Question

a. place the numbers 24 through 32 in the circles so that the sum of three numbers in a row are the same.
b. how many different numbers can be placed in the middle to obtain a solution?
a. choose the correct figure below

Explanation:

Step1: Calculate the sum of numbers from 24 to 32

The sum of an arithmetic - series $S_n=\frac{n(a_1 + a_n)}{2}$, where $n = 32−24 + 1=9$, $a_1 = 24$, $a_n=32$. So $S=\frac{9\times(24 + 32)}{2}=9\times28 = 252$.

Step2: Let the number in the middle be $x$ and the sum of each row be $k$.

Since there are 4 rows (counting the rows passing through the middle number) and the sum of all the numbers from 24 to 32 plus 3 times the middle - number (because the middle - number is counted 3 times in the row - sums) should be equal to 4 times the sum of each row. Let the sum of each row be $k$. Then $252+3x = 4k$.

Step3: Analyze the possible values of $x$

We know that $x$ must be such that $252 + 3x$ is divisible by 4. Since $252\div4 = 63$ (remainder 0), $3x$ must be divisible by 4. Among the numbers 24, 25, …, 32, the numbers that satisfy this condition are the multiples of 4 in this range. The multiples of 4 in the range 24 - 32 are 24, 28, 32.

Step4: Check the arrangements for part a

We can try to arrange the numbers around the middle number for each of the possible middle - numbers. For example, if $x = 24$:
Let the sum of each row be $k=\frac{252+3\times24}{4}=\frac{252 + 72}{4}=\frac{324}{4}=81$.
We can then arrange the remaining numbers to get the correct row - sums.

Answer:

a. Without seeing the full details of the figures in options A, B, C, D, we can't directly choose the correct figure. But we know that the middle number can be 24, 28 or 32.
b. 3