m^2+15m+9=9m^2+18m+9 \Rightarrow m^3-2m^2-3m=0 \Rightarrow m(m-3)(m+1)=0\).
Since \(m\) is a positive integer, \(m=3\).

Step 4: Check:
The integers are 3, 4, 5: \(3+16+125=144=12^2\) and \(3+4+5=12\). Correct.

Step 5: Pick the range:
\(m=3\) lies in \(1\le m\le3\).

Final Answer:
The correct option is (3): \(1 \le m \le 3\).