Question

Six of King Arthur’s knights are sitting around the Round Table eating peanuts. Since each knight suspected the others of not sharing fairly, each one secretly counted his two neighbors’ peanuts and summed them up, with the following results (in order): 20, 28, 36, 44, 52, 60. How many peanuts does the knight who counted 52 have?

Answer 1

You must add the numbers of the neigbors from the one who counted 52, subtract the number of the one who is opposide and divided by 2.

So i calculate

(44 + 60 - 28) = 38

The knight who counted 52 has 38 peanuts.

Answer 2

38

Explanation:

Let the six knights be represented by the variables A, B, C, D, E and F. (A figure is attached.)

We know the neighbors of A, F and B, sum to 20; this gives us

B+F = 20

The neighbors of B, A and C, sum to 28:

A+C = 28

The neighbors of C, B and D, sum to 36:

B+D = 36

The neighbors of D, C and E, sum to 44:

C+E = 44

The neighbors of E, D and F, sum to 52:

D+F = 52

The neighbors of F, E and A, sum to 60:

E+A = 60

We are concerned with the number of peanuts the knight who counted 52 has. The one with a sum of 52 is the one whose neighbors are D and F, which is knight E.

We will use the equations with the variable E. First we use

E+A = 60

Subtract A from each side:

E+A-A = 60-A

E = 60-A

Substitute this into the other equation with E:

C+E = 44

C+60-A = 44

Subtract 60 from each side:

C+60-A-60 = 44-60

C-A = -16

The other equation we have with C and A is

A+C = 28

This gives us the system

-A+C = -16

A+C = 28

We will eliminate A by adding the two equations:

-A+A+C+C = -16+28

2C = 12

Divide both sides by 2:

2C/2 = 12/2

C = 6

Substitute this into

C+E = 44

6+E = 44

Subtract 6 from each side:

6+E-6 = 44-6

E = 38

Knight E had 38 peanuts.