Place the numbers 1 to 5 in the circles above, one number per circle, so that the sum of numbers on each of the two diagonal lines is 9.

What number goes in the middle?

We'll spend the next few questions consider the puzzle below:

Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.

When we calculate all the sums described in the puzzle, what is the most important difference between the circles marked in red and those marked in green?

**which equation is true?** (Remember we're using the numbers 1 through 6.)

ORIGINAL QUESTION: Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.

Before trying to solve the puzzle, let's think of the sums together. There are three sums—one on each line and each equals 12—so the total of the sums is \( 12 \times 3 = 36 .\)

Which equation must be true?

Based on the previous problems, we now know that \(R + G = 21 \) and \( 2R + G = 36 .\) We can use algebra to combine these statements and figure out what \(R\) and \(G\) are.

What is \(R?\) (Hint: Subtract the first equation from the second.)

Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.

We have now learned that \(R\) sums to 15. What number must go in the top left circle if 4 is in the top right?

Fill the grid above with the numbers 1 through 6 (one in each square). There will be five sums: two made by adding up the numbers in each row, and three by adding up the numbers in each column.

There is a way to fill the grid so **four of the sums are the same and one will be different.** What is the sum that is different?

\((\)Hint: \( 1 + 2 + 3 + 4 + 5 + 6 = 21 .)\)

Let's change up the goal slightly!

Suppose you want to fill the puzzle above with the numbers 1 through 6 (one each circle) so that every set of three circles next to each other add up to the same sum.

Is this possible to do?

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