Question

use reasoning to solve a problem
example 3
place the numbers in the circles such that each straight line adds up to the same total.

Explanation:

Step1: Let the center - number be \(x\) and the sum of each line be \(S\).

The sum of the numbers \(1 + 2+3 + 4+5+6 + 7=\frac{7\times(7 + 1)}{2}=28\). When we sum up the three - line sums (counting the center number three times), we get \(3S=28 + 2x\).

Step2: Try values for \(x\).

We know that \(3S=28 + 2x\). Since \(S\) is an integer, we try \(x = 1\), then \(3S=28+2\times1 = 30\), so \(S = 10\); try \(x = 2\), then \(3S=28+2\times2=32\), \(S=\frac{32}{3}\) (not an integer); try \(x = 3\), then \(3S=28 + 2\times3=34\), \(S=\frac{34}{3}\) (not an integer); try \(x = 4\), then \(3S=28+2\times4 = 36\), so \(S = 12\); try \(x = 5\), then \(3S=28+2\times5=38\), \(S=\frac{38}{3}\) (not an integer); try \(x = 6\), then \(3S=28+2\times6 = 40\), \(S=\frac{40}{3}\) (not an integer); try \(x = 7\), then \(3S=28+2\times7=42\), so \(S = 14\).

Step3: For \(x = 1\) and \(S = 10\).

The pairs of numbers that add up to \(10 - 1=9\) are \((2,7)\), \((3,6)\), \((4,5)\).

Step4: For \(x = 4\) and \(S = 12\).

The pairs of numbers that add up to \(12 - 4 = 8\) are \((1,7)\), \((2,6)\), \((3,5)\).

Step5: For \(x = 7\) and \(S = 14\).

The pairs of numbers that add up to \(14 - 7=7\) are \((1,6)\), \((2,5)\), \((3,4)\).

One possible solution (with \(x = 4\)): Place \(4\) in the center. On one line, place \(1\) and \(7\); on another line, place \(2\) and \(6\); on the third line, place \(3\) and \(5\).

Answer:

One possible arrangement: Center circle: 4; First - line circles: 1, 7; Second - line circles: 2, 6; Third - line circles: 3, 5.