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?
b. how many different numbers can be placed in the middle? select the correct answer below and fill in the answer box to complete your choice (type a whole number. use commas to separate answers as needed.)
a. one number can be placed in the middle. it is

b. two numbers can be placed in the middle. they are

c. three numbers can be placed in the middle. they are

d. four numbers can be placed in the middle. they are

Explanation:

Step1: Find 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 middle number be $x$ and the sum of each row be $k$.

Since the figure has a central number and 4 pairs of numbers forming rows, the sum of all the numbers can be written as $4k=252 + 3x$ (the middle number is counted in each of the 4 rows). So $k=\frac{252 + 3x}{4}=63+\frac{3x}{4}$.

Step3: Determine the possible values of $x$

For $k$ to be an integer, $x$ must be such that $3x$ is divisible by 4. The numbers from 24 to 32 are considered. When $x = 24$, $k=\frac{252+3\times24}{4}=\frac{252 + 72}{4}=\frac{324}{4}=81$. When $x = 28$, $k=\frac{252+3\times28}{4}=\frac{252+84}{4}=\frac{336}{4}=84$. When $x = 32$, $k=\frac{252+3\times32}{4}=\frac{252 + 96}{4}=\frac{348}{4}=87$.

Answer:

C. Three numbers can be placed in the middle. They are 24, 28, 32